Question

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. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (A)

Consider, the horizontal ellipse
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where $a^{2}>\,b^{2}$

Since, $B B'S$ is an equilateral triangle.
$\therefore \, B S=B B'$
$\Rightarrow \,\sqrt{(a e-0)^{2}+(0-b)^{2}}=\sqrt{0+(2 b)^{2}}$
$\Rightarrow \, a^{2} e^{2}+b^{2}=4 b^{2}$
$\Rightarrow \,b^{2}=\frac{a^{2} e^{2}}{3}$
Now, $ e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{e^{2}}{3}}$
$\Rightarrow \, e^{2}=1-\frac{e^{2}}{3}$
$\Rightarrow \,\frac{4 e^{2}}{3}=1$
$\Rightarrow \, e=\frac{\sqrt{3}}{2}$