Question

Write a polar equation of a conic with the focus at the origin and the given data.

\({\rm{Hyperbola, eccentricity 3, directrix}}\)\({\rm{r = - 6csc\theta }}\)

Short answer

The polar equation is \({\rm{ = }}\frac{{{\rm{18}}}}{{{\rm{1 - 3sin\theta }}}}\).

Step-by-step solution

Hyperbola and directrix.

Hyperbola e=3

Directrix\({\rm{r = - 6csc\theta }}\)

\(\begin{aligned}{l}{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 \pm ecos\theta }}}}{\rm{,}}\;\;\\{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 \pm esin\theta }}}}\end{aligned}\)

\(\begin{aligned}{c}{\rm{r = - 6csc\theta }}\\ \Rightarrow {\rm{r}}\\{\rm{ = - }}\frac{{\rm{6}}}{{{\rm{sin\theta }}}}\\ \Rightarrow {\rm{rsin\theta }}\\{\rm{ = - 6}}\end{aligned}\)

Directrix and polar equation.

Hence,

\({\rm{y = rsin\theta }}\)

\({\rm{y = - 6}}\)

Directrix is parallel to the polar axis and it is below the pole. And d= 6.

Polar equation formula.

\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - esin\theta }}}}\)

Then,

\(\begin{aligned}{c}{\rm{r = }}\frac{{{\rm{3}} \cdot {\rm{6}}}}{{{\rm{1 - 3sin\theta }}}}\\{\rm{ = }}\frac{{{\rm{18}}}}{{{\rm{1 - 3sin\theta }}}}\end{aligned}\)

Then the polar equation is \({\rm{ = }}\frac{{{\rm{18}}}}{{{\rm{1 - 3sin\theta }}}}\)