Question

If latus rectum of ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1(0<b<4)$ subtends angle $2\theta$ at farthest vertex s.t $\text{cosec}\theta =5$ then $e=?$

Answer 1

Distance between latus rectum of ellipse and the farthest point along x-axis $=(ae+a)=(4e+4)$ units.

$\therefore$ $\sin \theta =\frac{1}{5}$

$\frac{\text{half of length of latus rectum}}{4e+4}=\frac{1}{5}$

$\Rightarrow$ Half of length of latus rectum $=\frac{4e+4}{5}$

$\therefore$ $\frac{b^2}{a^2}$ $=\frac{4(e+1)}{5}$

$\Rightarrow e=\frac{4}{5}$

$\therefore$ eccentricity $e=\frac{4}{5}$.