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Q13E

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Essential Calculus: Early Transcendentals
Found in: Page 535
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

(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 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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