Describes all three conics with a single polar equation by placing the pole at a focus. Identifies the conic from its eccentricity, reads the directrix's position from the form, converts between polar and rectangular equations, and applies the result to orbital motion.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 10 — Analytic Geometry
§10.5 Conic Sections in Polar Coordinates, pp. 1266-1280
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1266-1280 — the pages these objectives are drawn from
Warm-up
Their rectangular equations looked like three unrelated things.
Discussion prompt
The ellipse, hyperbola and parabola had quite different standard forms. What do all three share?
Hint: What appeared in every definition?
Answer:
Every one was defined using a focus. The ellipse and hyperbola had two, and the parabola had one.
But only the parabola's definition also used a directrix — the other two were defined without one, using their second focus instead.
It turns out all three can be defined by one focus and a directrix, with a ratio of distances that varies. When that ratio is the organising idea, one equation covers all three, and this section is that equation.
Concept
With the pole at a focus, every conic satisfies the same polar equation, and the eccentricity alone decides whether it is an ellipse, a parabola or a hyperbola.
\[ r=\frac{ep}{1\pm e\cos\theta} \quad\text{or}\quad r=\frac{ep}{1\pm e\sin\theta} \]
The unification depends on the pole being at a focus rather than at the centre, because the parabola has no centre. Choosing the focus is what makes a common form possible at all.
Figure (svg): A card giving the single polar equation for all three conics, with the eccentricity marked as the parameter that selects between them
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1266-1270
Section
Section 1
Concept
All three conics are the sets of points whose distance to a focus is a fixed multiple of their distance to a directrix. That multiple is the eccentricity.
Section 10.3's parabola definition was the special case with the ratio equal to one, which is why it used a plain equality. Allowing the ratio to differ from one is the generalisation that brings the other two conics into the same framework.
Figure (svg): A contrast between the rectangular convention of centring a conic and the polar convention of placing a focus at the pole
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1266-1271
Picture it
The parabola is what forces the choice.
Figure (svg): A contrast between the rectangular convention of centring a conic and the polar convention of placing a focus at the pole
The second row on each side is the decisive one. A centred convention cannot include the parabola, and a focal one can include everything.
Worked example
The eccentricity is one.
\[ \text{What does the ratio definition give when the eccentricity is one?} \]
Set the ratio to one
Why: The two distances are equal.
Compare with §10.3
Why: The focus-directrix definition.
Conclude
Why: The parabola.
\[ e = 1 \]
Note the generalisation
Why: Other ratios give other conics.
Figure (svg): A contrast between the rectangular convention of centring a conic and the polar convention of placing a focus at the pole
\[ e=1: \text{the parabola} \]
Verify: check the boundary role
Why: One is the boundary between the ellipse case and the hyperbola case, which matches the parabola being the boundary shape — a curve that is unbounded but has only one branch. The numerical boundary and the geometric boundary agree.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1267-1269
Prediction
The eccentricity is 0.6.
Predict first
Which conic is it?
Correct: An ellipse.
Why: Values below one give ellipses, one gives the parabola, and values above one give hyperbolas. A circle would need an eccentricity of exactly zero.
Worked example
The ellipse's eccentricity was defined differently in §10.1.
\[ \text{Is this eccentricity the same as the one from } \S 10.1? \]
Recall the earlier definition
Why: Focal distance over semi-major axis.
Recall its range
Why: Between zero and one.
Compare with this one
Why: Also between zero and one.
Conclude
Why: The same quantity.
Figure (svg): The solution to Worked example connect to the earlier eccentricity shown as a ladder of expressions, one row per legal move
\[ \text{the same eccentricity} \]
Verify: check the extreme case
Why: Both give zero for a circle: the earlier because the foci coincide, and this one because the directrix moves infinitely far away. Two definitions agreeing at the extremes and having the same range is what identifies them as the same quantity.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1269-1271
Trap
\[ \text{put the pole at the conic's centre, as in the rectangular forms} \]
Carry over the centred convention
Why: The origin is placed where the rectangular standard forms put it.
The parabola has no centre, so the convention cannot cover all three conics.
The pole goes at a focus. Every conic has one, including the parabola.
That is what makes a single equation possible — a centred convention could never include the parabola at all.
It is also the physically natural choice for orbits, where the massive body sits at a focus rather than at the centre.
Sorting
One is the boundary.
Sort into buckets
Sort each value.
Faded example
A conic with eccentricity 1.6.
Fill in the blanks
e=1.6>1 \;\Longrightarrow\; \texthyperbola___
Why: Any eccentricity above one gives a hyperbola, whatever the value. The magnitude affects how open the branches are but not which conic it is.
Explain it to yourself
The rectangular forms centred each conic instead.
Discussion prompt
Explain why the focal choice unifies and the centred one cannot.
Hint: Which conic lacks something?
Answer:
A parabola has no centre, so any convention built on centring cannot describe it at all. That is why the rectangular forms came out as three unrelated equations.
Every conic has a focus, so placing the pole there is a convention all three can share — and one equation can then cover them.
It is also the physically right choice, since the massive body in an orbit sits at a focus. A good explanation notes that the mathematical convenience and the physical relevance coincide here, which is unusual and worth noticing.
Section
Section 2
Concept
The eccentricity can only be read directly when the denominator begins with a one. If it does not, dividing through by the leading constant puts it into standard form first.
This is the single commonest error in the section: an equation whose denominator starts with a three has an eccentricity a third of what it appears to be. Dividing first costs one line and prevents misclassifying the conic entirely.
Figure (svg): A card giving the single polar equation for all three conics, with the eccentricity marked as the parameter that selects between them
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1271-1275
Picture it
The eccentricity is readable only in this shape.
Figure (svg): A card giving the single polar equation for all three conics, with the eccentricity marked as the parameter that selects between them
The three cases in the middle assume the equation is already standardised. Applying them to an unstandardised equation misreads the eccentricity and can name the wrong conic.
Worked example
The denominator does not start with one.
\[ \text{Identify the conic } r=\frac{12}{3+2\cos\theta}. \]
Note the leading constant
Why: Three, not one.
Divide numerator and denominator
Why: By three.
\[ \frac{4}{1 + (\frac{2}{3}) \cos} \]
Read the eccentricity
Why: The coefficient.
\[ \frac{2}{3} \]
Classify
Why: Below one.
Figure (svg): A card giving the single polar equation for all three conics, with the eccentricity marked as the parameter that selects between them
\[ e=\tfrac{2}{3}: \text{an ellipse} \]
Verify: check what reading without standardising would give
Why: Reading the coefficient as 2 directly would suggest a hyperbola — the wrong conic entirely. The division changed the classification, which is why it cannot be skipped.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1272-1274
Prediction
The denominator begins with a 3.
Predict first
What must you do before reading the eccentricity?
Correct: Divide numerator and denominator by 3.
Why: The eccentricity is only readable when the denominator's leading term is one. Dividing achieves that and scales the trigonometric coefficient to its true value, which can change the classification.
Worked example
From the numerator, after standardising.
\[ \text{Find the directrix distance for that conic.} \]
Use the standardised form
Why: Numerator four.
\[ e p = 4 \]
Recall the eccentricity
Why: Two thirds.
\[ e = \frac{2}{3} \]
Divide
Why: To isolate the distance.
\[ \frac{4}{\frac{2}{3}} \]
Compute
Why: Multiply by the reciprocal.
\[ p = 6 \]
Figure (svg): The solution to Worked example find the directrix distance shown as a ladder of expressions, one row per legal move
\[ p=6 \]
Verify: check at a convenient angle
Why: At an angle of zero the radius is 12 over 5, about 2.4 — the vertex nearest the directrix. That is less than the directrix distance of 6, which it must be since the vertex lies between the focus and the directrix.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1274-1275
Error analysis
A student identifies a conic from its polar equation.
Annotate
On: \( r=\frac{12}{3+2\cos\theta} \;\Longrightarrow\; e=2, \text{ a hyperbola} \)
The division takes one line and changes the classification. Checking that the denominator starts with a one before reading anything is what makes the rest of the analysis valid.
Faded example
Dividing through by the leading constant.
Fill in the blanks
\frac43=\frac___}______}\cos\theta}
Why: Dividing both numerator and denominator by three leaves the equation unchanged in value while putting it in standard form. Only then does the cosine's coefficient give the eccentricity.
Sorting
The denominator must lead with a one.
Sort into buckets
Sort each denominator.
Step zero
You are given a polar conic equation to identify.
Discussion prompt
What do you check before reading anything?
Hint: One number in the denominator.
Answer:
Whether the denominator begins with a one. The eccentricity is only readable in that form.
If it does not, divide both numerator and denominator by the leading constant. That changes nothing about the curve and everything about what the coefficients mean.
Skipping it misstates the eccentricity by exactly that factor, which is often enough to name the wrong conic. The check costs a glance and the fix costs a line.
Section
Section 3
Concept
A cosine means a vertical directrix and a sine a horizontal one. A plus puts it on the positive side of the pole and a minus on the negative side.
The conic always opens away from the directrix, so knowing where the directrix is also tells you which way the curve extends. That makes the sign worth reading even when only a rough sketch is wanted.
| form | directrix |
|---|---|
| one plus a cosine term | vertical, to the right |
| one minus a cosine term | vertical, to the left |
| one plus a sine term | horizontal, above |
| one minus a sine term | horizontal, below |
Figure (svg): Four cards giving the four polar conic forms and the directrix position each corresponds to
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1275-1278
Picture it
Two choices, four directrix positions.
Figure (svg): Four cards giving the four polar conic forms and the directrix position each corresponds to
The caption records the structure: the trigonometric function chooses horizontal or vertical, and the sign chooses which side. Neither choice affects the eccentricity.
Worked example
Read the function and the sign.
\[ \text{Locate the directrix of } r=\frac{4}{1+\tfrac{2}{3}\cos\theta}. \]
Read the function
Why: A cosine.
Read the sign
Why: Plus.
Find the distance
Why: Numerator over eccentricity.
\[ p = 6 \]
State the line
Why: Six units right of the pole.
\[ x = 6 \]
Figure (svg): Four cards giving the four polar conic forms and the directrix position each corresponds to
\[ x=6 \]
Verify: check the curve opens away from it
Why: At an angle of zero — towards the directrix — the radius is smallest, and at a straight angle it is largest. So the conic extends away from the directrix, as it must, and the near vertex sits between the focus and the line.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1276-1277
Matching
Function and sign together.
Match the pairs
Why: The trigonometric function chooses between horizontal and vertical, and the sign chooses which side of the pole. Two independent choices give exactly four forms.
Worked example
The directrix is horizontal now.
\[ \text{Locate the directrix of } r=\frac{6}{1-\sin\theta}. \]
Read the eccentricity
Why: The sine's coefficient.
\[ e = 1 \]
Classify
Why: A parabola.
Read the function and sign
Why: Sine, minus.
Find the distance
Why: Numerator over eccentricity.
\[ p = 6 \]
Figure (svg): The solution to Worked example a sine form shown as a ladder of expressions, one row per legal move
\[ y=-6, \text{ a parabola} \]
Verify: check the denominator's behaviour
Why: At an angle of a quarter turn the denominator is zero and the radius is undefined, so the curve never reaches that direction — consistent with an unbounded parabola opening upward, away from the directrix below. A denominator that can vanish is the signature of an unbounded conic.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1277-1278
Trap
\[ 1+e\cos\theta \;\Longrightarrow\; \text{directrix to the LEFT} \]
Associate the plus sign with the negative direction
Why: The sign convention is reversed.
The conic is then drawn opening the wrong way.
A plus puts the directrix on the positive side — to the right for a cosine, above for a sine.
Checking the radius at an angle of zero settles it: a plus makes the denominator largest there, so the radius is smallest, meaning the curve is closest to the pole in that direction — towards the directrix.
The curve always opens away from the directrix, so getting the side right also gets the opening direction right.
Prediction
The directrix is the vertical line to the right of the pole.
Predict first
Which way does the conic open?
Correct: To the left, away from the directrix.
Why: Every conic opens away from its directrix, since points far from the focus must be even further from the directrix to keep the ratio constant. Locating the directrix therefore fixes the opening direction too.
Faded example
With numerator 4 and eccentricity two thirds.
Fill in the blanks
ep=4 \;\Longrightarrow\; p=\frac36}}=___
Why: The numerator is the eccentricity times the directrix distance, so dividing by the eccentricity isolates the distance. This only works once the equation is in standard form.
Explain it
There are exactly four, not more.
Discussion prompt
Explain to a classmate why four.
Hint: How many independent choices?
Answer:
Two independent choices: which trigonometric function appears, and which sign it carries.
The function decides whether the directrix is vertical or horizontal, and the sign decides which side of the pole it sits on.
Two choices with two options each gives exactly four, which is why the list is complete. A good explanation notes that neither choice affects the eccentricity — the conic's type is independent of its orientation.
Section
Section 4
Concept
Multiplying through by the denominator produces terms that the §8.3 conversion formulas can replace, after which squaring removes the remaining radius.
Squaring can introduce extraneous points, as always, so the resulting rectangular equation may describe slightly more than the polar one did. For conics the difference is usually a branch that the polar form's radius restriction excluded.
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1278-1280
Picture it
All sharing the focus at the pole.
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
Converting any of these to rectangular form recovers one of §10.1 to §10.3's standard equations, but translated — because those forms centre the conic and this one puts a focus at the origin.
Worked example
Clear, substitute, square.
\[ \text{Convert } r=\frac{6}{1-\sin\theta} \text{ to rectangular form.} \]
Clear the fraction
Why: Multiply through.
\[ r - r \sin(\theta) = 6 \]
Substitute
Why: The product is the vertical coordinate.
\[ r - y = 6 \]
Isolate the radius
Why: Add.
\[ r = y + 6 \]
Square both sides
Why: The radius becomes a sum of squares.
\[ x ^{2} + y ^{2} = y ^{2} + 12 y + 36 \]
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
\[ x^2=12y+36 \]
Verify: check the conic type
Why: The rectangular form has one squared variable, which identifies a parabola — matching the eccentricity of one read from the polar form. Rearranged, it is the square of the horizontal variable equalling twelve times the vertical plus three, a parabola opening upward with vertex three below the origin.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1278-1279
Prediction
You are converting a polar conic to rectangular form.
Predict first
What comes first?
Correct: Multiply through by the denominator.
Why: The conversion formulas replace the radius multiplied by a trigonometric function, and clearing the fraction is what creates those products. Substituting before that leaves a radius stuck inside the denominator.
Worked example
One centres and one focuses.
\[ \text{Why does the converted equation not match } \S 10.1\text{'s standard form?} \]
Recall the polar convention
Why: Pole at a focus.
Recall the rectangular convention
Why: Origin at the centre.
Compare
Why: Different points at the origin.
Conclude
Why: The same conic, shifted.
Figure (svg): The solution to Worked example why the forms look different shown as a ladder of expressions, one row per legal move
\[ \text{translated, not different} \]
Verify: identify the translation
Why: The shift is exactly the focal distance, since that is how far the focus sits from the centre. Completing the square on the converted equation reveals a centre offset by that amount, confirming the two forms describe the same conic in different positions.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1279-1280
Error analysis
A student converts a polar conic.
Annotate
On: \( r=\frac{6}{1-\sin\theta} \;\Longrightarrow\; \sqrt{x^2+y^2}=\frac{6}{1-y/r} \)
The conversion formulas replace products of the radius with a trigonometric function, not the function alone. Clearing the fraction first is what creates those products.
Faded example
The product of the radius and the sine.
Fill in the blanks
r-r\sin\theta=6 \;\Longrightarrow\; r-y=6
Why: The radius times the sine is the vertical coordinate, by §8.3's conversion formula. The remaining radius is then isolated and squared, which replaces it with the sum of the squared coordinates.
Sorting
The conversion has a fixed order.
Sort into buckets
Sort each task.
Explain it to yourself
The converted equation does not match §10.1's.
Discussion prompt
Explain what accounts for the difference.
Hint: What is at the origin in each?
Answer:
The polar convention puts a focus at the origin; the rectangular standard forms put the centre there.
Those are different points, separated by the focal distance, so the two equations describe the same conic in different positions.
Completing the square on the converted equation reveals a centre offset by exactly that distance. A good explanation notes that neither form is more correct — they answer to different conventions, chosen for different purposes.
Section
Section 5
Concept
Kepler's first law places the sun at a focus of each planet's elliptical orbit, which is exactly where the polar form puts the pole — making this the natural equation for orbital motion.
The boundary at an eccentricity of one is physically meaningful, not just a change of curve name: it separates bodies bound to the central mass from those with enough energy to escape. Comets on parabolic or hyperbolic paths pass once and never return.
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1279-1280
Picture it
One equation covering bound and unbound paths.
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
The three curves share a focus, which in an orbital reading is the central body. Increasing eccentricity past one is the transition from a returning orbit to an escape.
Worked example
The radius is largest and smallest along the axis.
\[ \text{An orbit has } r=\frac{4}{1+0.5\cos\theta}. \text{ Find the nearest and furthest distances.} \]
Find where the denominator is largest
Why: The cosine is one.
Compute the nearest distance
Why: Divide.
\[ \frac{4}{1.5} = \frac{8}{3} \]
Find where it is smallest
Why: The cosine is negative one.
Compute the furthest
Why: Divide.
\[ \frac{4}{0.5} = 8 \]
Figure (svg): Three polar conics drawn from the same equation with different eccentricities, sharing a focus at the pole
\[ \tfrac{8}{3}\text{ and }8 \]
Verify: check against the semi-major axis
Why: The two extremes average to about 5.33, which is the semi-major axis — and the focal distance is half their difference, about 2.67. Their ratio gives an eccentricity of 0.5, matching the equation. Three independent checks agree.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1279-1280
Prediction
A trajectory has eccentricity 0.7.
Predict first
Is the path closed?
Correct: Yes, it is an ellipse and repeats.
Why: An eccentricity below one gives a closed elliptical orbit that returns to its starting point. At one or above the path is unbounded and the body never comes back.
Worked example
The eccentricity settles it.
\[ \text{A comet's path has eccentricity } 1.2. \text{ Will it return?} \]
Compare with one
Why: Above.
\[ e > 1 \]
Identify the conic
Why: A hyperbola.
Interpret physically
Why: The path never closes.
Conclude
Why: A single pass.
Figure (svg): The solution to Worked example bound or escaping shown as a ladder of expressions, one row per legal move
\[ \text{hyperbolic: it escapes} \]
Verify: check the denominator
Why: For an eccentricity above one the denominator vanishes at some angle, so the radius grows without bound in that direction — the comet recedes forever. For an eccentricity below one the denominator never vanishes and the radius stays finite, which is a closed orbit.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1280-1280
Trap
\[ \text{the sun is at the centre of the ellipse} \]
Put the massive body at the geometric centre
Why: The centre is assumed to be where the mass sits.
The nearest and furthest distances then come out equal, contradicting observation.
The central body is at a focus, which is Kepler's first law and exactly where the polar form's pole sits.
That is why an orbit has a nearest and a furthest point at all, differing by twice the focal distance.
The centre of an orbital ellipse is empty space, and the polar form's convention reflects the physics rather than merely being convenient.
Faded example
Where the cosine is one.
Fill in the blanks
r=\frac11.5})}=\frac______}
Why: The radius is smallest where the denominator is largest, which happens when the cosine reaches one. That direction points from the focus towards the near vertex.
Sorting
The eccentricity decides.
Sort into buckets
Sort each trajectory.
Explain it
An eccentricity of one separates two kinds of path.
Discussion prompt
Explain to a classmate what changes there.
Hint: What can the denominator do?
Answer:
Below one, the denominator never vanishes, so the radius stays finite in every direction and the path closes into an orbit.
At one and above, the denominator can vanish, and the radius grows without bound in that direction — the body recedes forever.
So the mathematical boundary is a physical one: bound orbits on one side and escapes on the other. A good explanation notes that this is why the eccentricity of a newly discovered comet is the first thing computed — it says whether the object will ever be seen again.
Comparison
Fill the blanks from memory. The conventions differ and so does what they unify.
Comparison matrix
| rectangular forms | polar form | |
|---|---|---|
| origin at | the centre | a focus |
| number of equations | three, one per conic | one |
| what distinguishes them | the signs and number of squared terms | the eccentricity |
| covers the parabola | with its own separate form | as the case with eccentricity one |
The last row is the point of the section. The parabola having no centre is what kept the rectangular forms separate, and moving the origin to a focus is what brings it into the family.
Pattern
Five steps, and the first must not be skipped.
Step 1 changes what every later step reads. An unstandardised equation misstates the eccentricity by the leading constant's factor, which is often enough to name the wrong conic.
OpenStax Algebra and Trigonometry 2e, §12.5 Conic Sections in Polar Coordinates §12.5
Check
Classification.
Check your understanding
A polar conic has eccentricity 1.4. Which conic is it?
Answer: A
Why: Values above one give hyperbolas, one gives the parabola, and values below one give ellipses. The eccentricity alone determines the type, which is what makes the single polar equation possible.
Check
Standard form.
Check your understanding
The denominator begins with a 3. What must you do first?
Answer: A
Why: The eccentricity is only readable when the denominator's leading term is one. Dividing puts the equation in standard form without changing the curve, and skipping it misstates the eccentricity by a factor of three.
Check
The pole's position.
Check your understanding
Where is the pole in the polar form of a conic?
Answer: A
Why: Placing the pole at a focus is what allows one equation to cover all three conics, since the parabola has no centre. It is also where the central body sits in an orbital reading.
Real world
Mission planners use this equation directly.
Discussion prompt
A spacecraft's trajectory is described in this polar form. What does the engineer read off first?
Hint: One number answers the most important question.
Answer:
The eccentricity, because it says whether the craft is on a closed orbit or an escape trajectory — the single most consequential fact about a path.
Below one the craft returns and can be tracked over repeated passes. At or above one it leaves and the encounter is a single opportunity.
And a gravity assist is precisely a manoeuvre that changes the eccentricity: a craft arrives on a hyperbolic path relative to a planet and departs on a different one, with its heliocentric orbit altered. The equation is the working tool, not an illustration of one.
Commit first
State your confidence along with your answer.
Predict first
Why does the polar form unify the three conics when the rectangular forms do not?
Correct: The pole is at a focus, and every conic has one.
Why: The rectangular forms centre each conic, and a parabola has no centre — so that convention could never include it. Every conic has a focus, which is why the focal convention supports a single equation.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate read the eccentricity straight off an equation whose denominator started with a 4. Explain the problem.
Hint: What does standard form require?
Answer:
The eccentricity is the trigonometric term's coefficient only when the denominator begins with a one. Otherwise the coefficient is scaled by that leading constant.
Dividing numerator and denominator by the 4 puts it into standard form without changing the curve at all, and only then is the coefficient meaningful.
The consequence is not small: the unstandardised reading is four times too large, which will usually name a hyperbola where the conic is an ellipse. A good explanation stresses that the check is a glance and the fix is a line.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where nearly all the errors occur and it can change the conic's identity. The fourth is where the chapter's geometry connects to something that had to be discovered rather than defined.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the polar equation with the three eccentricity cases beneath it, and note that the pole sits at a focus. Beside it, list the four forms with the directrix position each implies. Underneath, standardise one equation whose denominator does not begin with a one, and read off the conic, the eccentricity and the directrix.
If your note explains that the pole is at a focus because the parabola has no centre, the section's unifying idea is on the page rather than just its formula.
Recap
Five things, and the first is the one that makes the rest valid.
| if you remember one thing | it should be this |
|---|---|
| about the form | the denominator must begin with a one before anything is read |
| about classification | below one an ellipse, one a parabola, above one a hyperbola |
| about the pole | it sits at a focus, which is why one equation suffices |
| about orbits | an eccentricity of one separates returning from escaping |
Chapter 11 leaves geometry for sequences and series, where the patterns are lists of numbers rather than curves — and where the sums of infinitely many terms turn out to be finite surprisingly often.
OpenStax, Precalculus, §10.5 Conic Sections in Polar Coordinates §10.5, pp. 1266-1280 — everything on these slides traces back here
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