The three conics as slices of a cone and as focus-directrix loci, their standard rectangular equations, eccentricity as the single parameter distinguishing them, the unified polar equation, and identifying rotated conics from the discriminant.
Subject: Calculus II · 70 slides · symbolic lesson
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Title
Calculus II · Section 7.5
Three curves, one ratio
Objectives
Parabolas, ellipses and hyperbolas look like three unrelated curves with three unrelated equations. This lesson shows they are one family: slices of one cone, and points with one fixed ratio of distances to a point and a line.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 671-693 — learning objectives 7.5.1 to 7.5.6
You have met parabolas since algebra, and ellipses and hyperbolas probably in precalculus, each with its own formula to memorise. This lesson ties the three together, and that is the real payoff: once you see them as one family, the formulas stop being separate facts and start being consequences of a single definition.
The first three parts build each curve from a distance rule and turn the rule into a standard equation. The fourth part names the one number, the eccentricity, that tells the three apart. The fifth writes all three as a single polar equation, which is where this chapter on polar coordinates has been heading. The last part handles conics whose axes are tilted.
Completing the square is the workhorse skill throughout. If it feels rusty, the warm-up is your chance to shake the rust off before it matters.
Warm-up
Discussion prompt
Without looking anything up: complete the square on x squared minus 6x, then write the distance from a point (x, y) to the point (0, 2), and the distance from (x, y) to the line y = −2.
Write your answers down before revealing. Completing the square is how every messy equation in this lesson becomes a readable standard form, so it is worth being fast at it: take half the coefficient of x, square it, add it and subtract it.
The two distance formulas are even more fundamental, because they are the definitions. The distance to a point needs the square root of a sum of squares. The distance to a horizontal line is only the vertical gap, an absolute value, and that asymmetry is exactly why a parabola ends up with one squared variable and one unsquared one.
If you got the line distance wrong by writing a square root, notice why it is unnecessary: the closest point on a horizontal line is straight above or below you, so only the heights differ.
Section
Part 1
Concept
Figure (svg): Four side views of a double cone whose edges are the lines y equals 2x and y equals minus 2x about the vertical axis. A horizontal cut gives a circle, a tilted cut through one nappe gives an ellipse, a cut parallel to the edge gives a parabola, and a vertical cut through both nappes gives a hyperbola.
\[ \text{cone: revolve } y = 2x \text{ about the } y\text{-axis} \]
A double cone has two nappes meeting at a point. Tilt a flat plane through it and the angle of the cut alone decides the curve: level gives a circle, a gentle tilt an ellipse, parallel to the edge a parabola, and steeper still, through both nappes, a hyperbola.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 671-672 — Figures 7.43 and 7.44
Each panel is a side view of the same double cone, the shape you get by spinning a line through the origin about the vertical axis. The book spins the line y equals 3x; the panels use a slope of 2, and any slope gives a cone. The coloured stroke is the part of a flat cutting plane that lies inside the cone.
Look at how the angle of the cut changes the result. A level cut meets every edge line of one nappe at the same height, so the curve is a circle. Tilt it a little and it still crosses all the edge lines of that nappe, but at different heights: a closed oval, the ellipse. Tilt it until it is exactly parallel to one edge line and it never closes up: a parabola. Tilt past that and the plane reaches the other nappe too, giving two separate pieces: the hyperbola.
Notice that the parabola is the borderline case between the other two. Hold on to that; the eccentricity will make it precise.
Concept
parabola — The set of all points whose distance from a fixed point, the focus, equals their distance from a fixed line, the directrix. The vertex is the point halfway between them.
Figure (svg): The parabola x squared equals 4y, with focus F at (0, 1) and directrix y equals minus 1. A point P at (3, 2.25) is joined to F and dropped straight down to Q on the directrix; both segments have length 3.25.
\[ d(F, P) = d(P, Q) \]
Q is the foot of the perpendicular from P to the directrix. Put the focus at (0, p) and the directrix at height minus p, and the vertex lands at the origin.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 672 — definition and Figure 7.45
This definition replaces the cone with something you can compute with. A point belongs to the parabola exactly when it is as far from the focus as it is from the directrix. In the figure, the yellow and green segments from P both measure 3.25.
The vertex is the point of the parabola closest to both, which puts it halfway between the focus and the line. That is why the standard set-up places the focus at height p and the directrix at height minus p: the vertex then lands at the origin and the algebra is as clean as possible.
Keep the meaning of p clear from the start. It is the distance from the vertex to the focus, and also from the vertex to the directrix. Almost every mistake with parabolas comes from confusing p with some other number in the equation.
Concept
Write both distances with the distance formula, then square to clear the roots.
\[ \sqrt{(x-0)^2 + (y-p)^2} = \sqrt{(x-x)^2 + (y+p)^2} \]
\[ x^2 + (y-p)^2 = (y+p)^2 \]
\[ x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2 \]
\[ x^2 - 2py = 2py \]
\[ x^2 = 4py \]
The squares of y and of p cancel, which is why a parabola has only one squared variable. Test it on the point where y equals p: then x is 2p, and that point is 2p from the focus and 2p above the directrix.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 672 — derivation of x² = 4py
Follow the lines one at a time. The first line is the definition, written with the distance formula on both sides. Since both sides are distances, and so never negative, squaring them loses nothing, and the second line has no roots at all.
Expanding both squares shows why the parabola has only one squared variable: y squared and p squared appear on both sides and cancel, leaving only the terms in py. Collecting them gives x squared equals 4py.
The last sentence is a check you can always do: pick a convenient point and test the definition directly. At height p, x squared is 4p squared, so x is 2p. That point is 2p to the side of the focus and 2p above the directrix, exactly as the definition demands.
Concept
\[ y = \frac{1}{4p}(x-h)^2 + k \qquad \text{vertex } (h,k),\; \text{focus } (h, k+p) \]
Figure (svg): Four parabolas with vertex at the origin and p equal to 1: x squared equals 4y opens up, x squared equals minus 4y opens down, y squared equals 4x opens right, y squared equals minus 4x opens left. Each focus is marked one unit from the origin in the opening direction.
| opens | standard form | focus | directrix |
|---|---|---|---|
| up | (x − h)² = 4p(y − k) | (h, k + p) | y = k − p |
| down | (x − h)² = −4p(y − k) | (h, k − p) | y = k + p |
| right | (y − k)² = 4p(x − h) | (h + p, k) | x = h − p |
| left | (y − k)² = −4p(x − h) | (h − p, k) | x = h + p |
\[ \text{general form: } ax^2 + bx + cy + d = 0 \;\text{ or }\; ay^2 + bx + cy + d = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 673-674 — Theorem 7.8, Figure 7.46, general form
Moving the vertex to (h, k) replaces x by x minus h and y by y minus k; nothing else changes. Solving for y gives Theorem 7.8, and the number in front of the square is one over 4p.
The figure shows the four orientations with p equal to 1. The squared variable tells you the axis of symmetry: if x is squared, the parabola is symmetric about a vertical line and opens up or down. The sign on the other side tells you which way. The table collects where the focus and directrix go in each case; you do not need to memorise it if you remember that the focus is always p inside the curve and the directrix p outside.
The general form at the bottom is what problems usually hand you. The next example shows how completing the square turns it back into a standard form you can read.
Worked example
Put the equation into standard form and read off the vertex, focus and directrix.
\[ x^2 - 4x - 8y + 12 = 0 \]
Decide which variable to solve for
Why: Only x is squared, so the parabola opens up or down: isolate the y term.
\[ 8y = x^2 - 4x + 12 \]
Complete the square
Why: Half of minus 4 is minus 2, and its square is 4. Add it and subtract it.
\[ 8y = (x^2 - 4x + 4) + 12 - 4 \]
Factor the perfect square
Why: The bracket is a square; the constants combine.
\[ 8y = (x-2)^2 + 8 \]
Divide by 8
Why: Now it has the shape of Equation 7.11.
\[ y = \frac18(x-2)^2 + 1 \]
Match with Theorem 7.8
Why: The coefficient is one over 4p.
\[ h = 2,\quad k = 1,\quad \frac{1}{4p} = \frac18 \;\Rightarrow\; p = 2 \]
Read off the features
Why: The focus is p above the vertex and the directrix p below it.
\[ \text{vertex } (2,1),\quad \text{focus } (2,3),\quad \text{directrix } y = -1 \]
Figure (svg): The parabola y equals one eighth of (x minus 2) squared plus 1, opening upward, with vertex (2, 1), focus (2, 3), the dashed directrix y equals minus 1 and the dashed axis of symmetry x equals 2.
Check with a point
Why: At x equal to 6 the curve gives y equal to 3. That point satisfies the original equation, and it is 4 from the focus and 4 above the directrix.
\[ 36 - 24 - 24 + 12 = 0, \qquad \sqrt{(6-2)^2 + 0^2} = 4 = 3 - (-1) \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 674-675 — Example 7.19
The first decision is what to solve for. Only x is squared, so the answer will have the shape y equals a square plus a constant, and you isolate the y term first.
Completing the square is then routine: half of minus 4 is minus 2, its square is 4, and you add 4 inside the bracket while subtracting 4 outside so the equation is unchanged. After factoring and dividing by 8 you can match the equation against Theorem 7.8 term by term.
The most important line is the one where one over 4p equals one eighth, giving p equal to 2. The focus is two above the vertex, at (2, 3), and the directrix two below, at height minus 1. The check tests a point two ways: it satisfies the original equation, and its distances to the focus and to the directrix really are equal.
Step zero
\[ 2y^2 - x + 12y + 16 = 0 \]
Discussion prompt
Checkpoint 7.18 asks for the standard form. Before any algebra: which variable is squared, which directions can this parabola open, and which variable should you solve for?
Answer before revealing. This thirty-second step decides the whole shape of the solution, and getting it wrong means solving for the wrong variable and ending with a mess.
The rule is simple: the squared variable is the one you complete the square in, and the other variable is the one you solve for. Here y is squared, so you will write x as a function of y, and the parabola opens sideways.
The sign of the y squared coefficient then tells you which side. It is positive, so x increases as y moves away from the vertex in either direction: the parabola opens to the right.
Worked example
\[ 2y^2 - x + 12y + 16 = 0 \]
Solve for x
Why: The unsquared variable goes on its own.
\[ x = 2y^2 + 12y + 16 \]
Factor the 2 out of the y terms
Why: Completing the square needs a leading coefficient of 1 inside the bracket.
\[ x = 2(y^2 + 6y) + 16 \]
Complete the square inside
Why: Half of 6 is 3 and its square is 9. Adding 9 inside the bracket adds 18, so subtract 18.
\[ x = 2(y^2 + 6y + 9) + 16 - 18 \]
Factor
Why: The bracket is a perfect square.
\[ x = 2(y+3)^2 - 2 \]
Find p
Why: The coefficient is one over 4p.
\[ 2 = \frac{1}{4p} \;\Rightarrow\; p = \frac18 \]
Read off the features
Why: Positive coefficient: it opens right, so the focus is p to the right of the vertex.
\[ \text{vertex } (-2,-3),\quad \text{focus } \left(-\tfrac{15}{8}, -3\right),\quad x = -\tfrac{17}{8} \]
Figure (svg): The parabola x equals 2 times (y plus 3) squared minus 2, opening to the right, with vertex (minus 2, minus 3), focus (minus 1.875, minus 3) just inside it, and the dashed directrix x equals minus 2.125 just outside it.
Check with a point
Why: At y equal to minus 2 the standard form gives x equal to 0; put (0, minus 2) into the original equation.
\[ 2(-2)^2 - 0 + 12(-2) + 16 = 8 - 24 + 16 = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 675 — Checkpoint 7.18
The new wrinkle is the 2 in front of y squared. Completing the square needs a leading coefficient of 1, so you factor the 2 out of the y terms first. Then adding 9 inside the bracket really adds 18 to the right-hand side, because of the 2 outside, and you must subtract 18 to compensate. This is the step people most often get wrong.
The standard form x equals 2 times y plus 3 squared, minus 2, puts the vertex at (minus 2, minus 3). Setting 2 equal to one over 4p gives p equal to one eighth: a narrow parabola whose focus sits very close to its vertex, as the figure shows.
The check picks y equal to minus 2, where the square is 1 and x comes out 0, and confirms that (0, minus 2) satisfies the original equation.
Worked example
Find the equation of the parabola with focus (2, 3) and directrix the vertical line x equal to minus 2.
Write the definition as an equation
Why: Distance to the focus equals distance to the line, and the distance to a vertical line is a horizontal gap.
\[ \sqrt{(x-2)^2 + (y-3)^2} = x + 2 \]
Square both sides
Why: Both sides are non-negative, so nothing is lost.
\[ (x-2)^2 + (y-3)^2 = (x+2)^2 \]
Expand the x terms
Why: Leave the y part as a square: it is already in the form you want.
\[ x^2 - 4x + 4 + (y-3)^2 = x^2 + 4x + 4 \]
Cancel and collect
Why: The squares of x and the 4s cancel.
\[ (y-3)^2 = 8x \]
Read it against the standard form
Why: Four p is 8, and the vertex is halfway between focus and directrix.
\[ 4p = 8 \;\Rightarrow\; p = 2, \quad \text{vertex } (0, 3), \text{ opens right} \]
Figure (svg): The parabola (y minus 3) squared equals 8x opening right, with vertex (0, 3), focus (2, 3), dashed directrix x equals minus 2, and the test point (2, 7) joined to the focus and to the directrix by segments of length 4.
Check with a point
Why: At y equal to 7 the equation gives x equal to 2. That point is 4 from the focus and 4 from the directrix.
\[ (7-3)^2 = 16 = 8(2), \qquad \sqrt{0^2 + 4^2} = 4 = 2 + 2 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 692 — Exercise 258
This runs the definition directly instead of quoting a formula, and that is the safest approach whenever the focus and directrix are given. The distance to the vertical line x equals minus 2 is the horizontal gap, x plus 2, for every point to the right of the line.
Squaring and expanding only the x terms is a small but useful choice: the y part is already a perfect square in the shape you want, so leave it alone. The x squared terms and the 4s cancel, and you are left with y minus 3, squared, equals 8x.
Reading that against the standard form confirms the picture: 4p is 8, so p is 2, which is exactly half the distance from the focus to the directrix. The check uses the point (2, 7), which sits 4 above the focus and 4 to the right of the directrix.
Trap
Finding the focus of Example 7.19:
\[ y = \frac18(x-2)^2 + 1 \]
\[ p = 8 \;\Rightarrow\; (2, 9) \]
Wrong. Neither 8 nor one eighth is p.
The coefficient in front of the square is one over 4p. Set it equal and solve.
\[ \frac{1}{4p} = \frac18 \;\Rightarrow\; p = 2 \]
So the focus is (2, 3). A small coefficient means a wide parabola and a focus far from the vertex; a large coefficient means a narrow one with the focus close in.
The number in front of the square is not p, and it is not something simple like p over 4 either. It is one over 4p. Reading p as 8 puts the focus at (2, 9), far above the true focus at (2, 3).
A sanity check catches the error. A small coefficient like one eighth makes a wide, flat parabola, and a wide parabola has its focus a fair distance from the vertex. So p should be moderately large here: 2, not 8, and certainly not one eighth. Whenever you find p, glance back at the graph's width and make sure the two agree.
Picture it
Figure (svg): A parabolic dish y equals x squared over 8 with focus (0, 2). Five vertical rays come straight down, hit the dish, and each reflected ray runs to the focus.
Rays arriving parallel to the axis all reflect through the focus. A receiver at the focus collects signal from the whole dish; a bulb at the focus of a headlight sends a parallel beam out.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 676 — the reflective property
The orange rays arrive parallel to the axis, the way signals from a satellite far away do. Each one strikes the dish and reflects, and the yellow reflected rays all pass through the same point: the focus. That is the reflective property of the parabola.
It works both ways. A receiver at the focus collects energy from the whole face of the dish, which is why a small receiver can pick up a weak signal. Run it backwards and a light bulb at the focus of a headlight mirror sends out a parallel beam.
Proving the reflective property needs the tangent line, which you could do with derivatives, but the practical consequence is what matters here: designing a dish means knowing where its focus is.
Real world
Discussion prompt
Exercise 313: a satellite dish is 12 feet across at its opening and 4 feet deep at its centre. Put the vertex at the origin, opening up, so its cross-section is x squared = 4py. Where should the receiver be placed?
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 693 — Exercise 313
Write down your set-up before revealing. The key idea is to choose coordinates that make the equation as simple as possible: vertex at the origin, opening upward, so the cross-section is x squared equals 4py with only one unknown.
The dish is 12 feet across, so its rim is 6 feet either side of the axis, and it is 4 feet deep, so the rim is at height 4. The point (6, 4) must lie on the parabola, which gives 36 equals 16p and p equals 2.25.
So the receiver sits on the axis 2.25 feet above the bottom of the dish. Notice that p came from the shape alone. A shallower dish of the same width would have a larger p and a receiver farther out.
Section
Part 2
Concept
ellipse — The set of all points for which the sum of the distances to two fixed points, the foci, is constant.
Figure (svg): The ellipse x squared over 25 plus y squared over 9 equals 1 with foci F at (4, 0) and F prime at (minus 4, 0). A point P on the upper left is joined to both foci; the segments measure 2.65 and 7.35, which add to 10.
\[ d(P, F) + d(P, F') = 2a \]
Why call the constant 2a? Take P at the vertex (a, 0). It is a minus c from one focus and a plus c from the other:
\[ (a - c) + (a + c) = 2a \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 676-677 — definition and Figure 7.48
The ellipse has two foci instead of a focus and a line, and the rule is about a sum. In the figure the green and orange lengths are 2.65 and 7.35. Move P anywhere and the two lengths change, but they always add to 10.
The constant is called 2a, and the vertex shows why. At the vertex (a, 0), the near focus is a minus c away and the far focus is a plus c away, and the c's cancel in the sum. So the constant sum equals the length of the major axis.
A physical picture helps: pin a loop of string at the two foci and pull it taut with a pencil. The string length is fixed, so the pencil traces an ellipse. The closer the pins, the rounder the curve.
Concept
Put the foci at (c, 0) and (minus c, 0). Move one root across so each squaring removes one.
\[ \sqrt{(x+c)^2 + y^2} = 2a - \sqrt{(x-c)^2 + y^2} \]
\[ (x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2 \]
\[ 4cx = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} \]
\[ a\sqrt{(x-c)^2 + y^2} = a^2 - cx \]
\[ a^2(x^2 - 2cx + c^2 + y^2) = a^4 - 2a^2cx + c^2x^2 \]
The terms in cx cancel on the two sides. One root is gone after the first squaring, the other after the second.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 677-678 — the derivation, rearranged
This derivation is longer than the parabola's because there are two square roots, and each squaring can only remove one. The trick is to move one root to the other side first, so that squaring clears the root on the left and leaves exactly one root on the right.
After the first squaring, most terms cancel: the x squared, y squared and c squared terms appear on both sides. What survives is a single root, which you isolate and divide by 4. Then square again.
The book isolates the roots in the opposite order and arrives at the same place. What matters is the strategy: isolate one root, square, simplify, repeat.
Concept
\[ a^2x^2 + a^2c^2 + a^2y^2 = a^4 + c^2x^2 \]
\[ (a^2 - c^2)x^2 + a^2y^2 = a^2(a^2 - c^2) \]
\[ \frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1 \]
Figure (svg): The ellipse x squared over 25 plus y squared over 9 equals 1 with the right triangle joining the centre, the focus (4, 0) and the top of the minor axis (0, 3): legs c equal 4 and b equal 3, hypotenuse a equal 5.
\[ b^2 = a^2 - c^2 \;\Longrightarrow\; \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The number a squared minus c squared is positive because the foci lie inside the ellipse, and the triangle shows what it is: the square of the half-height b.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 678 — the b-c-a triangle
After the second squaring and a little collecting, you reach a line with a squared minus c squared in two places. Dividing by a squared times that number gives x squared over a squared plus y squared over a squared minus c squared, equal to 1.
The figure explains what a squared minus c squared means. The top of the minor axis, Q, is equally far from both foci, and its two distances must add to 2a, so each is a. The right triangle with legs b and c and hypotenuse a gives b squared equal to a squared minus c squared.
That is the source of the ellipse relation between a, b and c. It is Pythagoras, with a on the hypotenuse, which is why a is always the largest of the three.
Notation
Annotate
On: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \quad c^2 = a^2 - b^2, \quad x = h \pm \frac{a^2}{c} \)
\[ \text{vertical: } \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1, \quad \text{foci } (h, k \pm c), \quad y = k \pm \frac{a^2}{c} \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 678-679 — Theorem 7.9, Equations 7.12 and 7.13
Step through the notes and connect each piece to the picture. The centre is (h, k), and a is always the larger semi-axis, so the major axis goes with whichever denominator is larger. In the displayed form that is the x term, and the ellipse is horizontal.
The foci sit on the major axis, c from the centre, and c comes from subtracting squares. The directrices are new: two lines at distance a squared over c from the centre. Because a is bigger than c, that distance is bigger than a, so the directrices lie outside the ellipse, one beyond each vertex.
The vertical form below simply swaps the roles: a squared sits under the y term, and the foci and directrices move to the vertical axis. You never need two memorised versions if you always ask which denominator is larger.
Worked example
\[ 9x^2 + 4y^2 - 36x + 24y + 36 = 0 \]
Move the constant and group
Why: x terms together, y terms together.
\[ 9x^2 - 36x + 4y^2 + 24y = -36 \]
Factor out the leading coefficients
Why: Each bracket must start with a plain square.
\[ 9(x^2 - 4x) + 4(y^2 + 6y) = -36 \]
Complete both squares
Why: Adding 4 inside the first bracket adds 36; adding 9 inside the second adds 36. Add both to the right side.
\[ 9(x^2 - 4x + 4) + 4(y^2 + 6y + 9) = -36 + 36 + 36 \]
Factor
Why: Both brackets are perfect squares.
\[ 9(x-2)^2 + 4(y+3)^2 = 36 \]
Divide by 36
Why: Make the right side 1.
\[ \frac{(x-2)^2}{4} + \frac{(y+3)^2}{9} = 1 \]
Read the axes
Why: The larger denominator, 9, is under y: a vertical ellipse.
\[ h = 2,\quad k = -3,\quad a = 3,\quad b = 2 \]
Locate the foci
Why: Subtract the squares for an ellipse.
\[ c = \sqrt{9 - 4} = \sqrt5, \quad \text{foci } (2, -3 \pm \sqrt5) \]
Figure (svg): The vertical ellipse (x minus 2) squared over 4 plus (y plus 3) squared over 9 equals 1, centre (2, minus 3), with foci at (2, minus 3 plus and minus root 5) and dashed directrices y equals about 1.02 and y equals about minus 7.02.
Check with a point
Why: The end of the minor axis, (4, minus 3), must satisfy the original equation. (The book says compare with Equation 7.14; the vertical form is 7.13.)
\[ 9(16) + 4(9) - 36(4) + 24(-3) + 36 = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 679-680 — Example 7.20
The method is the parabola's, done twice. Group the x terms and the y terms, factor out the leading coefficients so each bracket starts with a plain square, and complete both squares.
The bookkeeping is where marks are lost. Adding 4 inside a bracket that has 9 in front adds 36, and adding 9 inside a bracket with 4 in front adds another 36, so the right side goes from minus 36 to 36. Dividing by 36 then gives denominators 4 and 9.
Since 9 is under the y term, this ellipse is vertical: a is 3, b is 2, and c is the square root of 5. The figure shows the foci above and below the centre and the two directrices outside. The check substitutes the minor-axis endpoint (4, minus 3) into the original equation and gets 0.
Prediction
\[ 9x^2 + 16y^2 + 18x - 64y - 71 = 0 \]
Predict first
The y-squared term carries the bigger coefficient, 16. Once this is in standard form, is the ellipse wider than it is tall, or taller than it is wide?
Correct: Wider than tall
Why: Dividing by the right-hand side puts the 16 under the x term and the 9 under the y term. A large coefficient on y squared makes y expensive: y can move only a little before the sum reaches its limit, so the ellipse is short and wide.
Commit to an answer before revealing. The trap is to think that the bigger coefficient on y squared means the ellipse is bigger in the y direction.
It is the other way around. A large coefficient makes the y term grow quickly, so y can only move a little before the total reaches its limit. When you divide through to get the standard form, the 16 ends up underneath the x term, not the y term.
So the ellipse is wider than it is tall. Keep this in mind whenever you read a general form: large coefficients mean short axes.
Worked example
\[ 9x^2 + 16y^2 + 18x - 64y - 71 = 0 \]
Group and factor
Why: Move 71 across and pull out the coefficients.
\[ 9(x^2 + 2x) + 16(y^2 - 4y) = 71 \]
Complete both squares
Why: Adding 1 inside adds 9; adding 4 inside adds 64.
\[ 9(x^2 + 2x + 1) + 16(y^2 - 4y + 4) = 71 + 9 + 64 \]
Factor
Why: The right side is 144.
\[ 9(x+1)^2 + 16(y-2)^2 = 144 \]
Divide by 144
Why: 144 over 9 is 16 and 144 over 16 is 9: the coefficients swap places.
\[ \frac{(x+1)^2}{16} + \frac{(y-2)^2}{9} = 1 \]
Read the features
Why: The larger denominator is under x: horizontal.
\[ \text{centre } (-1, 2),\quad a = 4,\quad b = 3,\quad c = \sqrt{16 - 9} = \sqrt7 \]
Figure (svg): The horizontal ellipse (x plus 1) squared over 16 plus (y minus 2) squared over 9 equals 1, centre (minus 1, 2), with foci at (minus 1 plus and minus root 7, 2) and dashed directrices x equals about 5.05 and x equals about minus 7.05.
Check with a vertex
Why: The right vertex is (3, 2); it must satisfy the original equation.
\[ 9(9) + 16(4) + 18(3) - 64(2) - 71 = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 680 — Checkpoint 7.19
Completing both squares, adding 1 inside the x bracket adds 9, and adding 4 inside the y bracket adds 64, so the right side becomes 144. Dividing by 144 swaps the coefficients: 144 over 9 is 16 and 144 over 16 is 9.
With 16 under the x term the ellipse is horizontal, centred at (minus 1, 2), with a equal to 4, b equal to 3 and c equal to the square root of 7. The figure confirms the prediction from the previous slide: wider than tall.
The check uses the right vertex, which is a units to the right of the centre, at (3, 2). Substituting it into the original equation gives exactly 0.
Error analysis
Annotate
On: \( \frac{x^2}{9} + \frac{y^2}{25} = 1: \quad c = 4, \quad \text{foci } (\pm 4, 0) \)
Look for the mistake before revealing the notes. The arithmetic for c is fine: 25 minus 9 is 16, so c is 4.
The error is the direction. The 25 is under y, so the long axis is vertical, and the foci must lie on it at (0, 4) and (0, minus 4). Putting them on the x-axis is the single most common ellipse mistake.
The last note gives a test that catches it every time: a focus is always inside an ellipse. Substituting (4, 0) into the left side gives 16 over 9, which is more than 1, so that point is outside the curve and cannot be a focus.
Worked example
A whispering gallery has an elliptical ceiling 120 feet long, with foci on the floor 30 feet from the centre. How high is the ceiling at the centre?
Set up the ellipse
Why: Centre on the floor at the origin, major axis along the floor.
\[ 2a = 120 \;\Rightarrow\; a = 60, \qquad c = 30 \]
Use the ellipse relation
Why: The height at the centre is the semi-minor axis b.
\[ b^2 = a^2 - c^2 = 3600 - 900 = 2700 \]
Take the root
Why: 2700 is 900 times 3.
\[ b = \sqrt{2700} = 30\sqrt3 \approx 51.96 \text{ ft} \]
Figure (svg): The upper half of an ellipse 120 feet long and about 52 feet high, with foci on the floor 30 feet either side of the centre. Four sound rays leave the left focus, bounce off the ceiling, and all arrive at the right focus.
Check with the definition
Why: The top of the ceiling is equally far from both foci, and the two distances must add to 2a.
\[ 2\sqrt{30^2 + 2700} = 2\sqrt{3600} = 120 = 2a \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 680, 693 — Figure 7.50 and Exercise 316
Real problems rarely hand you a, b and c directly, so the first job is translation. The length of the room is the major axis, so a is 60. The foci are 30 feet from the centre, so c is 30. The height at the centre is the semi-minor axis b.
The ellipse relation then does the work: b squared is 3600 minus 900, which is 2700, and b is 30 times the square root of 3, about 52 feet.
The figure shows why the room works: sound leaving one focus reflects off the ceiling toward the other, and every path has the same length, 120 feet, so all the echoes arrive together. The check confirms it at the top of the ceiling, where each distance is 60.
Section
Part 3
Concept
hyperbola — The set of all points where the difference between their distances from two fixed points, the foci, is constant.
Figure (svg): The hyperbola x squared over 16 minus y squared over 9 equals 1 with foci at (5, 0) and (minus 5, 0), dashed asymptotes y equals plus and minus three quarters x, and the dashed box with corners (plus or minus 4, plus or minus 3). A point P on the right branch is joined to both foci: 11.17 minus 3.17 equals 8.
\[ \big|d(P, F_1) - d(P, F_2)\big| = 2a \]
At the right vertex (a, 0) the two distances are c plus a and c minus a, so the constant is again 2a. The absolute value lets one rule cover both branches.
\[ (c + a) - (c - a) = 2a \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 681 — definition and Figure 7.51
Change the ellipse's sum to a difference and you get a hyperbola. In the figure, P is 11.17 from the left focus and 3.17 from the right one, a difference of 8, and that stays true all along the right branch.
The absolute value in the definition covers both branches. On the right branch P is closer to the right focus; on the left branch it is closer to the left one. Either way the larger distance exceeds the smaller by the same constant.
Once again the constant is 2a, and the vertex shows it: at (a, 0) the distances are c plus a and c minus a. For c minus a to be a positive distance, c must be bigger than a, so for a hyperbola the foci lie outside the vertices.
Concept
Run the ellipse derivation with a difference in place of the sum. Every squaring step is identical, and you land on the same line:
\[ (a^2 - c^2)x^2 + a^2y^2 = a^2(a^2 - c^2) \]
But now the foci lie outside the vertices, so c is bigger than a and a squared minus c squared is negative. Name its opposite b squared.
\[ b^2 = c^2 - a^2 \;\Longrightarrow\; \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, \quad c^2 = a^2 + b^2 \]
\[ \text{asymptotes } y = k \pm \frac ba(x-h), \quad \text{directrices } x = h \pm \frac{a^2}{c} \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 681-683 — Theorem 7.10, Equations 7.14 and 7.15
You do not need a new derivation. Squaring the difference equation twice produces exactly the same line as the ellipse did. The only difference is the sign of a squared minus c squared.
For the ellipse that number was positive and became b squared. For the hyperbola c is larger than a, so the number is negative, and b squared is defined as its opposite. The minus sign then moves into the equation, and the sum becomes a difference.
That gives Theorem 7.10 with the relation c squared equals a squared plus b squared. Two small warnings about the printed theorem: the directrix line is written with k where it should say h, and a few lines call the hyperbola an ellipse. The formulas on this slide are the corrected ones.
Concept
\[ y = \pm\frac ba\sqrt{x^2 - a^2} \]
\[ y = \pm\frac ba x\sqrt{1 - \frac{a^2}{x^2}} \]
\[ \sqrt{1 - \frac{a^2}{x^2}} \to 1 \quad\Longrightarrow\quad y \approx \pm\frac ba x \]
Figure (svg): The upper right branch of x squared over 16 minus y squared over 9 equals 1 plotted from x equals 4 to 40 together with its asymptote y equals three quarters x; vertical bars show the gap 1.5 at x equals 5, 0.63 at 10, 0.30 at 20 and 0.15 at 40.
Solve the standard form for y and factor x out of the root. Far from the centre the root is almost exactly 1, so the branches run alongside the two lines through the centre with slopes plus and minus b over a. An ellipse is bounded, so it has nothing to approach.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 682 — asymptotes in Theorem 7.10
Solving the standard form for y gives b over a times the square root of x squared minus a squared, with a plus or minus. Factor x out of the root, and what remains inside is 1 minus a squared over x squared.
As x grows, that fraction vanishes and the root approaches 1, so y approaches plus or minus b over a times x. Those two lines are the asymptotes, and the red bars in the figure measure how close the branch gets: 1.5 at x equal to 5, then 0.63, 0.30, and 0.15 at 40, roughly halving each time x doubles.
An ellipse has nothing like this. It is bounded, so it never runs off along a line. Algebraically, a sum of squares does not factor into two linear pieces, while the difference of squares does, and those linear pieces are the asymptotes.
Worked example
\[ 9x^2 - 16y^2 + 36x + 32y - 124 = 0 \]
Group and factor
Why: Move 124 across; factor minus 16 out of the y terms, so the sign changes inside.
\[ 9(x^2 + 4x) - 16(y^2 - 2y) = 124 \]
Complete both squares
Why: Adding 4 inside the first bracket adds 36; adding 1 inside the second SUBTRACTS 16.
\[ 9(x^2 + 4x + 4) - 16(y^2 - 2y + 1) = 124 + 36 - 16 \]
Factor
Why: The right side is 144.
\[ 9(x+2)^2 - 16(y-1)^2 = 144 \]
Divide by 144
Why: Standard form, x term positive: horizontal.
\[ \frac{(x+2)^2}{16} - \frac{(y-1)^2}{9} = 1 \]
Read the features
Why: For a hyperbola the squares add to give c.
\[ h = -2,\; k = 1,\; a = 4,\; b = 3,\; c = \sqrt{16 + 9} = 5 \]
Write the asymptotes
Why: Slopes plus and minus b over a, through the centre.
\[ y = 1 \pm \frac34(x+2) \]
Figure (svg): The hyperbola (x plus 2) squared over 16 minus (y minus 1) squared over 9 equals 1, centre (minus 2, 1), branches opening left and right through the vertices (2, 1) and (minus 6, 1), foci (3, 1) and (minus 7, 1), dashed asymptotes y equals 1 plus or minus three quarters of (x plus 2), and dashed directrices x equals 1.2 and x equals minus 5.2.
Check with a vertex
Why: The right vertex is (2, 1). (The book compares with Equation 7.15; the horizontal form is 7.14.)
\[ 9(4) - 16(1) + 36(2) + 32(1) - 124 = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 683-684 — Example 7.21
The only new hazard is the minus sign. Factoring minus 16 out of the y terms turns plus 32y into minus 2y inside the bracket, and completing that square with plus 1 subtracts 16 from the left side, not adds it. So the right side is 124 plus 36 minus 16, which is 144.
After dividing by 144, the positive term is the x term, so the hyperbola opens left and right, with a equal to 4 and b equal to 3. For a hyperbola c comes from adding: 16 plus 9 is 25, so c is 5 and the foci are 5 either side of the centre.
The asymptotes pass through the centre with slopes plus and minus three quarters. The figure also shows the directrices, which lie between the centre and the vertices, the reverse of an ellipse. The check uses the right vertex, and the book's reference to Equation 7.15 should be to 7.14, the horizontal form.
Worked example
\[ 4y^2 - 9x^2 + 16y + 18x - 29 = 0 \]
Group and factor
Why: Keep the positive squared term first.
\[ 4(y^2 + 4y) - 9(x^2 - 2x) = 29 \]
Complete both squares
Why: Adding 4 inside adds 16; adding 1 inside the second bracket subtracts 9.
\[ 4(y^2 + 4y + 4) - 9(x^2 - 2x + 1) = 29 + 16 - 9 \]
Factor
Why: The right side is 36.
\[ 4(y+2)^2 - 9(x-1)^2 = 36 \]
Divide by 36
Why: The y term is the positive one: vertical.
\[ \frac{(y+2)^2}{9} - \frac{(x-1)^2}{4} = 1 \]
Read the features
Why: Here a goes with y.
\[ \text{centre } (1, -2),\quad a = 3,\quad b = 2,\quad c = \sqrt{13} \]
Write the asymptotes
Why: For a vertical hyperbola the slopes are a over b.
\[ y = -2 \pm \frac32(x-1) \]
Figure (svg): The hyperbola (y plus 2) squared over 9 minus (x minus 1) squared over 4 equals 1, centre (1, minus 2), branches opening up and down through the vertices (1, 1) and (1, minus 5), foci at (1, minus 2 plus and minus root 13), and dashed asymptotes y equals minus 2 plus or minus three halves of (x minus 1).
Check with a vertex
Why: The upper vertex is (1, 1).
\[ 4(1) - 9(1) + 16(1) + 18(1) - 29 = 0 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 684 — Checkpoint 7.20
Here the y squared term is the positive one, so the hyperbola opens up and down, and a belongs to y: a is 3 and b is 2. That choice is not about which denominator is larger, as it was for the ellipse, but about which term is positive.
The asymptotes need care. For a vertical hyperbola, solving for y gives slopes a over b, which is three halves, not b over a. The figure shows the steep asymptotes and the branches between them.
The check uses the upper vertex, 3 above the centre at (1, 1), and it satisfies the original equation exactly.
Trap
Locating the foci of a hyperbola:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
\[ c^2 = 16 - 9 = 7 \]
Wrong. That is the ellipse relation.
A hyperbola's foci lie beyond its vertices, so c must be bigger than a. For a hyperbola the squares add.
\[ c^2 = a^2 + b^2 = 25 \]
\[ c = 5 > 4 = a \]
This mix-up is natural because the two derivations are nearly identical. But the relations point in opposite directions: an ellipse's foci are inside, so c is less than a and the squares subtract; a hyperbola's foci are outside, so c is more than a and the squares add.
A quick check catches the error. With c equal to the square root of 7, about 2.65, the supposed foci would sit inside the vertices at 4, between the two branches, which is impossible for a hyperbola. The correct value 5 puts them beyond the vertices, where they belong.
Discrimination
Sort into buckets
None of these has an xy term. Sort each by the squared terms alone.
Sort each equation before checking. With no xy term, you only need the squared terms: how many are there, and do their coefficients have the same sign?
One squared variable means a parabola. Two squares with the same sign mean an ellipse, and equal coefficients make it a circle. Two squares with opposite signs mean a hyperbola, and it does not matter which one comes first; the item that starts with 4y squared is a hyperbola too.
Most of these are equations you have already worked on this lesson. Classifying before completing the square tells you what shape to expect, so you notice at once if your algebra drifts somewhere impossible.
Counterexample
Discussion prompt
The book calls Ax² + By² + Cx + Dy + E = 0 an ellipse when A and B have the same sign. Find an equation of exactly that form whose graph is not an ellipse at all.
Try to find one before revealing. The point of this exercise is that the sign test tells you the type of equation, not whether it has a graph.
Completing the square turns the example into a sum of squares equal to minus 4. No real point makes a sum of squares negative, so the graph is empty. Change the constant and the right side can become 0, leaving only the single point where both squares vanish.
These are called degenerate conics, and they correspond to cutting planes that miss the cone or pass only through its tip. In practice, completing the square always reveals them: an ellipse needs a positive number on the right-hand side.
Section
Part 4
Concept
eccentricity — The distance from a point on the conic to a focus, divided by its perpendicular distance to the nearest directrix. It is the same number at every point of the curve.
\[ e = \frac{d(P, F)}{d(P, \text{directrix})} \]
Figure (svg): Three conics sharing the focus at the origin and the directrix x equals 1: an ellipse with e equal to one half (closed, from x equals minus 1 to one third), a parabola with e equal to 1 opening left with vertex at one half, and a hyperbola with e equal to 2 whose near branch has vertex at two thirds and whose far branch lies right of the directrix.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 684-685 — definition of eccentricity and Figure 7.54
The parabola's definition said the distance to the focus equals the distance to the directrix, a ratio of one. Eccentricity generalises that: allow the ratio to be any fixed positive number e.
The figure keeps one focus and one directrix and changes only e. With e equal to one half, points must be twice as far from the line as from the focus, which keeps them huddled near the focus: a closed ellipse. With e equal to 1 you get the parabola, and with e equal to 2 points must be closer to the line than to the focus, which allows two branches, one on each side of the directrix.
So the three curves are one family, controlled by one number, and the parabola is the boundary case, just as the cone picture suggested.
Concept
Measure both distances from the right-hand vertex (a, 0) of a horizontal ellipse: to the focus (c, 0), and to the directrix at a squared over c.
\[ d(\text{vertex}, F) = a - c, \qquad d(\text{vertex}, \text{directrix}) = \frac{a^2}{c} - a \]
\[ e = \frac{a - c}{a^2/c - a} = \frac{c(a - c)}{a^2 - ac} \]
\[ e = \frac{c(a - c)}{a(a - c)} = \frac ca \]
For an ellipse the foci are inside, so c is less than a and e is less than one. The same computation for a hyperbola, whose foci are outside, gives e equal to c over a again, now bigger than one.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 685 — the eccentricity of an ellipse
The eccentricity is the same at every point, so you may measure it at the most convenient one, the vertex. From the vertex (a, 0), the focus is a minus c away and the directrix is a squared over c minus a away.
Multiply the top and bottom of the ratio by c and factor: the bottom becomes a times a minus c, which cancels with the top's a minus c and leaves c over a. The printed version of this step writes the second distance as a squared over c minus c, which is a slip; the formula line is correct.
Now e equal to c over a explains everything. The foci of an ellipse are inside, so e is less than 1; those of a hyperbola are outside, so e is more than 1. As a check, measure at the top of the minor axis instead: it is a from the focus and a squared over c from the directrix, and the ratio is again c over a.
Picture it
Figure (svg): The ellipse x squared over 25 plus y squared over 9 equals 1 with focus (4, 0) and directrix x equals 6.25. Two points on the ellipse are joined to the focus and horizontally to the directrix: 1.94 over 2.43 and 7.67 over 9.58, both equal to 0.8.
Two very different points of the same ellipse, measured to the right focus and the right directrix. The distances change completely; their ratio does not.
This figure checks the claim numerically on the ellipse with a equal to 5 and c equal to 4. For each marked point, the solid yellow segment runs to the focus and the dashed green one runs across to the directrix at 6.25.
The point near the focus has distances 1.94 and 2.43; the point on the far side has 7.67 and 9.58. Completely different lengths, but both ratios equal 0.8, which is c over a.
This is what makes the focus-directrix definition powerful: a single fixed ratio, checked anywhere on the curve, captures the whole shape.
Worked example
\[ \frac{(x-3)^2}{16} + \frac{(y+2)^2}{25} = 1 \]
Identify a and b
Why: The larger denominator is a squared; it sits under y, so the ellipse is vertical.
\[ a^2 = 25,\; b^2 = 16 \;\Rightarrow\; a = 5,\; b = 4 \]
Find c
Why: For an ellipse, subtract.
\[ c^2 = a^2 - b^2 = 9 \;\Rightarrow\; c = 3 \]
Divide
Why: Eccentricity is c over a.
\[ e = \frac ca = \frac35 = 0.6 \]
Check with the definition
Why: Use the top vertex (3, 3), the top focus (3, 1), and the top directrix y equal to minus 2 plus 25 over 3, which is 19 over 3.
\[ \frac{3 - 1}{\tfrac{19}{3} - 3} = \frac{2}{10/3} = 0.6 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 686 — Example 7.22
Finding the eccentricity is a three-step routine: identify a and b, find c with the correct relation, then divide c by a. The only judgement is which denominator is a squared, and for an ellipse it is the larger one, 25.
Subtracting gives c squared equal to 9, so c is 3 and e is three fifths. The book stops there.
The check goes back to the definition, which is worth doing once to see that it is not just a formula. The top vertex is at (3, 3), the focus above the centre at (3, 1), and the directrix at height 19 thirds. The two distances are 2 and 10 thirds, and their ratio is 0.6.
Worked example
\[ \frac{(y-3)^2}{49} - \frac{(x+2)^2}{25} = 1 \]
Identify a and b
Why: For a hyperbola, a goes with the POSITIVE term, whatever its size.
\[ a^2 = 49,\; b^2 = 25 \;\Rightarrow\; a = 7,\; b = 5 \]
Find c
Why: For a hyperbola, add.
\[ c^2 = a^2 + b^2 = 74 \;\Rightarrow\; c = \sqrt{74} \approx 8.602 \]
Divide
Why: Bigger than one, as a hyperbola must be.
\[ e = \frac{\sqrt{74}}{7} \approx 1.229 \]
Check with the definition
Why: Top vertex (minus 2, 10); top focus at height 3 plus root 74, about 11.602; top directrix at height 3 plus 49 over root 74, about 8.696.
\[ \frac{11.602 - 10}{10 - 8.696} = \frac{1.602}{1.304} \approx 1.229 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 686 — Checkpoint 7.21
Two things change for a hyperbola. First, a goes with the positive term, even if its denominator were the smaller one. Here the y term is positive, so a squared is 49. Second, c comes from adding: 49 plus 25 is 74.
Dividing the square root of 74 by 7 gives about 1.229, more than 1, as it must be for a hyperbola. If your eccentricity for a hyperbola ever comes out below 1, you have used the ellipse relation.
The check again measures at the top vertex: about 1.602 to the focus and 1.304 to the directrix, a ratio of about 1.229.
Intuition
Figure (svg): Four ellipses centred at the origin, all with semi-major axis 4: eccentricity 0 is a circle of radius 4; 0.6 has semi-minor axis 3.2; 0.8 has 2.4; 0.95 has about 1.25, a thin oval.
\[ b = a\sqrt{1 - e^2} \]
Double every length of an ellipse and a and c both double, so their ratio does not move: every ellipse with the same eccentricity is the same shape at a different scale. What e controls is how flattened the shape is, and the flattening is slow at first and sudden near one.
All four curves have the same semi-major axis, 4, and different eccentricities. The formula b equals a times the square root of 1 minus e squared tells you how much the minor axis shrinks.
The shrinkage is slow at first. At e equal to 0.6, b is still 3.2, eighty percent of a. It takes e equal to 0.95 to squash the ellipse to a thin oval. That is why orbits with modest eccentricities still look almost circular.
Because e is a ratio of two lengths, scaling a picture up or down does not change it. Two ellipses with the same eccentricity are the same shape, one a magnified copy of the other, just as all circles are the same shape.
Ranking
Put in order
Put these in order from the smallest eccentricity to the largest.
Why: Circle 0; Earth's orbit 0.0167; the first ellipse has c = 3, so e = 0.6; the second has c = 4, so e = 0.8; every parabola has e = 1; the hyperbola has c = √2, so e = √2 ≈ 1.414.
Order the cards before checking. You will need to compute two of them and remember the rest.
The circle has eccentricity 0 and Earth's orbit only 0.0167. For the two ellipses with a equal to 5, c is 3 when b is 4 and c is 4 when b is 3, giving 0.6 and 0.8: the flatter ellipse has the larger eccentricity. Every parabola has eccentricity exactly 1, whatever its size.
The hyperbola x squared minus y squared equals 1 has a and b both equal to 1, so c is the square root of 2 and e is about 1.414. Hyperbolas with equal a and b are called rectangular, and they all have this eccentricity.
Estimation
\[ e_{\oplus} = 0.0167, \qquad b = a\sqrt{1 - e^2} \]
Predict first
Draw Earth's orbit 10 cm across. How much shorter than its width is its height?
Correct: About a hundredth of a millimetre
Why: b = a√(1 − 0.0167²) ≈ 0.99986a, so 100 mm across means 99.986 mm tall: a difference of 0.014 mm, thinner than a hair. Yet the Sun is off-centre by c = ae ≈ 0.8 mm, which you could see. A small eccentricity moves the focus long before it squashes the shape.
\[ 100\text{ mm} \times (1 - 0.99986) \approx 0.014\text{ mm}, \qquad c = 50\text{ mm} \times 0.0167 \approx 0.84\text{ mm} \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 680 — Kepler's first law and Figure 7.50
Make your guess first. Most people guess far too high, because diagrams of Earth's orbit are usually drawn as obvious ovals.
The formula for b shows the truth. With e equal to 0.0167, the square root of 1 minus e squared is 0.99986, so a drawing 100 millimetres across would be about 99.986 millimetres tall. No printer could show the difference.
Yet the Sun sits at a focus, c from the centre, and c equals a times e, about 0.8 millimetres in the same drawing. You could see that. A small eccentricity shifts the focus noticeably long before it visibly squashes the shape, which is why Earth's distance from the Sun varies by about three percent over a year while its orbit looks perfectly round.
Comparison
Comparison matrix
| ellipse | parabola | hyperbola | |
|---|---|---|---|
| defining distances | sum = 2a | equal | difference = 2a |
| eccentricity | e < 1 | e = 1 | e > 1 |
| number of foci | 2 | 1 | 2 |
| c from a and b | c² = a² − b² | (uses p) | c² = a² + b² |
| asymptotes | none | none | y − k = ±(b/a)(x − h) |
Fill each blank before revealing. The table collects the lesson so far into one place, and the blanks are the facts most often confused.
Read the rows as a comparison. The distance rule is a sum for the ellipse, equality for the parabola and a difference for the hyperbola. The eccentricity is below, at and above one. The relation for c subtracts for the ellipse and adds for the hyperbola.
Only the hyperbola has asymptotes, because only its branches run off to infinity along lines. A parabola also runs off to infinity, but it bends away ever more steeply and approaches no line.
Section
Part 5
Concept
focal parameter — The distance p from a focus to its nearest directrix.
| conic | e | focal parameter p |
|---|---|---|
| ellipse | 0 < e < 1 | (a² − c²)/c = a(1 − e²)/e |
| parabola | e = 1 | 2a, where a is the vertex-to-focus distance |
| hyperbola | e > 1 | (c² − a²)/c = a(e² − 1)/e |
Check the ellipse row on the ellipse with a equal to 5 and c equal to 4. The printed table divides by c at the end of that row; it must be e, as the numbers show.
\[ \frac{a^2}{c} - c = \frac{25}{4} - 4 = 2.25 = \frac{5(1 - 0.64)}{0.8} \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 686 — Table 7.7
The polar equation needs one more length: the focal parameter p, the distance from a focus to the directrix on the same side. For the parabola this is 2a, where a is the vertex-to-focus distance, since the directrix is as far beyond the vertex as the focus is before it.
For an ellipse, the focus is at c and the directrix at a squared over c, so p is their difference. Writing c as a times e turns that into a times 1 minus e squared, all over e. The printed table ends that entry with c instead of e; the check on this slide, with a equal to 5 and e equal to 0.8, shows that e is correct, since both sides come out to 2.25.
The hyperbola's entry is the same with the subtraction reversed. You will use these formulas when you know a and e, as with a planet's orbit, and need the polar equation.
Concept
Figure (svg): An ellipse with focus at the origin and directrix x equals p. A point P is joined to the origin by a segment labelled r at angle theta; a horizontal segment from P to the directrix is labelled p minus r cos theta; the foot of P on the x-axis marks the distance r cos theta, which here is negative.
Put a focus at the pole and the directrix at x equal to p. The definition says the distance to the focus is e times the distance to the directrix.
\[ r = e\,(p - r\cos\theta) \]
\[ r + er\cos\theta = ep \]
\[ r = \frac{ep}{1 + e\cos\theta} \]
Nothing in the algebra cared whether e was below, at or above one. One equation, one parameter, all three conics.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 686-687 — Theorem 7.11
Polar coordinates suit conics because r is the distance from the origin, so if a focus sits at the origin, one of the two distances in the definition is simply r. The other is the horizontal gap from P to the directrix at x equal to p, which is p minus the x-coordinate of P, and that coordinate is r cos theta.
The definition says the first distance is e times the second. Expand, move the r cos theta term across, factor out r, and divide. Three lines of algebra give the whole theorem.
The striking thing is that nothing in the derivation cares about the value of e. The same equation describes the ellipse, parabola and hyperbola, which the rectangular forms could never do: a sum of squares cannot slide continuously into a difference, but a number e can slide through 1.
Notation
Annotate
On: \( r = \frac{ep}{1 \pm e\cos\theta} \qquad\text{or}\qquad r = \frac{ep}{1 \pm e\sin\theta} \)
Step through the notes in order, because that is also the order in which you read a polar conic. First, a focus is at the pole. Second, make the constant in the denominator equal to 1, dividing top and bottom if necessary. Third, read the eccentricity as the coefficient of the trig function.
Then read the orientation. Cosine means the axis of the conic is horizontal and the directrix is a vertical line; sine means a vertical axis and a horizontal directrix. The sign tells you which side of the pole the directrix is on: plus for the positive side, minus for the negative side.
Finally, the numerator is e times p, so once you know e you can recover the focal parameter and the position of the directrix.
Tweak it
Parameter explorer
The focus is at the origin and the directrix is x = 1. The curve is the upper half of every point whose distance to the focus is e times its distance to the directrix. Slide e through 1.
\[ r = \frac{{e}}{1 + {e}\cos\theta} \]
\[ x^2 + y^2 = e^2(1 - x)^2 \;\Longrightarrow\; y = \sqrt{e^2(1-x)^2 - x^2} \]
The widget draws the upper half of the conic with focus at the origin and directrix x equal to 1; the lower half is its mirror image. Start at 0.6 and move e slowly upward.
Below 1, the curve is a closed arc, the top of an ellipse, and it stretches farther to the left as e grows. Watch what happens as e reaches 1: the left end never comes back, and the curve becomes a parabola. Push past 1 and a second piece appears to the right of the directrix. That is the second branch of a hyperbola.
Notice that the change is continuous. Nothing jumps; the ellipse stretches until it opens. That continuity is what the single polar equation captures and the three separate rectangular equations hide.
Worked example
\[ r = \frac{3}{1 + 2\cos\theta} \]
Read the eccentricity
Why: The constant in the denominator is already 1.
\[ e = 2 > 1 \;\Rightarrow\; \text{a hyperbola, horizontal axis} \]
Find the focal parameter
Why: The numerator is e times p.
\[ ep = 3 \;\Rightarrow\; p = \tfrac32, \quad \text{directrix } x = \tfrac32 \]
Tabulate the axis angles
Why: At theta equal to pi the value of r is negative, so the point plots on the opposite side.
\[ r(0) = 1, \quad r\left(\tfrac{\pi}{2}\right) = 3, \quad r(\pi) = -3 \]
Tabulate the diagonals
Why: These give the table's two irrational values.
\[ r\left(\tfrac{\pi}{4}\right) = \frac{3}{1 + \sqrt2} \approx 1.2426, \quad r\left(\tfrac{3\pi}{4}\right) = \frac{3}{1 - \sqrt2} \approx -7.2426 \]
Locate the vertices and centre
Why: Theta equal to 0 gives (1, 0); r equal to minus 3 at theta equal to pi gives (3, 0).
\[ V_1 = (1, 0),\; V_2 = (3, 0) \;\Rightarrow\; \text{centre } (2, 0),\; a = 1 \]
Find c and b
Why: The focus at the origin is 2 from the centre.
\[ c = 2, \quad b^2 = c^2 - a^2 = 3 \;\Rightarrow\; (x-2)^2 - \frac{y^2}{3} = 1 \]
Figure (svg): The hyperbola r equals 3 over (1 plus 2 cos theta) with focus at the origin: the near branch passes through (1, 0) and opens left around the origin, the far branch passes through (3, 0) and opens right. The eight table points are marked, the directrix x equals 1.5 is dashed, and the asymptotes through the centre (2, 0) have slopes plus and minus root 3.
Check two ways
Why: The ratio c over a must reproduce e, and the point (0, 3) from theta equal to pi over 2 must satisfy the rectangular equation.
\[ \frac ca = \frac21 = 2 = e, \qquad (0-2)^2 - \frac{3^2}{3} = 4 - 3 = 1 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 687-688 — Example 7.23 and Figure 7.55
The constant in the denominator is already 1, so the eccentricity can be read at once: 2, a hyperbola. The numerator 3 is e times p, so p is three halves and the directrix is the line x equals three halves.
The table of values needs care at theta equal to pi, where r is minus 3. A negative r means you go 3 units in the opposite direction, so the point is (3, 0), not (minus 3, 0). That point is the vertex of the far branch; theta equal to 0 gives the near vertex (1, 0). The figure colours the two branches by the sign of r.
From the two vertices, the centre is (2, 0) and a is 1. The focus at the origin is 2 from the centre, so c is 2 and b squared is 3. The check confirms that c over a reproduces e, and that the point (0, 3) satisfies the rectangular equation.
Worked example
\[ r = \frac{4}{1 - 0.8\sin\theta} \]
Read the eccentricity
Why: The constant is 1 and the trig function is sine.
\[ e = 0.8 < 1 \;\Rightarrow\; \text{an ellipse, vertical axis} \]
Find the focal parameter
Why: Minus with sine puts the directrix below the pole.
\[ ep = 4 \;\Rightarrow\; p = 5, \quad \text{directrix } y = -5 \]
Find the vertices
Why: Sine is 1 at the top and minus 1 at the bottom.
\[ r\left(\tfrac{\pi}{2}\right) = \frac{4}{0.2} = 20, \quad r\left(\tfrac{3\pi}{2}\right) = \frac{4}{1.8} = \frac{20}{9} \]
Find a
Why: The vertices are (0, 20) and (0, minus 20 over 9).
\[ 2a = 20 + \tfrac{20}{9} = \tfrac{200}{9} \;\Rightarrow\; a = \tfrac{100}{9} \approx 11.11 \]
Find the centre, c and b
Why: The centre is the midpoint of the vertices; the pole is a focus.
\[ \text{centre } \left(0, \tfrac{80}{9}\right), \quad c = \tfrac{80}{9}, \quad b = \sqrt{a^2 - c^2} = \tfrac{20}{3} \]
Figure (svg): The ellipse r equals 4 over (1 minus 0.8 sin theta), tall and vertical, with one focus at the origin and the other at (0, 17.8), vertices at (0, 20) and (0, minus 2.2), centre (0, 8.9), and the dashed directrix y equals minus 5 below it.
Check the eccentricity two ways
Why: c over a must be 0.8, and so must the ratio at the lower vertex: 20 over 9 to the focus, 25 over 9 to the directrix.
\[ \frac ca = \frac{80}{100} = 0.8, \qquad \frac{20/9}{25/9} = 0.8 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 688 — Checkpoint 7.22
Again the constant is 1, so the eccentricity is 0.8, an ellipse. Sine means the axis is vertical, and the minus sign puts the directrix below the pole: ep is 4, so p is 5 and the directrix is y equals minus 5.
The vertices come from the angles where sine is 1 and minus 1. At the top, r is 4 over 0.2, which is 20; at the bottom, r is 4 over 1.8, which is 20 ninths. The major axis is their sum, so a is 100 ninths, about 11.1, and the centre is halfway between the vertices, at height 80 ninths.
The distance from the centre to the focus at the pole is c, also 80 ninths, and b comes out to 20 thirds. The check computes the eccentricity twice, as c over a and as a ratio of distances at the lower vertex, and both give 0.8.
Worked example
Find a polar equation for the conic with a focus at the origin, directrix the line x equal to minus 4, and eccentricity 5.
Choose the form
Why: The directrix is vertical and on the negative side: cosine with a minus sign.
\[ r = \frac{ep}{1 - e\cos\theta} \]
Substitute
Why: The focal parameter is the focus-to-directrix distance, 4.
\[ e = 5,\; p = 4 \;\Rightarrow\; r = \frac{20}{1 - 5\cos\theta} \]
Name the conic
Why: Eccentricity above one.
\[ e = 5 > 1 \;\Rightarrow\; \text{a hyperbola} \]
Check a point against the definition
Why: At theta equal to pi, r is 20 over 6, the point (minus 10 over 3, 0). It is 10 over 3 from the focus and 2 over 3 from the directrix.
\[ \frac{10/3}{\left|-\tfrac{10}{3} + 4\right|} = \frac{10/3}{2/3} = 5 = e \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 692 — Exercise 286
This runs Theorem 7.11 in reverse. The directrix x equals minus 4 is vertical, which means cosine, and it lies on the negative side of the pole, which means a minus sign in the denominator.
The focal parameter is the distance from the focus to the directrix, which is 4, and e is 5, so the numerator is 20. The eccentricity is above 1, so the conic is a hyperbola.
The check picks theta equal to pi, where the cosine is minus 1 and r is 20 over 6. The point is 10 thirds to the left of the pole, 10 thirds from the focus and 2 thirds from the directrix, a ratio of 5, as required.
Trap
Exercise 281, read too fast:
\[ r = \frac{5}{2 + \sin\theta} \]
\[ e = 1 \;\Rightarrow\; \text{parabola} \]
Wrong. The denominator starts with 2.
The coefficient is the eccentricity only when the constant term is 1. Divide top and bottom by 2 first.
\[ r = \frac{5/2}{1 + \tfrac12\sin\theta} \]
\[ e = \tfrac12 \;\Rightarrow\; \text{ellipse} \]
The coefficient of the sine is the eccentricity only after the denominator's constant has been made 1. Here the denominator starts with 2, so dividing top and bottom by 2 turns the coefficient 1 into one half.
The difference is not cosmetic: the fast reading says parabola, the correct one says ellipse. Build the habit of looking at the constant first, every time, before you look at the trig coefficient.
Matching
Match the pairs
Why: Normalise each constant to 1 first. (a) already has it: e = 1, and the negative numerator only reflects the curve. (b) divide by 2: e = 1/2. (c) divide by 2: e = 6/2 = 3. (d) divide by −4: r = (−3/4)/(1 − (3/4) sin θ), so e = 3/4.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 692 — Exercises 279 to 284
Normalise before you match. Each equation has to be rewritten with 1 as the constant in the denominator, and only then can the coefficient be read.
The first already has a 1; its numerator is negative, which only turns the curve around, so it is still a parabola with eccentricity 1. The second and third divide by 2, giving one half and 3. The fourth is the subtle one: its constant is minus 4, and dividing by minus 4 flips the signs, giving eccentricity three quarters.
Eccentricity is never negative, so if you ever read a negative one, you have not yet divided by the constant correctly.
Real world
Figure (svg): The orbit of Halley's comet drawn from r equals 1.1645 over (1 plus 0.967 cos theta): a very long thin ellipse with the Sun at the pole near its right end, perihelion 0.59 AU to the right of the Sun and aphelion 35.3 AU to the left. Earth's nearly circular orbit, radius 1, is drawn around the Sun for scale.
Discussion prompt
Exercise 318: Halley's comet has a major axis of 35.88 AU and eccentricity 0.967. With the Sun at the pole, write the polar equation of its orbit and find its closest approach to the Sun.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 693 — Exercise 318
Try the set-up before revealing. The major axis gives a, and the focal-parameter table gives ep directly as a times 1 minus e squared, so the numerator is about 1.1645 AU.
With the Sun at the pole and the plus-cosine form, the comet is closest to the Sun at theta equal to 0, where r is 1.1645 over 1.967, about 0.592 AU. That agrees with a times 1 minus e, the distance from a focus to the nearer vertex, which is a good check.
The figure puts the orbit next to Earth's to scale. An eccentricity of 0.967 is close to 1, and the orbit is correspondingly extreme: it dips inside the orbit of Venus and swings out past Neptune before returning, about every 76 years.
Section
Part 6
Concept
\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
Every conic, in any position, has an equation of this shape. The cross term Bxy is the new ingredient: it appears exactly when the conic's axes are tilted away from the coordinate axes.
discriminant — The number 4AC minus B squared. It decides the conic without any rotating or completing of squares.
\[ B = 0: \quad 4AC > 0 \iff A, C \text{ same sign} \]
With no cross term the rule reduces to the sign test from Part 3: same signs for an ellipse, one square missing for a parabola, opposite signs for a hyperbola.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 688 — general equations of degree two
Every conic, wherever it sits and however it is tilted, satisfies an equation with six coefficients like this one. The x and y terms, D and E, only shift the centre, and completing the square handles them. The cross term is different: it appears precisely when the conic's axes are not parallel to the coordinate axes.
The discriminant, 4AC minus B squared, names the conic without any further work. Its sign alone decides: positive for an ellipse, zero for a parabola, negative for a hyperbola.
When there is no cross term, the discriminant is just 4AC, and its sign is positive exactly when A and C have the same sign. So the sign test from Part 3 is the special case B equal to 0 of this rule, not a separate fact.
Picture it
Figure (svg): The two branches of y equals one over x, with the dashed lines y equals x and y equals minus x as rotated axes. The vertices (1, 1) and (minus 1, minus 1) lie on the line y equals x, and the foci are farther out on the same line at (1.41, 1.41) and (minus 1.41, minus 1.41).
\[ x = \frac{x' - y'}{\sqrt2}, \quad y = \frac{x' + y'}{\sqrt2} \;\Longrightarrow\; xy = \frac{x'^2 - y'^2}{2} = 1 \]
Its discriminant is 0 minus 1, negative: a hyperbola. In the turned coordinates it is x prime squared over 2 minus y prime squared over 2 equal to 1, with a equal to b equal to root 2.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 688-689 — Figure 7.56
The graph of y equals one over x is familiar, but you may never have called it a hyperbola. Its discriminant is 0 minus 1, which is negative, so it must be one.
Rotating the axes by 45 degrees shows it plainly. Substitute the rotation formulas for x and y, multiply, and the product becomes half the difference of the new squares. In the turned coordinates the equation is a standard hyperbola with a and b both equal to the square root of 2.
The picture shows the new axes as dashed lines. The hyperbola opens along the line y equals x, its vertices are at (1, 1) and (minus 1, minus 1), and its asymptotes are the old coordinate axes. Its foci are at distance 2 from the origin along that line.
Notation
Annotate
On: \( \cot 2\theta = \frac{A - C}{B} \;\Longrightarrow\; B' = 0 \)
\[ A' = A\cos^2\theta + B\cos\theta\sin\theta + C\sin^2\theta \]
\[ C' = A\sin^2\theta - B\sin\theta\cos\theta + C\cos^2\theta \]
\[ D' = D\cos\theta + E\sin\theta, \quad E' = -D\sin\theta + E\cos\theta, \quad F' = F \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 689 — the rotated coefficients
The whole purpose of the rotation is to make the cross term vanish in the new coordinates, and the formula for cot 2 theta is exactly the condition that does it. You never have to derive it here; you have to know what it is for.
Notice the special case in the second note. When A equals C, the cotangent is 0, so 2 theta is 90 degrees and theta is 45 degrees, whatever B is. That is why xy equals 1, and any equation whose squared coefficients match, is turned by 45 degrees.
The formulas below compute the new coefficients. A prime and C prime mix all three of A, B and C; the linear terms rotate like a vector; the constant is unchanged. Only A prime and C prime need real work in the examples that follow.
Worked example
\[ 13x^2 - 6\sqrt3\,xy + 7y^2 - 256 = 0 \]
Read the coefficients
Why: D and E are zero.
\[ A = 13,\quad B = -6\sqrt3,\quad C = 7,\quad F = -256 \]
Compute the discriminant
Why: B squared is 36 times 3.
\[ 4AC - B^2 = 364 - 108 = 256 > 0 \;\Rightarrow\; \text{ellipse} \]
Find the angle
Why: Cotangent of minus one over root 3 is 120 degrees.
\[ \cot 2\theta = \frac{13 - 7}{-6\sqrt3} = -\frac{1}{\sqrt3} \;\Rightarrow\; 2\theta = 120^\circ,\; \theta = 60^\circ \]
Compute A prime
Why: Cos 60 is one half, sin 60 is root 3 over 2.
\[ A' = 13\cdot\tfrac14 - 6\sqrt3\cdot\tfrac{\sqrt3}{4} + 7\cdot\tfrac34 = \tfrac{13 - 18 + 21}{4} = 4 \]
Compute C prime
Why: The cross term now enters with a plus sign.
\[ C' = 13\cdot\tfrac34 + 6\sqrt3\cdot\tfrac{\sqrt3}{4} + 7\cdot\tfrac14 = \tfrac{39 + 18 + 7}{4} = 16 \]
Rewrite and divide by 256
Why: D prime and E prime are zero; F prime is minus 256.
\[ 4x'^2 + 16y'^2 = 256 \;\Rightarrow\; \frac{x'^2}{64} + \frac{y'^2}{16} = 1 \]
Figure (svg): The ellipse 13x squared minus 6 root 3 xy plus 7y squared equals 256, drawn from x prime squared over 64 plus y prime squared over 16 equals 1 on axes turned 60 degrees: its long axis, length 16, runs along the dashed x prime line at 60 degrees, and its short axis, length 8, along the dashed y prime line.
Check with the invariant
Why: The discriminant does not change under rotation, so the new one must also be 256. (The printed C prime line has a stray equals sign and drops the 13; the value 16 is right.)
\[ 4A'C' - B'^2 = 4(4)(16) - 0 = 256 = 4AC - B^2 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, pp. 689-691 — Example 7.24 and Figure 7.57
Start with the discriminant, because it answers the question of what the curve is before any trigonometry. B squared is 36 times 3, which is 108, and 4AC is 364, so the discriminant is 256, positive: an ellipse.
The angle comes from cot 2 theta equal to 6 over minus 6 root 3, which is minus one over root 3. The angle between 0 and 180 degrees with that cotangent is 120 degrees, so the axes turn by 60 degrees. Then the new coefficients use cos 60 equal to one half and sin 60 equal to root 3 over 2; notice that the root 3 in B cancels against the root 3 in the sine, which is why the answers come out whole.
Dividing by 256 gives semi-axes 8 and 4 along the turned axes, as the figure shows. The check uses a fact worth remembering: the discriminant does not change under rotation. The new one, 4 times 4 times 16, is 256 again. The printed working for C prime contains a typo, an equals sign in place of a plus and a missing 13, but its final value 16 is correct.
Worked example
\[ 3x^2 + 5xy - 2y^2 - 125 = 0 \]
Compute the discriminant
Why: A is 3, B is 5, C is minus 2.
\[ 4AC - B^2 = 4(3)(-2) - 25 = -49 < 0 \;\Rightarrow\; \text{hyperbola} \]
Find the angle
Why: Cotangent 1 means 45 degrees.
\[ \cot 2\theta = \frac{3 - (-2)}{5} = 1 \;\Rightarrow\; 2\theta = 45^\circ,\; \theta = 22.5^\circ \]
Rewrite A prime with double angles
Why: Cos squared is one half of one plus cos 2 theta; sin times cos is half of sin 2 theta.
\[ A' = \frac{A + C}{2} + \frac{A - C}{2}\cos 2\theta + \frac B2\sin 2\theta \]
Evaluate A prime
Why: Cos 45 and sin 45 are both root 2 over 2.
\[ A' = \tfrac12 + \tfrac52\cdot\tfrac{\sqrt2}{2} + \tfrac52\cdot\tfrac{\sqrt2}{2} = \tfrac{1 + 5\sqrt2}{2} \approx 4.036 \]
Evaluate C prime the same way
Why: The last two signs flip.
\[ C' = \tfrac12 - \tfrac{5\sqrt2}{4} - \tfrac{5\sqrt2}{4} = \tfrac{1 - 5\sqrt2}{2} \approx -3.036 \]
Rewrite
Why: F prime is minus 125; there are no linear terms.
\[ 4.036\,x'^2 - 3.036\,y'^2 = 125 \]
Figure (svg): The hyperbola 3x squared plus 5xy minus 2y squared equals 125, drawn from its rotated form with x prime axis at 22.5 degrees: two branches opening along the tilted x prime axis with vertices about 5.57 from the origin, and dashed asymptotes through the origin.
Check with the invariant
Why: The product of the two new coefficients is one quarter minus fifty quarters.
\[ 4A'C' = 4\cdot\tfrac{1 - 50}{4} = -49 = 4AC - B^2 \]
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 691 — Checkpoint 7.23
The discriminant is 4 times 3 times minus 2, minus 25, which is minus 49: a hyperbola. The angle is also quick, since cot 2 theta is 5 over 5, which is 1, giving 2 theta equal to 45 degrees and theta equal to 22.5 degrees.
Evaluating A prime directly would need the cosine of 22.5 degrees, which is awkward. The double-angle identities avoid it: rewrite the squares of sine and cosine and their product in terms of 2 theta, whose cosine and sine are both root 2 over 2. Then A prime is one half plus 5 root 2 over 2, and C prime is the same with a minus sign.
One coefficient positive and one negative is exactly what a hyperbola should give. The invariant check confirms the arithmetic: 4 times A prime times C prime is 4 times one quarter minus fifty quarters, which is minus 49, the original discriminant.
Sorting
Sort into buckets
Each has a cross term. Compute 4AC − B² and sort.
OpenStax Calculus Volume 2, §7.5 Conic Sections §7.5, p. 693 — Exercises 306 to 311
Compute each discriminant before sorting. The squared coefficients alone will mislead you here, since several have the same sign and are still not ellipses.
The two hyperbolas come out at minus 1 and minus 24. The three ellipses come out at 3, 5000 and 10000, the last two being exercises whose tidy discriminants suggest they were built by rotating simple ellipses.
The last equation is the interesting one: its discriminant is exactly 0, and it is a parabola. You can see it directly, since the first three terms form x minus y, squared. Parabolas are the boundary case here too, sitting between the positive and negative discriminants, just as their eccentricity sits at the boundary value 1.
Section
Part 7
Pattern
Figure (svg): A three-lane decision diagram. Polar equation: make the constant in the denominator 1, read e, then e below 1 is an ellipse, e equal to 1 a parabola, e above 1 a hyperbola. General equation with an xy term: compute 4AC minus B squared; positive ellipse, zero parabola, negative hyperbola. No xy term: compare the squared terms; same sign ellipse, one missing parabola, opposite signs hyperbola.
The diagram sorts every conic problem by the form of the equation you are given. Each form has one question that names the curve, and the answers are always the same three.
For a polar equation, the question is the eccentricity, read only after the constant is 1. For a rectangular equation with a cross term, it is the discriminant. Without a cross term, it is the signs of the squared terms, which is really the discriminant in disguise.
Naming the conic is the first step, not the last. After it, complete the square to reach the standard form and read off the features, using subtraction for an ellipse and addition for a hyperbola. And check a point from your answer in the original equation: it takes a minute and catches nearly every slip.
Check
Check your understanding
Where are the foci of (x − 1)²/9 + (y + 2)²/25 = 1?
Answer: A
Why: The larger denominator, 25, is under y, so the major axis is vertical and a = 5, b = 3. For an ellipse c² = a² − b² = 16, so c = 4, measured up and down from the centre (1, −2).
Work it out before choosing. Two decisions are needed: which axis the foci are on, and how far from the centre they are.
The 25 is under the y term, so the major axis is vertical and the foci are directly above and below the centre (1, minus 2). For an ellipse the squares subtract, so c squared is 16 and c is 4.
Each wrong option makes one specific mistake: the wrong axis, the hyperbola's addition, or the wrong sign for the centre. If you picked one, identify which, since that is the habit to fix.
Check
Check your understanding
What is r = 12/(3 − 4 cos θ)?
Answer: A
Why: Divide top and bottom by 3: r = 4/(1 − (4/3) cos θ). Now the constant is 1, so e = 4/3, which is more than 1: a hyperbola. Also ep = 4 gives p = 3, so the directrix is x = −3.
The denominator starts with 3, so divide top and bottom by 3 before reading anything. That leaves 4 over 1 minus four thirds cos theta.
Now the coefficient, four thirds, is the eccentricity, and since it is bigger than 1 the conic is a hyperbola. The numerator 4 is e times p, so p is 3 and the directrix is x equals minus 3.
Check
Check your understanding
What kind of conic is x² + 4xy + y² = 4?
Answer: B
Why: With a cross term the sign test no longer works; use the discriminant. 4AC − B² = 4(1)(1) − 16 = −12, which is negative, so this is a hyperbola, tilted at 45° because A = C.
The squared terms have the same sign and even equal coefficients, which tempts you toward an ellipse or a circle. But that test only applies when there is no cross term.
With B equal to 4, the discriminant is 4 minus 16, which is minus 12, so this is a hyperbola. Since A equals C, it is tilted at 45 degrees, just like xy equals 1.
Explain it to yourself
Discussion prompt
In two or three sentences: using only the focus-directrix definition, explain why a ratio below one gives a closed curve and a ratio above one gives a curve that runs off to infinity.
Write your explanation before revealing. The aim is to see why the number 1 is the dividing line, rather than just remembering that it is.
The first argument is a bound. Going from a point to the focus and on to the directrix is one way of reaching the line, so the point's distance to the line is at most p plus r. When e is less than 1, the definition then forces r to be at most ep over 1 minus e: the whole curve lies within a fixed distance of the focus, so it is closed.
The second argument uses the polar form. The curve reaches infinity exactly where the denominator is zero, which needs a cosine equal to minus one over e. That is possible only when e is at least 1. At e equal to 1 there is one such direction, the parabola's single opening; above 1 there are two, the directions of the hyperbola's asymptotes.
Exit ticket
\[ 9x^2 - 4y^2 - 18x - 16y - 43 = 0 \]
Discussion prompt
Name the conic, put it in standard form, and give its eccentricity and asymptotes.
This pulls together the rectangular half of the lesson: classify, complete the square, and read off the features. The squared terms have opposite signs, so expect a hyperbola.
Completing the squares adds 9 and subtracts 16, so the right side becomes 43 plus 9 minus 16, which is 36. Dividing by 36 gives denominators 4 and 9, with the x term positive: a horizontal hyperbola centred at (1, minus 2), with a equal to 2 and b equal to 3.
Adding squares gives c equal to the square root of 13, and the eccentricity is about 1.80. The asymptotes have slopes plus and minus three halves through the centre. The vertex (3, minus 2) checks in the original equation.
Recap
| conic | distance definition | standard form | c | e |
|---|---|---|---|---|
| parabola | d(P, F) = d(P, directrix) | x² = 4py | (p) | 1 |
| ellipse | sum of distances to foci = 2a | x²/a² + y²/b² = 1 | c² = a² − b² | c/a < 1 |
| hyperbola | difference of distances = 2a | x²/a² − y²/b² = 1 | c² = a² + b² | c/a > 1 |
\[ r = \frac{ep}{1 \pm e\cos\theta}, \qquad 4AC - B^2 \gtrless 0 \]
This closes Chapter 7. The conics return in Calculus III, where every slice of a quadric surface by a coordinate plane is one of these three curves.
Stewart, Calculus: Early Transcendentals 8e, §10.5 Conic Sections §10.5-10.6, pp. 674-687 — the same material in Stewart
Three curves, one idea. Each is the set of points with a fixed ratio of distances to a focus and a directrix, and that ratio, the eccentricity, is below 1 for an ellipse, exactly 1 for a parabola and above 1 for a hyperbola. The two-focus descriptions, a constant sum and a constant difference, lead to the standard forms, with c found by subtracting squares for the ellipse and adding them for the hyperbola.
The polar equation puts all three in one formula with a focus at the pole, which is why it is the natural language for orbits. The discriminant names a conic from its general equation, even when its axes are tilted.
The table is your reference. When you meet a conic in any form, the pattern slide tells you the one question to ask first.
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