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Precalculus : Polar Equations of Conic Sections

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Example Questions

Example Question #1 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{2}{1-2\sin(\theta)}$ $\displaystyle r=\frac{2}{1-2\sin(\theta)}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Hyperbola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

Now, for the given conic section, e>1 $\displaystyle e>1$ so it must be a hyperbola.

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Example Question #2 : Identify The Conic With A Given Polar Equation

Given the polar equation, determine the conic section:

$r=\frac{6}{1-\frac{1}{2}\sin(\theta)}$ $\displaystyle r=\frac{6}{1-\frac{1}{2}\sin(\theta)}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Ellipse

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

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

Report an Error

Example Question #1 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic sectioN:

$r=\frac{8}{4-\sin\theta}$ $\displaystyle r=\frac{8}{4-\sin\theta}$

Possible Answers:

Ellipse

Hyperbola

Parabola

Correct answer:

Ellipse

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 4$.

$r=\frac{8}{4-\sin\theta}=\frac{2}{1-\frac{1}{4}\sin\theta}$ $\displaystyle r=\frac{8}{4-\sin\theta}=\frac{2}{1-\frac{1}{4}\sin\theta}$

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

Report an Error

Example Question #2 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{6}{6-6\cos\theta}$ $\displaystyle r=\frac{6}{6-6\cos\theta}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Parabola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 6$.

$r=\frac{6}{6-6\cos\theta}=\frac{1}{1-\cos\theta}$ $\displaystyle r=\frac{6}{6-6\cos\theta}=\frac{1}{1-\cos\theta}$

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

Report an Error

Example Question #3 : Polar Equations Of Conic Sections

Given the polar equation, determine the conic section:

$r=\frac{12}{4-8\sin\theta}$ $\displaystyle r=\frac{12}{4-8\sin\theta}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Hyperbola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 4$.

$r=\frac{12}{4-8\sin\theta}=\frac{3}{1-2\sin\theta}$ $\displaystyle r=\frac{12}{4-8\sin\theta}=\frac{3}{1-2\sin\theta}$

Now, for the given conic section, e>1 $\displaystyle e>1$ so it must be a hyperbola.

Report an Error

Example Question #4 : Polar Equations Of Conic Sections

Given the polar equation, identify the conic section.

$r=\frac{2}{1+\sin(\theta+\frac{\pi}{4})}$ $\displaystyle r=\frac{2}{1+\sin(\theta+\frac{\pi}{4})}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Parabola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

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

Report an Error

Example Question #5 : Polar Equations Of Conic Sections

Given the polar equation, identify the conic section:

$r=\frac{6}{2+3\sin(\theta+\pi)}$ $\displaystyle r=\frac{6}{2+3\sin(\theta+\pi)}$

Possible Answers:

Hyperbola

Parabola

Ellipse

Correct answer:

Hyperbola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 2$.

$r=\frac{6}{2+3\sin(\theta+\pi)}=\frac{3}{1+\frac{3}{2}\sin(\theta+\pi)}$ $\displaystyle r=\frac{6}{2+3\sin(\theta+\pi)}=\frac{3}{1+\frac{3}{2}\sin(\theta+\pi)}$

Now, for the given conic section, e>1 $\displaystyle e>1$ so it must be a hyperbola.

Report an Error

Example Question #6 : Polar Equations Of Conic Sections

Given the polar equation, identify the conic section:

$r=\frac{10}{2+5\cos\theta}$ $\displaystyle r=\frac{10}{2+5\cos\theta}$

Possible Answers:

Ellipse

Parabola

Hyperbola

Correct answer:

Hyperbola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 2$.

$r=\frac{10}{2+5\cos\theta}=\frac{5}{1+\frac{5}{2}\cos\theta}$ $\displaystyle r=\frac{10}{2+5\cos\theta}=\frac{5}{1+\frac{5}{2}\cos\theta}$

Now, for the given conic section, e>1 $\displaystyle e>1$ so it must be a hyperbola.

Report an Error

Example Question #7 : Polar Equations Of Conic Sections

Given the polar equation, identify the conic section.

$r=\frac{36}{9+6\cos\theta}$ $\displaystyle r=\frac{36}{9+6\cos\theta}$

Possible Answers:

Ellipse

Hyperbola

Parabola

Correct answer:

Ellipse

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 9$.

$r=\frac{36}{9+6\cos\theta}=\frac{4}{1+\frac{2}{3}\cos\theta}$ $\displaystyle r=\frac{36}{9+6\cos\theta}=\frac{4}{1+\frac{2}{3}\cos\theta}$

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

Report an Error

Example Question #3 : Identify The Conic With A Given Polar Equation

Given the polar equation, identify the conic section.

$r=\frac{12}{6-6\cos\theta}$ $\displaystyle r=\frac{12}{6-6\cos\theta}$

Possible Answers:

Hyperbola

Ellipse

Parabola

Correct answer:

Parabola

Explanation:

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

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

$r=\frac{ed}{1\pm e \sin(\theta)}$ $\displaystyle r=\frac{ed}{1\pm e \sin(\theta)}$, where $\displaystyle e$ 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 $\displaystyle 0< e< 1$ will give an ellipse.

e=1 $\displaystyle e=1$ will give a parabola.

e>1 $\displaystyle e>1$ will give a hyperbola.

First, put the given polar equation into one of the forms seen above by dividing everything by $\displaystyle 6$.

$r=\frac{12}{6-6\cos\theta}=\frac{2}{1-\cos\theta}$ $\displaystyle r=\frac{12}{6-6\cos\theta}=\frac{2}{1-\cos\theta}$

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

Report an Error

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Precalculus : Polar Equations of Conic Sections

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Polar Equations Of Conic Sections

Example Question #2 : Identify The Conic With A Given Polar Equation

Example Question #1 : Polar Equations Of Conic Sections

Example Question #2 : Polar Equations Of Conic Sections

Example Question #3 : Polar Equations Of Conic Sections

Example Question #4 : Polar Equations Of Conic Sections

Example Question #5 : Polar Equations Of Conic Sections

Example Question #6 : Polar Equations Of Conic Sections

Example Question #7 : Polar Equations Of Conic Sections

Example Question #3 : Identify The Conic With A Given Polar Equation

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