Q. If for the ellipse $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{b^{2}} = 1$, $y$-axis is the minor axis and the length of the latusrectum is one half of the length of its minor axis, then its eccentricity is
A) $\frac{1}{\sqrt{2}}$
B) $\frac{1}{2}$
C) $\frac{\sqrt{3}}{2}$
D) $\frac{3}{4}$
Explanation
(c)
We know that length of latusrectum of an ellipse$= \frac{2b^{2}}{a}$ and length of its minor axis $= \ 2b$
Then, $\frac{2b^{2}}{a} = b\ \ \Rightarrow \ \ \ 2b = a\$
$\therefore\ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{b^{2}}{4b^{2}} =}\frac{\sqrt{3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.