Ellipse
Question

# $\mathrm{\alpha },\mathrm{\beta }$ are the eccentric angles of the extremities of a focal chord of the ellipse ${\mathrm{x}}^{2}/16+{\mathrm{y}}^{2}/9=1$ then $\mathrm{tan}\left(\mathrm{\alpha }/2\right)\mathrm{tan}\left(\mathrm{\beta }/2\right)=$

Moderate
Solution

## The eccentricity $e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$Let  be a focalchord of the ellipse passing through the focus at $\left(\sqrt{7},0\right)$then,$\begin{array}{l}⇒\frac{3\mathrm{sin}\mathrm{\beta }}{4\mathrm{cos}\mathrm{\beta }-\sqrt{7}}=\frac{3\mathrm{sin}\mathrm{\alpha }}{4\mathrm{cos}\mathrm{\alpha }-\sqrt{7}}\\ ⇒\frac{\mathrm{sin}\left(\mathrm{\alpha }-\mathrm{\beta }\right)}{\mathrm{sin}\mathrm{\alpha }-\mathrm{sin}\mathrm{\beta }}=\frac{\sqrt{7}}{4}\\ ⇒\frac{\mathrm{cos}\left[\left(\mathrm{\alpha }-\mathrm{\beta }\right)/2\right]}{\mathrm{cos}\left[\left(\mathrm{\alpha }+\mathrm{\beta }\right)/2\right]}=\frac{\sqrt{7}}{4}\\ ⇒\mathrm{tan}\left(\frac{\mathrm{\alpha }}{2}\right)\mathrm{tan}\left(\frac{\mathrm{\beta }}{2}\right)=\frac{\sqrt{7}-4}{\sqrt{7}+4}=\frac{23-8\sqrt{7}}{-9}\end{array}$

Get Instant Solutions
When in doubt download our app. Now available Google Play Store- Doubts App