Use the following results the polar equation of the hyperbolla $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ is $r ^ { 2 } = \frac { - b ^ { 2 } } { 1 - e ^ { 2 } \cos ^ { 2 } \theta }$ to write the polar form of the equation of the conic $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$ .
A) $r ^ { 2 } = \frac { 144 } { 16 \cos ^ { 2 } \theta + 25 }$
B) $r ^ { 2 } = \frac { 144 } { 25 \cos ^ { 2 } \theta - 16 }$
C) $r ^ { 2 } = \frac { 144 } { 25 \cos ^ { 2 } \theta + 16 }$
D) $r ^ { 2 } = \frac { - 144 } { 25 \cos ^ { 2 } \theta + 16 }$
E) $r ^ { 2 } = \frac { 144 } { 16 - 25 \cos ^ { 2 } \theta }$

Question

Use the following results the polar equation of the hyperbolla $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ is $r ^ { 2 } = \frac { - b ^ { 2 } } { 1 - e ^ { 2 } \cos ^ { 2 } \theta }$ to write the polar form of the equation of the conic $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$ . 
A) $r ^ { 2 } = \frac { 144 } { 16 \cos ^ { 2 } \theta + 25 }$
B) $r ^ { 2 } = \frac { 144 } { 25 \cos ^ { 2 } \theta - 16 }$
C) $r ^ { 2 } = \frac { 144 } { 25 \cos ^ { 2 } \theta + 16 }$
D) $r ^ { 2 } = \frac { - 144 } { 25 \cos ^ { 2 } \theta + 16 }$
E) $r ^ { 2 } = \frac { 144 } { 16 - 25 \cos ^ { 2 } \theta }$

Quizplus · Accepted Answer

The given hyperbola equation is $\frac{x^2}{16} - \frac{y^2}{9} = 1$, where $a^2 = 16$ and $b^2 = 9$. The eccentricity $e$ is calculated as $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. Substituting into the polar form $r^2 = \frac{-b^2}{1 - e^2 \cos^2 \theta}$, we get $r^2 = \frac{-9}{1 - \left(\frac{5}{4}\right)^2 \cos^2 \theta} = \frac{-9}{1 - \frac{25}{16} \cos^2 \theta} = \frac{144}{25 \cos^2 \theta - 16}$.

Use the Following Results the Polar Equation of the Hyperbolla x2a2−y2b2=1\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1a2x2​−b2y2​=1

Use the Following Results the Polar Equation of the Hyperbolla $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$