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Precalculus : Identify the Conic With a Given Polar Equation

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Example Question #21 : Conic Sections

Given the polar equation, identify the conic section.

$r=\frac{14}{7-7\cos\theta}$

Possible Answers:

Parabola

Hyperbola

Ellipse

Correct answer:

Parabola

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

$r=\frac{14}{7-7\cos\theta}=\frac{2}{1-\cos\theta}$

Now, for the given conic section, e=1 so it must be a parabola.

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Example Question #22 : Conic Sections

Given the polar equation, identify the conic section.

$r=\frac{16}{8+4\cos\theta}$

Possible Answers:

Hyperbola

Ellipse

Parabola

Correct answer:

Ellipse

Explanation:

Recall that the polar equations of conic sections can come in the following forms:

$r=\frac{ed}{1\pm e \cos(\theta)}$

$r=\frac{ed}{1\pm e \sin(\theta)}$ , where is the eccentricity of the conic section.

To determine what conic section the polar graph depicts, look only at the conic section's eccentricity.

0<e<1 will give an ellipse.

e=1 will give a parabola.

e>1 will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by .

$r=\frac{16}{8+4\cos\theta}=\frac{2}{1+\frac{1}{2}\cos\theta}$

Now, for the given conic section, 0<e<1 so it must be an ellipse.

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Example Question #31 : Conic Sections

Which type of conic equation would have the polar equation $r = \frac{1}{ 3 + \sqrt8 \cos \theta}$ ?

Possible Answers:

Ellipse

Not a conic section

Circle

Parabola

Hyperbola

Correct answer:

Ellipse

Explanation:

This would be an ellipse.

The polar form of any conic is $r = \frac{ep }{ 1 \pm e \sin \theta}$ [or cosine], where e is the eccentricity. If the eccentricity is between 0 and 1, then the conic is an ellipse, if it is 1 then it is a parabola, and if it is greater than 1 then it is a hyperbola. Circles have eccentricity 0.

To figure out what the eccentricity is, we need to get our equation so that the denominator is in the form $1 + e \sin \theta$ . Right now it is $3 + a \sin \theta$ , so multiply top and bottom by $\frac{1}{3}$ :

$\frac{1}{3 + \sqrt 8 \cos \theta} * \frac{\frac{1}{3}}{\frac{1}{3}} = \frac{\frac{1}{3}}{1 + \frac{\sqrt8 }{3}\cos \theta }$ .

Now we can identify our eccentricity as $\frac {\sqrt 8 }{3} \approx 0.942$ which is between 0 and 1.

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Example Question #11 : Identify The Conic With A Given Polar Equation

Which type of conic section is the polar equation $r = \frac{6}{2 -2 \sin \theta}$ ?

Possible Answers:

Ellipse

Parabola

Circle

Hyperbola

Not a conic section

Correct answer:

Parabola

Explanation:

All polar forms of conic equations are in the form $r = \frac{ep}{ 1 \pm e \sin \theta}$ [or cosine] where e is the eccentricity.

If the eccentricity is between 0 and 1 the conic is an ellipse, if it is 1 then it is a parabola, if it is greater than 1 it is a hyperbola. Circles have an eccentricity of 0.

We want the denominator to be in the form of $1 - e \sin \theta$ , so we can multiply top and bottom by one half:

$\frac{6}{2 -2 \sin \theta} * \frac{\frac{1}{2}}{\frac{1}{2}} = \frac{3}{1 - 1 \sin \theta}$

The eccentricity is 1, so this is a parabola.

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Example Question #15 : Identify The Conic With A Given Polar Equation

Which type of conic section is the polar equation $r = \sin \theta$ ?

Possible Answers:

Circle

Parabola

Hyperbola

Ellipse

None of these

Correct answer:

Circle

Explanation:

Although it's not immediately obvious, this is a circle. One way we can see this is by converting from polar form to cartesian:

$r = \sin \theta$ multiply both sides by r

$r^2 = r \sin \theta$ we can now replace r^2 with x^2 + y^2 and $r \sin \theta$ with :

x^2 + y^2 = y We can already mostly tell this is a circle, but just to be safe we can put it all the way into standard form:

$x^2 + y^2 - y \enspace \enspace \enspace = 0$ complete the square by adding $(-\frac{1}{2})^2 = \frac{1}{4}$ to both sides

$x^2 + y^2 - y + \frac{1}{4} = \frac{1}{4}$ condense the left side

$x^2 + (y - \frac{1}{2}) ^ 2 = \frac{1}{4}$

Now this is clearly a circle.

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Precalculus : Identify the Conic With a Given Polar Equation

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Conic Sections

Example Question #22 : Conic Sections

Example Question #31 : Conic Sections

Example Question #11 : Identify The Conic With A Given Polar Equation

Example Question #15 : Identify The Conic With A Given Polar Equation

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