Question

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

\({\rm{ Hyperbola, eccentricity 1}}{\rm{.5, directrix y = 2}}\)

Short answer

Polar equation of a conic with the focus at the origin and the Hyperbola, eccentricity 1.5, directrix\({\rm{y = 2}}\) and \({\rm{r = }}\frac{{\rm{6}}}{{{\rm{2 + 3sin\theta }}}}\).

Step-by-step solution

Hyperbola and directrix.

Hyperbola with eccentricity\({\rm{e = 1}}{\rm{.5}}\)

Directrix\({\rm{y = 2}}\)

\({\rm{d = 2}}\)

Hyperbola and ellipse.

Half of the hyperbola and or ellipse that was closest to the focus at the origin.

Because the focus is at the origin and the directrix is at y = 2, the portion of the hyperbola closest to the focus expands downwards, and the equation to apply is:

\(\begin{aligned}{l}{\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 + e sin\theta }}}}\\{\rm{r = }}\frac{{{\rm{(1}}{\rm{.5)(2)}}}}{{{\rm{1 + (1}}{\rm{.5) sin\theta }}}}\\{\rm{r = }}\frac{{\rm{3}}}{{{\rm{1 + 1}}{\rm{.5 sin\theta }}}}\end{aligned}\)

Polar equation of a conic with the focus at the origin is.

\({\rm{r = }}\frac{{\rm{6}}}{{{\rm{2 + 3sin\theta }}}}\)