Use the following results the polar equation of the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ is $r ^ { 2 } = \frac { b ^ { 2 } } { 1 - e ^ { 2 } \cos ^ { 2 } \theta }$ to write the polar form of the equation of the conic $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$ .
A) $r ^ { 2 } = \frac { 25 } { 25 - 9 \cos ^ { 2 } \theta }$
B) $r ^ { 2 } = \frac { 400 } { 25 - 9 \cos ^ { 2 } \theta }$
C) $r ^ { 2 } = \frac { 400 } { 25 + 9 \cos ^ { 2 } \theta }$
D) $r ^ { 2 } = \frac { 400 } { 9 + 25 \cos ^ { 2 } \theta }$
E) $r ^ { 2 } = \frac { 400 } { 9 - 25 \cos ^ { 2 } \theta }$

Question

Use the following results the polar equation of the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ is $r ^ { 2 } = \frac { b ^ { 2 } } { 1 - e ^ { 2 } \cos ^ { 2 } \theta }$ to write the polar form of the equation of the conic $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$ . ​
A) $r ^ { 2 } = \frac { 25 } { 25 - 9 \cos ^ { 2 } \theta }$
B) $r ^ { 2 } = \frac { 400 } { 25 - 9 \cos ^ { 2 } \theta }$
C) $r ^ { 2 } = \frac { 400 } { 25 + 9 \cos ^ { 2 } \theta }$
D) $r ^ { 2 } = \frac { 400 } { 9 + 25 \cos ^ { 2 } \theta }$
E) $r ^ { 2 } = \frac { 400 } { 9 - 25 \cos ^ { 2 } \theta }$

Quizplus · Accepted Answer

Given the ellipse equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$, we identify $a^2 = 25$ and $b^2 = 16$. The eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{25} = \frac{9}{25}$. Substituting $a^2$, $b^2$, and $e^2$ into the polar equation $r^2 = \frac{b^2}{1 - e^2\cos^2\theta}$ yields $r^2 = \frac{16}{1 - \frac{9}{25}\cos^2\theta} = \frac{400}{25 - 9\cos^2\theta}$.

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