The third gap from the Algebra 2 pre-class diagnostic. Where all four conics come from, identifying any of them from the two squared coefficients, completing the square through a factored-out coefficient, and reading centre, radius, vertices, foci, directrix and asymptotes off standard form - with the ellipse and hyperbola relations kept firmly apart.
Subject: Algebra 2 · 61 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Algebra 2 · Session 1
Four curves, one cone, one procedure - and the two coefficients that tell you which is which
Objectives
The third diagnostic gap, and the one most often taught as four unrelated formula sheets. It is not four topics.
The last one is the fastest win. It takes about ten minutes to learn and it makes every other question easier to start.
Section
Section 1
Warm-up
Two things from earlier work that this whole deck is built on.
Discussion prompt
Write down the distance formula between two points, and say what completing the square does to a quadratic. You will use both on every slide from here.
Hint: One is Pythagoras rearranged; the other turns a sum into a perfect square.
Answer:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
\[ x^2 - 6x \;\longrightarrow\; x^2 - 6x + 9 = (x - 3)^2 \]
That is the entire toolkit. Every conic formula in this deck is one of these two applied to a geometric condition.
Concept
Take a double cone - two cones tip to tip - and slice it with a flat plane. The shape of the cut depends only on how the plane is tilted.
conic section — The curve formed where a plane cuts a double cone. There are exactly four non-degenerate cases.
This is why they share one general equation. They are not four separate discoveries; they are four views of one object.
Picture it
Same cone in all four pictures. Watch only the plane.
Figure (svg): One double cone sliced four different ways, producing a circle, an ellipse, a parabola and a hyperbola
Level cut gives a circle. Tilt it a little and the circle stretches into an ellipse. Tilt until the plane matches the slope of the cone and it opens forever into a parabola. Tilt past that and it catches both cones, giving two branches.
Concept
\[ Ax^2 + Cy^2 + Dx + Ey + F = 0 \]
Every conic in this course is this equation with particular values of A and C. Nothing else changes.
So identifying a conic is not pattern-matching against four memorised templates. It is reading two numbers.
| what A and C do | what D, E and F do |
|---|---|
| decide which conic it is | decide where it sits and how big it is |
Prediction
Commit before you compute anything at all.
\[ 4x^2 - 9y^2 + 8x + 36y - 68 = 0 \]
Predict first
Which conic is this?
Correct: Hyperbola.
\[ A = 4, \quad C = -9 \;\Longrightarrow\; \text{opposite signs} \]
Why: A is 4 and C is negative 9. Both squared terms are present, so it is not a parabola, and they carry opposite signs, which is the hyperbola signature. Equal coefficients would mean a circle and same-sign-but-different values would mean an ellipse. None of the linear terms matter for this decision - they only tell you where the centre is.
Picture it
Four rows. This is the whole of conic identification.
Figure (svg): A table matching the relationship between the two squared coefficients to the conic it produces
Learn this table before anything else in the deck. It turns every conic question into a known problem instead of an unknown one.
Matching
Match each coefficient pattern to the curve it produces.
Match the pairs
Why: Equal coefficients mean the curve stretches identically in both directions, which is a circle. Same sign but unequal means it stretches more one way, which is an ellipse. Opposite signs mean one direction opens outward while the other closes, which produces two branches. A missing squared term means one variable appears only to the first power, which is a parabola.
Section
Section 2
Concept
A circle is every point at one fixed distance from one fixed point. That sentence is the definition, and the equation is just that sentence written down.
\[ (x - h)^2 + (y - k)^2 = r^2 \]
standard form of a circle — Centre at the point (h, k) and radius r. The right-hand side is r squared, not r.
Picture it
Draw the radius as a hypotenuse and the equation appears on its own.
Figure (svg): A circle with a radius drawn to a general point, forming a right triangle whose legs are the horizontal and vertical differences
There is no circle formula to memorise separately. It is the distance formula with the distance held constant and both sides squared.
Intuition
The equation contains x minus h, so a centre at positive three shows up as x minus three. That reversal catches everyone at least once.
The reason is that the expression measures a difference, not a position. To be three units right of the centre you need x minus three to equal three.
The practical rule: read the opposite of what you see inside each bracket, and remember that a plus sign inside means a negative coordinate.
Worked example
\[ x^2 + y^2 - 6x + 8y - 11 = 0 \]
Group the x terms and the y terms, and move the constant across
Why: Keep the two variables apart. The constant goes to the right so both squares can be completed on the left.
\[ (x^2 - 6x) + (y^2 + 8y) = 11 \]
Complete the square on x: half of -6 is -3, and -3 squared is 9
Why: Add 9 inside the x bracket, and add the same 9 to the right so the equation stays balanced.
\[ (x^2 - 6x + 9) + (y^2 + 8y) = 11 + 9 \]
Complete the square on y: half of 8 is 4, and 4 squared is 16
Why: Add 16 inside the y bracket and 16 on the right.
\[ (x - 3)^2 + (y + 4)^2 = 11 + 9 + 16 = 36 \]
Read the centre and the radius
Why: The x bracket gives 3 and the y bracket, which contains a plus, gives negative 4. The right side is 36, so the radius is its square root, 6.
\[ \text{centre } (3, -4), \quad r = 6 \]
Verify: test one point
Why: The point (9, -4) should be on the circle since it is 6 to the right of the centre. Substituting: 81 plus 16 minus 54 minus 32 minus 11 equals 0. It checks.
Picture it
Drawn from the standard form.
Figure (svg): The circle from the worked example, centred at three, negative four, with radius six
The most common slip is reporting the radius as 36. The right-hand side is always r squared, so the last thing you do is take a square root.
Picture it
Why the number you add is the square of half the coefficient.
Figure (svg): A square of side x with two rectangles attached and one small square filling the missing corner, showing how a constant completes the square
Half the middle coefficient splits the rectangle into two equal strips, and the missing corner is that half, squared. The picture is where the rule comes from.
Trap
\[ x^2 + y^2 - 6x + 8y - 11 = 0 \]
Group and complete both squares
Why: Half of -6 squared is 9 and half of 8 squared is 16, so those go in.
\[ (x - 3)^2 + (y + 4)^2 = 11 \]
Leave the right side alone
Why: The 9 and the 16 were added inside brackets, so it feels like nothing left the equation.
Report the radius
Why: The right side is 11, so the radius comes out as the root of 11, about 3 point 32 - a completely different circle from the real one.
\[ x^2 + y^2 - 6x + 8y - 11 = 0 \]
Group and complete both squares, adding to the right as you go
Why: Adding 9 to the left means adding 9 to the right. Same for the 16. An equation only stays true if both sides change together.
\[ (x - 3)^2 + (y + 4)^2 = 11 + 9 + 16 \]
\[ (x - 3)^2 + (y + 4)^2 = 36 \]
Verify: test a point that must be on it
Why: The point (3, 2) is 6 above the centre. Substituting into the original gives 9 plus 4 minus 18 plus 16 minus 11, which is 0. The radius really is 6.
Error analysis
One line here is wrong. Name it and fix it.
Annotate
On: \( x^2 + y^2 + 10x - 4y + 13 = 0 \;\Longrightarrow\; (x + 5)^2 + (y - 2)^2 = -13 + 25 + 4 = 16 \)
The habit worth building is verifying rather than eyeballing. Substitute one point and you know for certain.
Section
Section 3
Step zero
No algebra yet.
\[ y^2 - 8x + 4y + 12 = 0 \]
Discussion prompt
Only one variable is squared. What does that tell you about the shape, and which way does it open?
Hint: Think about which variable is free to grow without limit.
Answer:
Only y is squared, so it is a parabola. The squared variable is the one that runs across the opening, so a squared y means the parabola opens sideways, left or right.
To find which way, complete the square in y and see whether the x side ends up positive or negative. Positive opens right; negative opens left.
Rule of thumb: whichever variable is squared, the parabola opens along the other one.
Concept
A parabola is every point equally far from a fixed point and a fixed line.
focus — The fixed point. Every point on the curve is the same distance from it as from the directrix.
directrix — The fixed line, sitting the same distance on the other side of the vertex as the focus.
\[ (x - h)^2 = 4p(y - k) \]
Picture it
The defining property, drawn on a real curve.
Figure (svg): A parabola with its focus, directrix and vertex marked, and two equal distances drawn from a point on the curve
The vertex is the only point where those two distances are both exactly p. Everywhere else they are equal but larger.
Intuition
The value of p is the distance from the vertex to the focus, and it carries a sign.
A positive p puts the focus above or to the right of the vertex, so the curve opens that way. A negative p flips it.
A large p means a wide, shallow curve because the focus is far away. A small p means a narrow, steep one. Satellite dishes use a small p on purpose.
Worked example
\[ (x - 2)^2 = 8(y + 1) \]
Read the vertex from the two brackets
Why: The x bracket gives 2 and the y bracket, which contains a plus, gives negative 1.
\[ \text{vertex } (2, -1) \]
Match the coefficient against 4p
Why: The number multiplying the linear bracket is 8, and that number is 4p, so p is 2.
\[ 4p = 8 \;\Longrightarrow\; p = 2 \]
Use the sign and the squared variable to place the focus
Why: X is squared, so the parabola opens vertically, and p is positive, so it opens upward. The focus is 2 above the vertex.
\[ \text{focus } (2, 1), \quad \text{directrix } y = -3 \]
Verify: check the equal-distance property at one point
Why: At x equal to 6 the equation gives 16 equals 8 times y plus 1, so y is 1. The point (6, 1) is 4 units from the focus (2, 1), and 4 units above the directrix y equals -3. The distances agree.
Picture it
Vertex, focus and directrix, with the defining distances measured.
Figure (svg): The parabola from the worked example, with vertex at two negative one, focus at two one, and directrix at y equals negative three
The directrix is always the same distance on the opposite side of the vertex from the focus. If you know one, you know the other.
Fill the middle
Start and finish are given. Supply the completed square.
\[ y^2 + 4y - 8x + 12 = 0 \]
Fill in the blanks
y^2 + 4y + 4 = 8x - 12 + 4 \;\Longrightarrow\; (y + 2)^2 = 8(x - 1)
Why: Move the x and constant terms to the right, then complete the square on y. Half of 4 is 2 and 2 squared is 4, so add 4 to both sides. The left becomes (y + 2) squared and the right becomes 8x minus 8, which factors as 8(x - 1). The vertex is (1, -2), 4p is 8 so p is 2, and since y is squared with p positive the parabola opens to the right.
Discrimination
Do not solve. Just sort.
Sort into buckets
Sort each parabola by the direction it opens.
Section
Section 4
Concept
An ellipse is every point whose two distances to a pair of fixed points add to the same total.
\[ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \]
major axis — The longer of the two axes. The foci always lie on it, and a is always the semi-major length.
The denominators are the squares of the semi-axis lengths, so a and b come from square-rooting them.
Picture it
This is how gardeners actually draw an ellipse.
Figure (svg): Two pins and a loop of string tracing an ellipse, showing that the sum of the two distances stays constant
The constant total is 2a, the whole length of the long axis. Move the pins closer together and the ellipse rounds toward a circle; a circle is an ellipse whose two foci have merged.
Socratic
One sentence answer.
Discussion prompt
If a circle is an ellipse whose foci coincide, what does that force to be true about a, b and c for a circle?
Hint: Where is c when the two foci are at the same place?
Answer:
If the foci coincide they are both at the centre, so c is zero.
\[ a^2 = b^2 + c^2 \;\Longrightarrow\; a^2 = b^2 \;\Longrightarrow\; a = b \]
Equal semi-axes is exactly what makes a circle, and it explains why a circle's equation has equal denominators - which is the same thing as A equalling C in the general form.
Concept
\[ a^2 = b^2 + c^2 \]
Here a is the largest of the three, always. It is the hypotenuse of the right triangle formed by the semi-minor axis and the focal distance.
So finding the foci is one subtraction: c squared is a squared minus b squared.
Picture it
You do not have to memorise the equation if you can draw this.
Figure (svg): An ellipse with semi-axes a and b marked and a right triangle showing that a squared equals b squared plus c squared
Draw b up from the centre, then swing a down to the axis. Where it lands is a focus, and the horizontal leg is c.
Worked example
\[ 9x^2 + 25y^2 - 36x + 50y - 164 = 0 \]
Group by variable and factor out the squared coefficients
Why: The coefficients must come out before completing the square, because the completion rule assumes a leading coefficient of one.
\[ 9(x^2 - 4x) + 25(y^2 + 2y) = 164 \]
Complete both squares, and multiply what you added by the factored-out coefficient
Why: Half of -4 squared is 4, but it sits inside a bracket multiplied by 9, so 36 was really added. Half of 2 squared is 1, inside a bracket multiplied by 25, so 25 was really added.
\[ 9(x - 2)^2 + 25(y + 1)^2 = 164 + 36 + 25 = 225 \]
Divide through by the right-hand side to get 1
Why: Standard form always has 1 on the right. Dividing by 225 gives denominators of 25 and 9.
\[ \dfrac{(x - 2)^2}{25} + \dfrac{(y + 1)^2}{9} = 1 \]
Read off the parts
Why: Centre (2, -1). The larger denominator 25 sits under x, so the major axis is horizontal and a is 5; b is 3. Then c squared is 25 minus 9, which is 16, so c is 4.
\[ a = 5, \quad b = 3, \quad c = 4, \quad \text{foci } (-2, -1) \text{ and } (6, -1) \]
Verify: check a vertex
Why: The vertex (7, -1) is 5 right of the centre. Substituting into the standard form gives 25 over 25 plus 0, which is 1. It checks.
Picture it
Same numbers, drawn on the grid.
Figure (svg): The ellipse from the worked example, centred at two negative one with a horizontal major axis of length ten
The foci are inside, on the long axis, and closer to the centre than the vertices are. If your c comes out bigger than your a, you have used the wrong relation.
Trap
\[ \dfrac{x^2}{9} + \dfrac{y^2}{25} = 1 \]
Call the first denominator a squared
Why: The formula was memorised with a under x, so 9 becomes a squared and a becomes 3.
\[ a = 3, \quad b = 5 \]
Compute c
Why: c squared is a squared minus b squared, which is 9 minus 25, or negative 16. A negative square is impossible, and the problem looks broken.
\[ \dfrac{x^2}{9} + \dfrac{y^2}{25} = 1 \]
Identify a as the square root of the LARGER denominator
Why: a is defined as the semi-major axis, so it always goes with the bigger number, wherever that number happens to sit.
\[ a = 5 \text{ (under } y\text{)}, \quad b = 3 \]
The major axis runs along the variable with the larger denominator
Why: Twenty-five is under y, so this ellipse is tall, and the foci sit on the vertical axis.
\[ c^2 = 25 - 9 = 16 \;\Longrightarrow\; c = 4, \quad \text{foci } (0, \pm 4) \]
Verify: sanity-check the sign
Why: c squared came out positive, which it always must. A negative c squared is the signal that a and b were swapped.
Two truths and a lie
Three claims about ellipses. Knock out the wrong one.
Eliminate the wrong options
Which statement is false?
Survives elimination: D
Why: The larger denominator sits under whichever variable the major axis runs along, which may be x or y. If it sits under y, the ellipse is taller than it is wide and the foci are vertical. The other three statements are all true: a is the semi-major length by definition, so it is the largest; c comes from a squared minus b squared, so it is smaller than a; and the foci are always on the long axis.
Section
Section 5
Concept
An ellipse fixes the sum of the two focal distances. A hyperbola fixes their difference. That single change produces two branches instead of one loop.
\[ \dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1 \]
The minus sign is the whole difference between this and the ellipse equation, and it changes almost everything downstream.
Hypothesis
Reason it out before checking.
\[ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \]
Predict first
In this equation, can x ever be zero?
Correct: No, because it would give a negative equal to one.
\[ -\dfrac{y^2}{b^2} = 1 \;\Longrightarrow\; y^2 = -b^2 \;\text{: impossible} \]
Why: Setting x to zero leaves negative y squared over b squared equal to 1, which needs y squared to be negative. No real y does that. This is exactly why the curve never crosses the y-axis and why there is a gap between the two branches - the equation itself forbids that whole strip.
Concept
\[ c^2 = a^2 + b^2 \]
For a hyperbola c is the largest, not a. The foci sit outside the vertices rather than inside them.
| ellipse | hyperbola | |
|---|---|---|
| relation | a squared equals b squared plus c squared | c squared equals a squared plus b squared |
| largest | a | c |
| foci sit | inside the curve | outside the vertices |
| how to find c | subtract | add |
One is a subtraction and the other is an addition. That is the only thing to keep straight, and it is worth over-rehearsing.
Intuition
For a hyperbola, a is always the denominator under the positive term, regardless of which is numerically larger.
That is a real change from the ellipse, where a was whichever was larger. Here the sign decides, not the size.
So a hyperbola can perfectly well have b bigger than a. The positive term names the transverse axis, and that is that.
Worked example
\[ 16x^2 - 9y^2 - 64x - 54y - 161 = 0 \]
Group and factor out the squared coefficients, keeping the signs
Why: Factoring negative 9 out of the y terms flips the sign inside the bracket, which is where sign errors start. Write it carefully.
\[ 16(x^2 - 4x) - 9(y^2 + 6y) = 161 \]
Complete both squares and adjust the right side by the true amounts added
Why: Adding 4 inside a bracket multiplied by 16 adds 64. Adding 9 inside a bracket multiplied by negative 9 adds negative 81, so 81 is subtracted from the right.
\[ 16(x - 2)^2 - 9(y + 3)^2 = 161 + 64 - 81 = 144 \]
Divide through by 144
Why: One hundred forty-four over 16 is 9, and 144 over 9 is 16, so the denominators are 9 and 16.
\[ \dfrac{(x - 2)^2}{9} - \dfrac{(y + 3)^2}{16} = 1 \]
Read the parts, using the sign to place a
Why: Centre (2, -3). The positive term is the x term, so the transverse axis is horizontal and a squared is 9, giving a equal to 3, while b is 4. Then c squared is 9 plus 16, so c is 5.
\[ \text{vertices } (-1, -3), (5, -3) \qquad \text{foci } (-3, -3), (7, -3) \]
Write the asymptotes through the centre with slope b over a
Why: The slopes are plus and minus 4 over 3, and both lines pass through the centre.
\[ y + 3 = \pm\dfrac{4}{3}(x - 2) \]
Verify: check a vertex
Why: The vertex (5, -3) is 3 right of the centre. Substituting gives 9 over 9 minus 0, which is 1. It checks, and notice c is 5, which is further out than the vertex at 3 - exactly as a hyperbola requires.
Picture it
The fastest way to sketch any hyperbola by hand.
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running through the rectangle's diagonals
Mark the centre, go a across and b up to make a box, draw its diagonals as asymptotes, then put the vertices at the midpoints of two sides and curve outward. No point-plotting needed.
Trap
\[ \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 \]
Reach for a squared equals b squared plus c squared
Why: The two relations look almost identical, so the wrong one gets used by reflex.
\[ 9 = 16 + c^2 \;\Longrightarrow\; c^2 = -7 \]
Get a negative
Why: There is no real c, so either the problem is broken or - much more likely - the wrong relation was used.
\[ \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 \]
Use the hyperbola relation: c squared equals a squared plus b squared
Why: The minus sign in the equation is the cue. Minus in the equation means plus in the relation.
\[ c^2 = 9 + 16 = 25 \;\Longrightarrow\; c = 5 \]
Sanity-check the geometry
Why: The vertices are 3 from the centre and the foci are 5 from it. The foci being further out is what a hyperbola always looks like.
Verify: compare against the ellipse
Why: For an ellipse the foci fall between the centre and the vertices, so c is less than a. For a hyperbola they fall beyond, so c is greater than a. If your c is on the wrong side of a, you used the wrong relation.
Comparison
Complete the blank cells. This table is the deck in miniature.
Comparison matrix
| circle | ellipse | hyperbola | |
|---|---|---|---|
| sign between squared terms | plus | plus | minus |
| relation for c | c = 0 | a^2 = b^2 + c^2 | c^2 = a^2 + b^2 |
| largest of a, b, c | a = b, c = 0 | a | c |
| how a is chosen | not applicable | the larger denominator | the positive term |
| asymptotes | none | none | slope b over a through the centre |
The row that trips people is how a is chosen. For an ellipse size decides; for a hyperbola sign decides. They are genuinely different rules and no amount of pattern-matching will merge them.
Ranking
Put the steps of converting a general conic to standard form in order.
Put in order
Why: Grouping first keeps the two variables from interfering. Factoring the coefficient out has to come before completing the square, because the halve-and-square rule assumes a leading coefficient of one. Completing the square is then mechanical. Adjusting the other side is the step most often skipped, and the amount added is the completed constant times the factored-out coefficient, not the constant alone. Dividing to make the right side 1 is only for ellipses and hyperbolas; circles and parabolas stop a step earlier.
Section
Section 6
Pattern
Two phases. Identify, then convert. Never start converting before you know what you are converting to.
Step five is the one that goes wrong most often, and step seven is where the ellipse and hyperbola relations get mixed up. Those two steps are worth slowing down for.
Picture it
Everything in this deck, at the same scale.
Figure (svg): All four conic sections drawn on one coordinate grid at the same scale for comparison
If you can sketch these four from memory and label one distinguishing feature on each, you can answer almost any conic question you will be asked.
Check
Complete both squares and watch the right-hand side.
Check your understanding
Find the centre and radius of x^2 + y^2 + 10x - 4y + 13 = 0.
Answer: A
Why: Group and move the constant: x squared plus 10x, plus y squared minus 4y, equals -13. Half of 10 squared is 25 and half of -4 squared is 4, so add both to each side. The left becomes (x + 5) squared plus (y - 2) squared and the right becomes -13 plus 25 plus 4, which is 16. Centre (-5, 2) and radius the square root of 16, which is 4.
Check
Identify a and b first, then subtract.
Check your understanding
Find the foci of x^2/16 + y^2/7 = 1.
Answer: A
Why: The larger denominator is 16, and it sits under x, so a is 4 and the major axis is horizontal. Then b squared is 7, and c squared is a squared minus b squared, which is 16 minus 7, or 9. So c is 3 and the foci are 3 units left and right of the centre at the origin.
Check
Identify from A and C, then convert.
Check your understanding
Identify the conic and give its centre: 4x^2 - 9y^2 + 8x + 36y - 68 = 0.
Answer: A
Why: A is 4 and C is -9, opposite signs, so it is a hyperbola. Grouping gives 4(x squared + 2x) - 9(y squared - 4y) = 68. Completing the squares adds 4 inside the first bracket, which is really 4, and 4 inside the second, which is really -36. So the right side becomes 68 + 4 - 36 = 36, giving 4(x + 1) squared - 9(y - 2) squared = 36, or (x + 1)^2/9 - (y - 2)^2/4 = 1. Centre (-1, 2).
Tweak it
Drag the parameters and watch the parabola change shape.
Parameter explorer
Set p small, then large. What happens to the width of the curve, and where does the focus go?
\[ y = \dfrac{(x - {h})^2}{4 \cdot {p}} \]
Real world
One concrete case, then generalise it yourself.
Discussion prompt
A satellite dish is a parabola rotated around its axis, and the receiver sits at the focus. If a dish is 4 feet across and 1 foot deep at the centre, where should the receiver go?
Hint: Put the vertex at the origin and use the point at the rim.
Answer:
Put the vertex at the origin so the equation is x squared equals 4py. The rim is 2 feet from the axis and 1 foot up, so the point (2, 1) is on the curve.
\[ 2^2 = 4p(1) \;\Longrightarrow\; 4p = 4 \;\Longrightarrow\; p = 1 \]
So the receiver goes 1 foot above the vertex, right at the rim level. This is the reason parabolas are used at all: every ray coming in parallel to the axis reflects to exactly that one point.
Ellipses do the analogous thing with two foci, which is how whispering galleries and lithotripsy machines work.
Missing information
The question as stated cannot be answered.
\[ \dfrac{(x - 3)^2}{16} + \dfrac{(y + 2)^2}{k} = 1 \]
Discussion prompt
Someone asks you for the foci of this ellipse. What do you need to know about k first, and how many different answers are possible?
Hint: Which denominator is larger decides more than you might think.
Answer:
You need to know whether k is less than, greater than, or equal to 16, because that decides which axis is major.
If k is less than 16 the major axis is horizontal, a is 4, and the foci are at 3 plus or minus the root of 16 minus k, with y equal to -2.
If k is greater than 16 the major axis is vertical, a is the root of k, and the foci are at x equal to 3 with y equal to -2 plus or minus the root of k minus 16.
If k equals 16 it is a circle and both foci sit on top of each other at the centre. So there are three genuinely different answers, and the question is incomplete without k.
Explain it
Two or three sentences, no formulas.
Discussion prompt
How would you explain to a classmate why circles, ellipses, parabolas and hyperbolas are considered one topic rather than four?
Hint: Start from the cone, or start from the general equation - either works.
Answer:
A model answer: all four are what you get by slicing the same double cone at different angles, so they are literally the same object viewed differently.
Algebraically, all four are the same second-degree equation, and the only thing that changes is the two coefficients on the squared terms. Learning one conversion procedure handles all four, because the procedure is just completing the square.
Exit ticket
Last commitment. This is what the session will open with.
Predict first
Which of these is shakiest right now?
Correct: Whichever you picked is the first drill.
Why: The second one is the most common on tests, because the amount added to the other side is the completed constant times the factored-out coefficient, and forgetting the multiplication silently produces a wrong but plausible answer. The third is pure memorisation and the triangle picture fixes it in about five minutes. The first is the cheapest to fix and makes everything else easier, so it is worth doing first regardless of what you picked.
Connect it up
Build it from memory, then check it against the deck.
Draw it
Draw a two-by-two grid. In each cell put one conic: sketch it, write its standard form, write how a is chosen, write the relation for c, and note anything unique to it such as the directrix or the asymptotes. Then across the bottom write the identification rule using A and C. Finally, circle the two places where the ellipse and hyperbola genuinely differ rather than merely look different.
Those two circled places - how a is chosen, and whether you add or subtract for c - are the whole difference. Everything else about the two curves follows the same procedure.
Recap
One cone, one general equation, one procedure, four curves.
| if you remember one thing | it should be |
|---|---|
| about identifying | look at A and C and nothing else |
| about completing the square | the amount added to the other side is the constant times the coefficient you factored out |
| about the circle | the right-hand side is r squared, so take the root at the end |
| about the ellipse | a is the larger denominator, and you subtract to get c |
| about the hyperbola | a is the positive term, and you add to get c |
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