Q. In an ellipse, if the lines joining focus to the extremities of the minor axis form an equilateral triangle with the minor axis, then the eccentricity of the ellipse is
A) $\frac{\sqrt{3}}{2}$
B) $\frac{\sqrt{3}}{4}$
C) $\frac{1}{\sqrt{2}}$
D) $\sqrt{\frac{2}{3}}$
Explanation
(a)
Since, $BB''S$ is an equilateral triangle
$\therefore BS = BB''$
[image:11_image34.emf]
$\Rightarrow \sqrt{(ae - 0)^{2} + (0 - b^{2}}) = \sqrt{0 + (2b)^{2}}$
$\Rightarrow \ \ \ a^{2}e^{2} + b^{2} = 4b^{2} \Rightarrow \ b^{2} = \frac{a^{2}e^{2}}{3}$
$\because\ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{e^{2}}{3}}$
$\Rightarrow \ \frac{4e^{2}}{3} = 1 \Rightarrow e = \frac{\sqrt{3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.