Question

Show that a conic with a focus at the origin, eccentricity\({\rm{e}}\), and directrix\({\rm{y = d}}\) has a polar equation.

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

Short answer

Showed\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + esin\theta }}}}\).

Step-by-step solution

Sketch the graph.

Make a calculation for the given circumstances.

To get,

\(\begin{aligned}{c}\left| {{\rm{PF}}} \right|{\rm{ = r}}\\\left| {{\rm{Pl}}} \right|{\rm{ = d - rsin\theta }}\end{aligned}\)

To gain an idea of what eccentricity is, look at the definition.

\(\begin{aligned}{c}\frac{{\left| {{\rm{PF}}} \right|}}{{\left| {{\rm{Pl}}} \right|}}{\rm{ = e}}\\\left| {{\rm{PF}}} \right|{\rm{ = e}}\left| {{\rm{Pl}}} \right|\end{aligned}\)

To show\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + esin\theta }}}}\).

After then, it becomes

\(\begin{aligned}{c}{\rm{r = e(d - rsin\theta )}}\\{\rm{r = ed - resin\theta }}\\{\rm{r + resin\theta = ed}}\\{\rm{r(1 + esin\theta ) = ed}}\\{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + esin\theta }}}}\end{aligned}\)

As a result,\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + esin\theta }}}}\).