One definition covering all four conics: the set of points whose distance to a fixed focus is a constant multiple of the distance to a fixed directrix. In polar coordinates with the focus at the pole, that definition becomes a single equation with one parameter, the eccentricity, whose value against 1 decides between ellipse, parabola and hyperbola. Covers the four standard forms and their directrices, reading a given equation by normalising it first, and the rotated general form that collapses all four into one.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.6 Hooked on Conics Again, pp. 981-986
Objectives
Five outcomes, and the third is the step that everything else depends on.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 981-986 — the pages these objectives are drawn from
Warm-up
Chapter 7 defined the conics one at a time, each with its own construction.
Discussion prompt
Recall how a parabola and an ellipse were each defined. What do the two definitions have in common, and what is different?
Hint: How many fixed objects does each use?
Answer:
A parabola was the set of points equidistant from a focus and a directrix. An ellipse was the set of points whose distances to two foci sum to a constant. Different constructions, different apparatus.
But the parabola's definition is already a ratio: the two distances are equal, so their quotient is 1. Nothing stops that quotient being something other than 1.
This lesson takes that seriously. One focus, one directrix, one ratio, and all four conics appear as the ratio varies — with the parabola sitting exactly at the boundary value.
Concept
Fix a point and a line not through it. The set of points whose distance to the point is a constant multiple of the distance to the line is a conic section, and the constant decides which one.
Definition 11.1 — Given a fixed line L, a point F not on L, and a positive number e, a conic section is the set of all points P for which the distance from P to F divided by the distance from P to L equals e. The line is the directrix, the point is a focus, and e is the eccentricity.
\[ \frac{\text{distance from } P \text{ to } F}{\text{distance from } P \text{ to } L} = e \]
Placing the focus at the pole and the directrix at a vertical line makes the two distances easy to write in polar coordinates, and solving for r gives a single equation covering every case.
Figure (svg): The focus-directrix definition of a conic: a point, a vertical line, and a point of the curve with its two distances marked and their ratio labelled as the eccentricity
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 981-982
Section
Section 1
Concept
With the focus at the pole and the directrix the vertical line at distance d to the left, both distances are one-line expressions in polar coordinates, and the definition becomes an equation solvable for r.
\[ r = \frac{ed}{1 - e\cos(\theta)} \]
The derivation assumes r positive, which is harmless: any point of the curve has such a representation, and the Fundamental Graphing Principle only asks for one.
Figure (svg): The focus-directrix definition of a conic: a point, a vertical line, and a point of the curve with its two distances marked and their ratio labelled as the eccentricity
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 981-982
Picture it
Same focus, same directrix, three values of the eccentricity.
Figure (svg): A family of conics sharing a focus and a directrix, drawn for eccentricities below, equal to, and above one, showing the ellipse, parabola and hyperbola
At exactly 1 the curve stops closing. Below it, the far branch curls back; above it, the denominator changes sign and a second branch appears.
Worked example
From the definition to the standard form, in four lines.
\[ \text{Derive } r = \frac{ed}{1 - e\cos(\theta)}. \]
Write the two distances
Why: To the focus and to the directrix.
Set the quotient to e
Why: That is the definition.
\[ \frac{r}{d + r \cos \theta} = e \]
Clear the fraction
Why: Multiply across.
\[ r = e d + e r \cos \theta \]
Collect the r terms and divide
Why: Factor r out on the left.
\[ r(1 - e \cos \theta) = e d \]
Figure (svg): The solution to Worked example the derivation shown as a ladder of expressions, one row per legal move
\[ r = \frac{ed}{1 - e\cos(\theta)} \]
Verify: test the vertex
Why: At theta equal to pi, the cosine is negative 1 and the denominator is 1 plus e, so r is ed over 1 plus e. That is the point of the curve nearest the directrix, and it lies between the focus and the directrix as it must.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 982-982
Faded example
Clear the fraction and solve for r.
Fill in the blanks
\frac1 - e cos thed = e \;\Longrightarrow\; r = ed + er\cos\theta \;\Longrightarrow\; r\left(___\right) = ___
Why: Moving the r cosine term to the left and factoring r out gives the bracket, and the numerator is what remains on the right. Dividing then produces the standard form, which is why the constant term in the denominator is 1 and the numerator is the product ed.
Worked example
The case where the eccentricity is exactly 1.
\[ \text{Show } r = \frac{d}{1 - \cos(\theta)} \text{ is a parabola with directrix } x = -d. \]
Set e to 1 in the derivation
Why: The equation before dividing.
\[ r = d + r \cos \theta \]
Convert to rectangular form
Why: Replace r cosine theta by x and r by the root of the sum of squares.
\[ \sqrt{x ^{2} + y ^{2}} = d + x \]
Square and expand
Why: The x squared terms cancel.
\[ y ^{2} = 2 \,dx + d ^{2} \]
Read the form
Why: Factor out 2d.
\[ y ^{2} = 2 d(x + \frac{d}{2}) \]
Figure (svg): The solution to Worked example recovering the parabola shown as a ladder of expressions, one row per legal move
\[ y^2 = 2d\left(x + \tfrac{d}{2}\right) \]
Verify: check the parameters
Why: Comparing with the standard form, four times the focal parameter is 2d so the parameter is d over 2. The vertex is at negative d over 2 comma 0, the focus is that far to the right, namely at the origin, and the directrix is the same distance to the left, at x equal to negative d. All three match the setup.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 982-982
Trap
\[ e = \frac{r}{d} \]
Take the distance from the point to the directrix as d
Why: The directrix is at distance d from the focus, so d seems like the right length.
But d is the distance from the focus to the directrix, not from the point to it. The point is somewhere else, and its horizontal displacement from the focus is r cosine theta.
\[ e = \frac{r}{d + r\cos(\theta)} \]
Add the point's horizontal displacement to d
Why: That gives the perpendicular distance from the point to the vertical line.
Read the picture rather than the parameters: the directrix is a line, and distance to a line means perpendicular distance from the point in question. The parameter d only tells you where the line is.
Prediction
A conic has eccentricity 1.
Predict first
What does the definition say about its points?
Correct: They are equidistant from the focus and the directrix.
Why: An eccentricity of 1 means the ratio of the two distances is 1, so the distances are equal. That is exactly the definition of a parabola given in Chapter 7, which is how the new definition reconciles with the old one at this particular value. The parabola is not a special case bolted on; it is the boundary between the ratio favouring the focus and favouring the directrix.
Sorting
Compare the eccentricity against 1.
Sort into buckets
Sort each value.
Socratic
The derivation places the focus at the origin rather than the centre.
Discussion prompt
Why is that the useful choice, given that Chapter 7 usually centred the conic?
Hint: Does a parabola have a centre?
Answer:
A parabola has no centre, so any scheme centred on the centre has to treat it as a separate case. Every conic has a focus, so centring on a focus is the choice that covers all four uniformly.
It also makes the distance to the focus simply r, which is what turns the definition into a one-line equation. Centring elsewhere would put a square root in the numerator.
And it matches the physics. A planet orbits with the sun at a focus, not at the centre, which is Kepler's first law — so the polar form with the focus at the pole is the form in which orbital mechanics is actually written. The mathematical convenience and the physical relevance point the same way here, which is worth noticing rather than treating as luck.
Section
Section 2
Concept
The denominator is 1 minus e cosine theta. Whether it can reach zero decides whether the curve closes, and that is governed entirely by whether e exceeds 1.
The old definitions from Chapter 7 all reappear. Working the conversion through shows that the focus really is at the origin in each case and that the number e really is the eccentricity as previously defined.
Figure (svg): A family of conics sharing a focus and a directrix, drawn for eccentricities below, equal to, and above one, showing the ellipse, parabola and hyperbola
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 982-983
Picture it
The same construction at three values.
Figure (svg): A family of conics sharing a focus and a directrix, drawn for eccentricities below, equal to, and above one, showing the ellipse, parabola and hyperbola
What happens at exactly 1 is worth dwelling on: it is the only value at which the denominator touches zero without crossing it, and that single tangency is the difference between one branch and two.
Worked example
Converting to rectangular form when the eccentricity is below 1.
\[ \text{Show } r = \frac{ed}{1 - e\cos\theta} \text{ is an ellipse when } 0 < e < 1. \]
Clear and convert
Why: Replace r cosine theta by x and square.
\[ x ^{2} + y ^{2} = e ^{2}(d + x) ^{2} \]
Expand and collect
Why: Gather the x terms.
\[ (1 - e ^{2}) x ^{2} - 2 e ^{2} \,dx + y ^{2} = e ^{2} d ^{2} \]
Complete the square in x
Why: The coefficient is positive since e is below 1.
\[ \text{centre at } e ^{2} d / (1 - e ^{2}) \]
Divide to standard form
Why: An ellipse with the stated semi-axes.
Figure (svg): The solution to Worked example the ellipse case shown as a ladder of expressions, one row per legal move
\[ \text{major axis } \frac{2ed}{1-e^2}, \quad \text{minor axis } \frac{2ed}{\sqrt{1-e^2}} \]
Verify: check the sign condition
Why: The step dividing by 1 minus e squared requires that quantity to be positive, which holds exactly when e is below 1. For e above 1 it is negative and the same algebra produces a hyperbola instead, with the minus sign appearing between the two terms.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 982-982
Prediction
An ellipse has eccentricity very close to 1.
Predict first
What does it look like?
Correct: Very elongated.
Why: As the eccentricity rises towards 1 the ellipse stretches, since the denominator comes close to vanishing at one angle and r grows large there while staying moderate elsewhere. At exactly 1 it fails to close. Near zero the ellipse is nearly circular instead. Nothing about the eccentricity says anything about overall size, which is set by the numerator.
Worked example
Watching the denominator rather than the algebra.
\[ \text{Explain the behaviour of } r \text{ as } \theta \text{ approaches } 0 \text{ for each range of } e. \]
Evaluate the denominator at zero
Why: The cosine is 1.
\[ 1 - e \]
For e below 1
Why: The denominator is positive.
For e equal to 1
Why: The denominator is zero.
For e above 1
Why: The denominator is negative, and zero nearby.
Figure (svg): The solution to Worked example why e 1 is the boundary shown as a ladder of expressions, one row per legal move
\[ 1 - e > 0: \text{ closed}; \quad 1 - e = 0: \text{ unbounded}; \quad 1 - e < 0: \text{ two branches} \]
Verify: match against the pictures
Why: The ellipse is bounded, the parabola escapes in one direction, and the hyperbola has two separate pieces. All three behaviours are visible in the denominator before any curve is drawn, which is the practical value of reading the equation this way.
Error analysis
A student compares two conics by their eccentricities.
Annotate
On: \( e_1 = 0.5, \; e_2 = 0.9 \;\Longrightarrow\; \text{the second is larger} \)
The two parameters do different jobs. The eccentricity sets the shape and the numerator sets the scale, and separating them is what makes the polar form so easy to read.
Matching
Real orbits span most of the range.
Match the pairs
Why: The Earth's orbit is so nearly circular that its eccentricity is under two percent. Halley's is a highly elongated ellipse that still closes, which is why the comet returns. Exactly 1 is the parabolic trajectory of a body at precisely escape speed, and anything above is hyperbolic and unbound. The mathematical boundary at 1 is a physical boundary too.
Sorting
The two parameters of the equation do different jobs.
Sort into buckets
Sort each property by which parameter controls it.
Edge cases
The definition requires the eccentricity to be a positive number.
Discussion prompt
What happens as the eccentricity approaches zero, and why does the definition struggle there?
Hint: What happens to the directrix distance if the numerator is held fixed?
Answer:
As the eccentricity approaches zero the ellipse becomes more and more circular, which is exactly what you would want. But the definition itself breaks down at the limit: with e equal to zero the ratio says the distance to the focus is zero, so the only point would be the focus itself.
The way out is to hold the numerator fixed rather than d. Writing the numerator as a single constant, the equation at e equal to zero becomes r equal to that constant — a circle centred at the pole, of exactly the right radius.
That is why the final theorem of the section is stated with the numerator as its own parameter rather than as a product. Reparametrising is what lets the circle join the family, and the price is that d has to be recovered as the numerator divided by the eccentricity, which is undefined for a circle — appropriately, since a circle has no directrix.
Section
Section 3
Concept
The four standard forms differ only in whether the denominator uses a cosine or a sine and in the sign. None of the parameters can be read until the constant term in the denominator has been made 1.
Skipping the normalisation and reading the coefficients off directly is the standard error, and it gets both the eccentricity and the classification wrong.
Figure (svg): The four polar forms of a conic and the position of the directrix in each
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 983-983
Picture it
Divide top and bottom by whatever the constant term happens to be.
Figure (svg): An equation being rewritten into the standard form by dividing numerator and denominator so the constant term of the denominator becomes one
After that single division every parameter is visible: the eccentricity, the product, and hence the directrix distance.
Worked example
Example 11.6.4, part 2. The equation as given hides everything.
\[ \text{Identify the conic } r = \frac{12}{3 - \cos(\theta)}. \]
Divide top and bottom by 3
Why: To make the constant term 1.
\[ r = \frac{4}{1 - (\frac{1}{3}) \cos \theta} \]
Read the eccentricity
Why: The coefficient of the cosine.
\[ e = \frac{1}{3} \]
Classify
Why: Below 1.
Find d
Why: The numerator is ed, and e is one third.
\[ d = 12,\text{ directrix } x = -12 \]
Figure (svg): The solution to Worked example normalising and reading shown as a ladder of expressions, one row per legal move
\[ e = \tfrac{1}{3}, \quad d = 12, \quad \text{directrix } x = -12 \]
Verify: check the unnormalised reading
Why: Reading the original equation directly would have suggested an eccentricity of 1 and a parabola, which is wrong on both counts. The division is not cosmetic; it is what makes the coefficients mean what the theorem says they mean.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 983-983
Faded example
Normalise r equal to 10 over 5 plus 3 sine theta.
Fill in the blanks
r = \frac210/3 = \frac___}______\sin\theta} \;\Longrightarrow\; e = \tfrac______, \; d = ___
Why: Dividing top and bottom by 5 gives a numerator of 2 and an eccentricity of three fifths, so the conic is an ellipse. Since the numerator is ed and e is three fifths, d is 2 divided by three fifths, namely ten thirds. The plus sign with a sine puts the directrix at y equal to ten thirds.
Worked example
Example 11.6.4, part 3. Already normalised, with an eccentricity above 1.
\[ \text{Identify } r = \frac{6}{1 + 2\sin(\theta)}. \]
Check the constant term
Why: It is already 1.
Read the eccentricity
Why: The coefficient of the sine.
\[ e = 2 \]
Classify
Why: Above 1.
Locate the directrix
Why: Sine with a plus sign, and ed is 6.
\[ d = 3,\text{ directrix } y = 3 \]
Figure (svg): The solution to Worked example a hyperbola shown as a ladder of expressions, one row per legal move
\[ e = 2, \quad d = 3, \quad \text{directrix } y = 3 \]
Verify: check the axis orientation
Why: The directrix is horizontal, and the directrix is always perpendicular to the transverse axis. So the transverse axis is vertical, along the y-axis, which means the vertices are found by evaluating r at pi over 2 and 3 pi over 2.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 984-984
Trap
\[ r = \frac{12}{3 - \cos\theta} \;\Longrightarrow\; e = 1, \; ed = 12 \;\Longrightarrow\; \text{a parabola} \]
Read the coefficient of the cosine as the eccentricity
Why: In the standard form that is exactly what it is.
But the standard form has a constant term of 1 in the denominator, and this one has 3. The coefficients only carry their stated meanings after that has been arranged.
\[ r = \frac{4}{1 - \tfrac{1}{3}\cos\theta} \;\Longrightarrow\; e = \tfrac{1}{3}, \; ed = 4 \;\Longrightarrow\; \text{an ellipse} \]
Divide numerator and denominator by the constant term first
Why: One division, then read everything.
The check that costs nothing: look at the denominator's constant term before reading anything else. If it is not 1, nothing in the equation means what it appears to mean.
Sorting
The function says horizontal or vertical; the sign says which side.
Sort into buckets
Sort each denominator.
Prediction
A polar conic's denominator involves a cosine.
Predict first
Which axis does the conic's major or transverse axis lie along?
Correct: The x-axis.
Why: A cosine in the denominator puts the directrix on a vertical line, and the directrix is always perpendicular to the major axis of an ellipse or the transverse axis of a hyperbola. Perpendicular to vertical is horizontal, so the axis lies along the x-axis. This means the vertices are found by evaluating r at theta equal to 0 and pi.
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about the polar form is wrong.
Survives elimination: B
Why: Statement B is false as written, because it omits the normalisation. In the equation with denominator 3 minus cosine theta the coefficient is 1, but the eccentricity is one third. The coefficient is the eccentricity only after the constant term has been made 1, and the same proviso applies to statement D, which states it.
Section
Section 4
Concept
Once the conic is classified, evaluate r at the two angles along the axis. Those give the vertices, whose midpoint is the centre, from which the second focus and the axis lengths follow.
A negative r among the evaluations is normal and must be interpreted, not discarded; it is how the far vertex of a hyperbola is reached.
Figure (svg): Two columns separating what can be read directly from a normalised polar conic equation from what requires a short calculation
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 983-985
Picture it
The left column costs one glance and the right column costs one substitution each.
Figure (svg): Two columns separating what can be read directly from a normalised polar conic equation from what requires a short calculation
In practice the left column is done first, because it tells you which two angles to substitute in the right column.
Worked example
Example 11.6.4, part 2, completed.
\[ \text{Sketch } r = \frac{12}{3 - \cos(\theta)}. \]
Evaluate at the axis angles
Why: Theta equal to 0 and to pi.
\[ r(0) = 6, r(\pi) = 3 \]
Convert to points
Why: Along the positive and negative x-axis.
\[ (6, 0)\text{ and } (-3, 0) \]
Find the centre
Why: Midpoint of the vertices.
\[ (\frac{3}{2}, 0) \]
Find the minor axis
Why: From the theorem's formula.
\[ 6 \sqrt{2} \]
Figure (svg): The solution to Worked example sketching an ellipse shown as a ladder of expressions, one row per legal move
\[ \text{vertices } (6,0), (-3,0); \; \text{centre } \left(\tfrac{3}{2}, 0\right); \; \text{minor axis } 6\sqrt{2} \]
Verify: check the major axis length
Why: The distance from negative 3 to 6 is 9, and the theorem gives 2ed over 1 minus e squared, which is 8 over one minus one ninth, namely 8 divided by eight ninths, which is 9. The two agree. The second focus is as far the other side of the centre as the pole, so at 3 comma 0.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 983-984
Faded example
For r equal to 4 over 1 minus sine theta, find the vertex.
Fill in the blanks
r\left(\tfrac2-2\right) = \frac______ = ___ \;\Longrightarrow\; \text___ (0, ___)
Why: At three pi over 2 the sine is negative 1, so the denominator is 2 and r is 2. That angle points straight down, so the point is 0 comma negative 2. At pi over 2 the denominator would be zero, which is the direction in which this parabola escapes — so there is only one vertex, as a parabola should have.
Worked example
Example 11.6.4, part 3, completed. One evaluation comes out negative.
\[ \text{Sketch } r = \frac{6}{1 + 2\sin(\theta)}. \]
Evaluate at the axis angles
Why: Pi over 2 and three pi over 2.
\[ r = 2\text{ and } r = -6 \]
Convert both to points
Why: The negative r means half a turn.
\[ (0, 2)\text{ and } (0, 6) \]
Find the centre and second focus
Why: Midpoint, then reflect the pole.
\[ \text{centre } (0, 4),\text{ focus } (0, 8) \]
Find the conjugate axis and slopes
Why: From the theorem's formula.
\[ 4 \sqrt{3},\text{ slopes } +- \sqrt{3} / 3 \]
Figure (svg): The solution to Worked example sketching a hyperbola shown as a ladder of expressions, one row per legal move
\[ \text{vertices } (0,2), (0,6); \; \text{centre } (0,4); \; \text{transverse axis } 4, \; \text{conjugate axis } 4\sqrt{3} \]
Verify: handle the negative r correctly
Why: At three pi over 2 the value is negative 6, and a negative r at that angle lands half a turn round, straight up, at 0 comma 6. Discarding it as impossible would lose one vertex and with it the centre, the second focus, and the whole sketch.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 984-984
Error analysis
A student evaluates a hyperbola at its axis angles.
Annotate
On: \( r\left(\tfrac{3\pi}{2}\right) = \frac{6}{1 - 2} = -6 \;\Longrightarrow\; \text{no point here} \)
This is the negative-r rule from Lesson 11.4a doing real work. On a hyperbola the two branches are reached with opposite signs of r, so treating a negative value as an error deletes exactly half the curve.
Prediction
You evaluate a polar conic at both axis angles and one denominator comes out zero.
Predict first
What does that tell you?
Correct: The conic is a parabola.
Why: The denominator vanishes exactly when the eccentricity is 1 and the trigonometric function is at the extreme value that cancels it, which happens at precisely one angle. That is the direction in which the parabola escapes to infinity, and it is why a parabola has only one vertex rather than two. An ellipse never has a vanishing denominator and a hyperbola has two vanishing angles, neither of them on the axis.
Sorting
The axis direction follows from the function in the denominator.
Sort into buckets
Sort each equation by which pair of angles to substitute.
Explain it to yourself
The centre is found as the midpoint of the two vertices, and the second focus by reflecting the pole in it.
Discussion prompt
Explain why that reflection works, and why the same construction is unavailable for a parabola.
Hint: What symmetry does an ellipse or hyperbola have?
Answer:
An ellipse and a hyperbola are both symmetric about their centre. That symmetry exchanges the two vertices and also exchanges the two foci, so reflecting one focus in the centre must give the other.
Since one focus is the pole, and the centre is known once the vertices are, the second focus is a single reflection away — no formula needed.
A parabola has no centre, because it has only one vertex and no central symmetry. It correspondingly has only one focus. The construction is unavailable precisely because the object it relies on does not exist, which is the same reason the polar form is centred on a focus rather than a centre in the first place.
Section
Section 5
Concept
Replacing theta by theta minus a constant rotates the whole graph by that constant. Since the four standard forms differ by quarter turns, all four collapse into one form with a rotation parameter.
Theorem 11.13 — For positive numerator, non-negative eccentricity and any rotation angle, this equation gives a conic with eccentricity e and a focus at the pole. If the eccentricity is zero the graph is a circle of radius equal to the numerator; otherwise the directrix contains the point with polar coordinates negative d comma the rotation angle, where d is the numerator divided by the eccentricity.
\[ r = \frac{\ell}{1 - e\cos(\theta - \varphi)} \]
Writing the numerator as its own parameter rather than as a product is what lets the eccentricity be zero, admitting circles into the family.
Figure (svg): A parabola in polar form shown before and after replacing theta by theta minus a constant, which rotates the whole curve counter-clockwise
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 985-985
Picture it
One substitution, and the whole curve turns.
Figure (svg): A parabola in polar form shown before and after replacing theta by theta minus a constant, which rotates the whole curve counter-clockwise
Rotation is cheap in polar form and expensive in rectangular form, which is exactly the reverse of the previous lesson. That contrast is the argument for having both.
Worked example
The book's illustration. The curve from the first example, turned.
\[ \text{Describe the graph of } r = \frac{4}{1 - \sin\left(\theta - \tfrac{\pi}{4}\right)}. \]
Recognise the unrotated version
Why: Set the rotation to zero.
\[ r = \frac{4}{1 - \sin \theta} \]
Identify that curve
Why: Eccentricity 1, directrix y = -4.
Apply the rotation
Why: Subtracting pi over 4 turns counter-clockwise.
\[ \text{rotate by } \frac{\pi}{4} \]
Describe the result
Why: Same parabola, axis turned.
Figure (svg): The solution to Worked example rotating a parabola shown as a ladder of expressions, one row per legal move
\[ \text{a parabola with focal diameter } 8, \text{ its axis at } \tfrac{\pi}{4} + \tfrac{\pi}{2} \]
Verify: check what did not change
Why: The eccentricity is still 1, the focal diameter is still 8, and the focus is still the pole. Rotation is a rigid motion about the pole, so every length and every ratio survives it, and only directions change.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 985-985
Prediction
You replace theta by theta minus pi over 2 in a polar conic whose denominator uses a cosine.
Predict first
What is the resulting equation equivalent to?
Correct: The corresponding sine form.
Why: The cosine of theta minus pi over 2 is the sine of theta, by the cofunction identity, so the substitution converts one standard form into another. Geometrically the conic has been turned through a quarter turn, which swaps a vertical directrix for a horizontal one. This is exactly why the four separate forms of the earlier theorem collapse into one with a rotation parameter.
Worked example
The reparametrisation that admits a circle.
\[ \text{Describe the graph when } e = 0. \]
Substitute zero for the eccentricity
Why: The trigonometric term vanishes.
\[ r = \frac{l}{1 - 0} \]
Simplify
Why: The denominator is 1.
\[ r = l \]
Recognise the equation
Why: A constant radius.
Note the directrix
Why: d would be the numerator over zero.
Figure (svg): The solution to Worked example the circle case shown as a ladder of expressions, one row per legal move
\[ e = 0 \;\Longrightarrow\; r = \ell, \text{ a circle of radius } \ell \]
Verify: explain why the directrix disappears
Why: The formula for d is the numerator divided by the eccentricity, which is undefined at zero. That is appropriate: a circle has no directrix, and the definition by ratio degenerates there. The circle joins the family through the equation rather than through the definition, which is why the theorem states it as a separate bullet.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 985-985
Trap
\[ r = \frac{4}{1 - \sin\left(\theta + \tfrac{\pi}{4}\right)} \;\Longrightarrow\; \text{rotated counter-clockwise by } \tfrac{\pi}{4} \]
Read the plus sign as a counter-clockwise turn
Why: Adding to an angle usually means increasing it, which is counter-clockwise.
But the rotation goes the other way. Adding to theta inside the function means the curve reaches a given shape at a smaller theta, so the graph turns clockwise.
\[ r = f(\theta - \varphi) \;\Longrightarrow\; \text{rotated counter-clockwise by } \varphi \]
Subtract to turn counter-clockwise
Why: The same convention as horizontal shifts of a graph.
This is the identical reversal as with shifting a function's graph: subtracting inside the argument moves the graph in the positive direction. If it was confusing there it will be confusing here, and the fix is the same — test one point rather than trusting the sign.
Faded example
Identify the conic r equal to 6 over 3 minus cosine of theta plus pi over 4.
Fill in the blanks
r = \frac+clockwise___\cos\left(\theta ___ \tfrac______\right)} \;\Longrightarrow\; e = \tfrac______, \text___ ___ \text___ \tfrac______
Why: Dividing by 3 normalises the equation and reveals an eccentricity of one third, so the curve is an ellipse. The plus sign inside the cosine means the rotation is clockwise, by pi over 4 — the opposite of what a minus sign would do.
Sorting
Rotation about the pole is a rigid motion.
Sort into buckets
Sort each quantity.
Real world
Kepler's first law says a planet moves in an ellipse with the sun at one focus. Orbital elements are published as a semi-major axis and an eccentricity, plus angles giving the orientation.
Discussion prompt
Explain why the polar form is the natural language for orbits, and identify what each parameter of the general equation corresponds to.
Hint: Where is the observer, and what is being measured?
Answer:
The sun sits at a focus, and the polar form is the only one of the two that puts a focus at the origin. Every distance in the problem — how far the planet is from the sun right now — is exactly r, measured from that focus.
The eccentricity is published directly and is the same number as in this lesson: it decides ellipse from parabola from hyperbola, which physically is bound orbit from escape trajectory from flyby. The numerator sets the scale, and it has its own name, the semi-latus rectum. The rotation angle is the orientation of the orbit within its plane, published as the argument of periapsis.
So a published orbit is literally a set of values for the general equation of this section, and the equation is how a position is computed from them. The conic sections were studied for two thousand years before anyone knew what they were for, and this is what they turned out to be for.
It is also why the parabolic case matters physically rather than only as a boundary. A comet at exactly escape speed traces a parabola, and the same value of the eccentricity that divides closed curves from open ones divides bodies that return from bodies that do not.
Comparison
Fill the blanks from memory. Every row assumes the equation has been normalised first.
Comparison matrix
| What you want | Where it comes from | Note |
|---|---|---|
| the eccentricity | the coefficient of the sine or cosine | only after normalising |
| which conic | compare the eccentricity against 1 | below, equal, above: ellipse, parabola, hyperbola |
| the directrix | the function gives its direction, the sign its side | d is the numerator divided by e |
| the vertices | evaluate r at the two angles along the axis | a negative r is normal, not an error |
| the second focus | reflect the pole in the centre | no such thing for a parabola |
The first three rows cost one glance each; the last two cost one substitution each. Doing the first three first is what tells you which two angles to use in the fourth.
Pattern
Five moves, and the first one is the one people skip.
If the argument is theta minus a constant, everything above applies to the unrotated curve, and the whole figure is then turned counter-clockwise by that constant.
OpenStax Algebra and Trigonometry 2e, §12.5 Conic Sections in Polar Coordinates §12.5
Check
Normalising.
Check your understanding
What is the eccentricity of the conic r equal to 15 over 5 plus 3 cosine theta?
Answer: B
Why: Dividing numerator and denominator by 5 gives r equal to 3 over 1 plus three fifths cosine theta. The coefficient of the cosine is then three fifths, which is the eccentricity, and since it is below 1 the conic is an ellipse.
Check
Locating the directrix.
Check your understanding
For r equal to 8 over 1 minus 2 sine theta, where is the directrix?
Answer: B
Why: The eccentricity is 2 and the numerator is 8, so d is 4. A sine in the denominator makes the directrix horizontal, and the minus sign places it on the negative side, at y equal to negative 4. The conic is a hyperbola, since the eccentricity exceeds 1.
Check
The general form.
Check your understanding
What is the graph of r equal to 7 over 1 minus 0 times cosine of theta minus pi over 3?
Answer: A
Why: With eccentricity zero the trigonometric term vanishes entirely and the equation reduces to r equal to 7. That is a circle of radius 7 centred at the pole, and the rotation parameter has nothing to act on since a circle is unchanged by rotation about its centre.
Real world
A mission planner is designing a spacecraft trajectory past a planet. The approach is hyperbolic, and firing the engine at closest approach changes the speed, which changes the eccentricity. The planner needs to know how much of a burn turns the hyperbola into a closed ellipse.
Discussion prompt
Explain what the planner is computing in the language of this lesson, and what happens at the boundary.
Hint: Which value of the eccentricity separates the two cases?
Answer:
The trajectory is a conic with the planet at a focus, exactly the configuration of this lesson. Arriving from far away it is a hyperbola, with eccentricity above 1, which is why it does not close and the spacecraft would otherwise leave.
The burn reduces the speed, which reduces the eccentricity. The planner is computing how much reduction brings the eccentricity below 1, at which point the trajectory closes into an ellipse and the spacecraft is captured into orbit.
The boundary at exactly 1 is the parabolic trajectory, corresponding precisely to escape speed. It is a genuine physical threshold and not merely a mathematical one: below it the spacecraft is bound, above it is not, and the crossing is instantaneous rather than gradual.
Everything in the previous lesson about the difficulty of the rectangular form is why this is done in polar coordinates. The numerator sets the size of the orbit and the eccentricity sets its shape, and the burn changes one of those cleanly — which would be invisible in a rectangular equation where both parameters are smeared across every coefficient.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
For the conic r equal to 12 over 4 plus 4 cosine theta, which conic is it?
Correct: A parabola.
\[ r = \frac{12}{4 + 4\cos\theta} = \frac{3}{1 + \cos\theta} \;\Longrightarrow\; e = 1 \]
Why: Dividing top and bottom by 4 gives r equal to 3 over 1 plus cosine theta, so the eccentricity is exactly 1 and the conic is a parabola. Reading the unnormalised equation might suggest an eccentricity of 4, and a hyperbola, which is wrong. The two coefficients being equal is the signature of a parabola in this form, whatever their common value happens to be.
Explain it
They have learned four separate conic definitions and four sets of standard forms, and find the whole topic a memory exercise with no through-line.
Discussion prompt
In no more than five sentences, give them the through-line.
Hint: How many definitions are actually needed?
Answer:
There is one definition, not four: fix a point and a line, and take every point whose distance to the point is a fixed multiple of its distance to the line. That multiple is the eccentricity, and it is the only thing that distinguishes the four shapes.
Below 1 the curve closes into an ellipse, at exactly 1 it is a parabola, above 1 it splits into a hyperbola, and at 0 it is a circle. One dial, four shapes, with the circle and the parabola sitting at the two special values.
Written in polar coordinates with the focus at the origin, the whole family is a single equation, and the eccentricity is a coefficient you can read off in one glance. The four sets of standard forms were four views of one object, which is worth knowing before memorising any of them.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Normalising is fixed by checking the constant term in the denominator before reading anything. The directrix is fixed by the two-part rule: the function gives the orientation and the sign gives the side. The vertices are fixed by substituting the two axis angles and interpreting a negative r rather than discarding it. The rotated form is fixed by identifying the unrotated curve first and turning it afterwards. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a focus, a vertical directrix, and a point of the curve with both its distances marked, and write the definition as a ratio beside it. Underneath, write the four-line derivation of the polar equation from that definition. In the middle of the page draw three small polar sketches for eccentricities of one half, one, and two, labelling each with the conic it is and with what the denominator does in each case. Beside them write the four standard forms with the position of the directrix for each. In the bottom left, take the equation with numerator 12 and denominator 3 minus cosine theta, normalise it, and find the eccentricity, the directrix, both vertices, the centre and the second focus. In the bottom right, write the general rotated form and note what happens when the eccentricity is zero. Finally, circle the one step without which nothing else on the page is valid.
The circled step is the normalisation. Every parameter in every rule assumes the constant term in the denominator is 1, and reading an unnormalised equation gets the eccentricity wrong, which gets the classification wrong, which invalidates everything after it.
Recap
Five things, and the first is what makes the other four one topic rather than four.
| If the question says | Your first move is |
|---|---|
| Identify this polar conic | Divide until the constant term is 1 |
| Where is the directrix? | Read the function for direction, the sign for side |
| Find the vertices | Evaluate r at the two angles along the axis |
| One value of r came out negative | It names a point half a turn round |
| The argument is theta minus a constant | Identify the unrotated curve, then turn it |
The conics have now been given a single definition and a single equation, with the focus at the pole and one parameter controlling the shape. The chapter turns next to complex numbers, where the same polar idea makes multiplication a rotation and roots evenly spaced points on a circle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 981-986 — everything on these slides traces back here
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