Question

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

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

Short answer

Showed\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - ecos\theta }}}}\).

Step-by-step solution

Definition.

Have a conic with foci\({\rm{F}}\) at the origin and eccentricity\(e\), as well as a directrix with the equation\({\rm{x = - d}}\), which indicates the directrix is parallel to the\(y\)-axis and perpendicular to the \(x\)-axis.

If our conic is an ellipse, \({\rm{e < 1}}\)our conic is parabolic if\({\rm{e = 1}}\). Our conic will be hyperbola if\({\rm{e > 1}}\).

Assume our conic is a parabola and\({\rm{P}}\) is the point on the parabola. Then you've got quirkiness by definition.

\(\frac{{{\rm{|PF|}}}}{{{\rm{|Pl|}}}}{\rm{ = e}}\)

Our directrix is\({\rm{l}}\).\(\left| {{\rm{P F}}} \right|{\rm{ = r}}\) since\({\rm{F}}\) is at the origin.

\(\left| {{\rm{P l}}} \right|{\rm{ = |F l| - |F A|}}\)

Because\({\rm{|F l|}}\)is the directrix's distance from the origin,

\({\rm{|F l| = d}}\)

To show\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - ecos\theta }}}}\).

When looking at the right triangle \({\rm{A F P}}\)we may conclude that the angle \({\rm{A F P}}\)equals\({\rm{\pi - \theta }}\). As a result of this and the definition of\({\rm{cos}}\),

\(\begin{aligned}{l}{\rm{cos(\pi - \theta ) = }}\frac{{{\rm{|FA|}}}}{{{\rm{|PF|}}}}\\{\rm{|FA| = |PF|cos(\pi - \theta )}}\\{\rm{|FA| = rcos(\pi - \theta )}}\\{\rm{|FA| = - rcos\theta }}\end{aligned}\)

Now

\(\begin{aligned}{l}{\rm{|Pl| = d - ( - rcos\theta )}}\\{\rm{ = d + rcos\theta }}\end{aligned}\)

To get,

\(\begin{aligned}{c}{\rm{e = }}\frac{{{\rm{|PF|}}}}{{{\rm{|Pl|}}}}\\{\rm{|PF| = e|Pl|}}\\{\rm{r = e(d + rcos\theta )}}\\{\rm{r = ed + recos\theta }}\\{\rm{r - recos\theta = ed}}\\{\rm{r(1 - ecos\theta ) = ed}}\\{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - ecos\theta }}}}\end{aligned}\)

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