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Expert-verified Found in: Page 535 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # To determine,a) The eccentricity of the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.b) To identify the conic which is represented by the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.c) An equation of the directrix of the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.d) To sketch the graph of the conic represented by the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.

a) The eccentricity of the polar equation is $$\frac{{10}}{3}$$.

b) The polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ represents as hyperbola.

c) The polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ has directrix $$x = - \frac{6}{5}$$.

d) It is observed that the hyperbola is parallel to $$y$$-axis.

See the step by step solution

## Step 1: Given data

The conic is given by the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.

## Step 2: Theorem of Polar equation

A polar equation of the form $$r = \frac{{ed}}{{{l_{ \pm e}}\sin \theta }}$$ or $$r = \frac{{ed}}{{{1_{ \pm e}}\cos \theta }}$$ represents a conic section with eccentricity $$e$$.

The conic is an ellipse if $$e < 1$$, a parabola if $$e = 1$$, or a hyperbola if $$e > 1$$.

## Step 3: Calculation of the eccentricity of polar equation

a)

Divide the numerator and denominator of polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ by $$3$$.

$$r = \frac{4}{{1 - \frac{{10}}{3}\cos \theta }}$$

Clearly, the above polar equation is in the form of $$r = \frac{{ed}}{{1 - e\cos \theta }}$$ where $$e = \frac{{10}}{3}$$.

Hence, the eccentricity of the polar equation is $$\frac{{10}}{3}$$.

## Step 4: Identification of the conic

b)

From part $$(a)$$, the eccentricity $$e = \frac{{10}}{3}$$.

Clearly, $$e > 1$$.

By the result (1), the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ represents hyperbola.

Hence, the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ represents as hyperbola.

## Step 5: Calculation for the equation of directrix

c)

From part (a), the eccentricity $$e = \frac{{10}}{3}$$ and $$ed = 4$$.

\begin{aligned}{c}ed = 4\\d = \frac{4}{e}\\d = \frac{4}{{\frac{{10}}{3}}}\\d = \frac{6}{5}\end{aligned}

By the result, the polar equation has the directrix $$x = - \frac{6}{5}$$.

Hence, the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ has directrix $$x = - \frac{6}{5}$$.

## Step 6: Calculation to obtain $$x$$-intercept

d)

Obtain the $$x$$-intercepts of the conic.

Substitute $$\theta = 0$$ in $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.

\begin{aligned}{l}r = \frac{{12}}{{3 - 10\cos (0)}}\\r = \frac{{12}}{{3 - 10}}\\r = \frac{{12}}{{ - 7}}\end{aligned}

That is, $$\left( { - \frac{{12}}{7},0} \right)$$.

Substitute $$\theta = \pi$$ in $$r = \frac{{12}}{{3 - 10\cos \theta }}$$.

\begin{aligned}{l}r = \frac{{12}}{{3 - 10\cos (\pi )}}\\r = \frac{{12}}{{3 + 10}}\\r = \frac{{12}}{{13}}\end{aligned}

That is, $$\left( {\frac{{12}}{{13}},\pi } \right)$$.

## Step 7: Sketch the graph of the conic section

Sketch the graph of the conic section represented by the polar equation $$r = \frac{{12}}{{3 - 10\cos \theta }}$$ as shown in Figure below. From figure, it is observed that the hyperbola is parallel to $$y$$-axis. ### Want to see more solutions like these? 