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Q10E

Expert-verifiedFound in: Page 535

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.

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

**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 **\(e > 1\)**.**

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 \(\frac{{10}}{3}\).

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.

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 \(x = - \frac{6}{5}\).

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

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.

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