Find a polar equation of the conic with the given characteristics and with one focus at the pole.
$\begin{array}{cl}
\text { Conic } & \text { Vertices } \
\text { Hyperbola } & \left(\frac{2}{3}, \frac{\pi}{2}\right),\left(-2, \frac{3 \pi}{2}\right)
\end{array}$

A) $r = \frac { 2 } { 1 - 2 \sin \theta }$
B) $r = \frac { 2 } { 1 + 2 \sin \theta }$
C) $r = \frac { 2 } { 1 + 2 \cos \theta }$
D) $r = \frac { 2 } { 1 - 2 \cos \theta }$
E) $r = \frac { 2 } { 1 + 1 \cos \theta }$

Question

Find a polar equation of the conic with the given characteristics and with one focus at the pole.
$\begin{array}{cl}
\text { Conic } & \text { Vertices } \
\text { Hyperbola } & \left(\frac{2}{3}, \frac{\pi}{2}\right),\left(-2, \frac{3 \pi}{2}\right)
\end{array}$

A) $r = \frac { 2 } { 1 - 2 \sin \theta }$
B) $r = \frac { 2 } { 1 + 2 \sin \theta }$
C) $r = \frac { 2 } { 1 + 2 \cos \theta }$
D) $r = \frac { 2 } { 1 - 2 \cos \theta }$
E) $r = \frac { 2 } { 1 + 1 \cos \theta }$

Quizplus · Accepted Answer

The vertices given are on the y-axis, indicating the transverse axis is aligned with the y-axis, and the hyperbola opens up and down. The correct equation for a hyperbola with a vertical transverse axis and one focus at the pole is $r = \frac{ed}{1 \pm e\sin\theta}$, where $e$ is the eccentricity and $d$ is the distance between the center and a vertex. The given vertices suggest $d = 2$, and since it's a hyperbola, $e > 1$. The negative vertex indicates the hyperbola opens downwards, which corresponds to the $+$ sign in the denominator. Thus, the correct equation is $r = \frac{2}{1 + 2\sin\theta}$.

Find a Polar Equation of the Conic with the Given