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Q13E

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Found in: Page 535

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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# (a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic.$${\rm{r = }}\frac{{\rm{3}}}{{{\rm{2 + 2cos\theta }}}}$$.

A. Eccentricity$${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}$$.

B. The conic is Ellipse.

C. The directrix $${\rm{x = }}\frac{{\rm{9}}}{{\rm{2}}}$$.

See the step by step solution

## Step 1: Answer of the eccentricity.

Equation of conic in polar $${\rm{r = }}\frac{{\rm{9}}}{{{\rm{6 + 2cos\theta }}}}$$

Divide the numerator and denominator by 6 to convert the polar curve equation to standard form.

\begin{aligned}{l}{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 \pm ecos\theta }}}}\\{\rm{r = }}\frac{{\frac{{\rm{9}}}{{\rm{6}}}}}{{\frac{{\rm{6}}}{{\rm{6}}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{6}}}{\rm{cos\theta }}}}\\{\rm{ = }}\frac{{\frac{{\rm{3}}}{{\rm{2}}}}}{{{\rm{1 + }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{cos\theta }}}}\end{aligned}

Eccentricity$${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}$$.

## Step 2: Identify the conic.

The conic is,

When that the form is, $${\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + ecos\theta }}}}$$

Then the given equation, $${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}$$

Hence, $${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ < 1}}$$ so the conic is ellipse.

## Step 3: Equation of the directrix.

From the given equation, $${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}$$.

$${\rm{ed = }}\frac{{\rm{3}}}{{\rm{2}}}$$

That is $${\rm{e = }}\frac{{\rm{1}}}{{\rm{3}}}$$

Then the values,

\begin{aligned}{c}\frac{{\rm{1}}}{{\rm{3}}}{\rm{d = }}\frac{{\rm{3}}}{{\rm{2}}}\\{\rm{d = }}\frac{{\rm{9}}}{{\rm{2}}}\end{aligned}

We may deduce from the equation that the directrix is parallel to the $${\rm{y}}$$-axis and to the right of the origin, hence the directrix equation is,

Directrix is $${\rm{x = }}\frac{{\rm{9}}}{{\rm{2}}}$$

## Step 4: Graph of the conic.

Calculate given points,

 $${\rm{\theta }}$$ $${\rm{r}}$$ $${\rm{0}}$$ $${\rm{1}}{\rm{.125}}$$ $$\frac{\pi }{{\rm{2}}}$$ $${\rm{1}}{\rm{.5}}$$ $$\pi$$ $${\rm{2}}{\rm{.25}}$$ $$\frac{{{\rm{3}}\pi }}{{\rm{2}}}$$ $${\rm{1}}{\rm{.5}}$$

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