Question

In an ellipse, if the lines joining focus to the extremities of the minor axis from 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. $\frac{\sqrt{2}}{3}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Finding the eccentricity of the ellipse:

In the given ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2>b^2$.

Since, $BB\text{'}S$ is an equilateral triangle, so $BS=BB\text{'}$.

$\begin{array}{rcl}\sqrt{{\left(ae-0\right)}^2+{\left(0-b\right)}^2}& =& \sqrt{{\left(0-0\right)}^2+{\left(b+b\right)}^2}\\ \Rightarrow \sqrt{{\left(ae\right)}^2+{\left(-b\right)}^2}& =& \sqrt{{\left(2b\right)}^2}\end{array}$

By squaring both sides we get,

$\begin{array}{rcl}{a}^2{e}^2+b^2& =& 4b^2\\ \Rightarrow a^2{e}^2& =& 3b^2\\ \Rightarrow b^2& =& \frac{a^2{e}^2}{3}\end{array}$

Now,

$\begin{array}{rcl}e& =& \sqrt{1-\frac{b^2}{a^2}}\\ & =& \sqrt{1-\frac{e^2}{3}}\end{array}$

By squaring both sides we get

$\begin{array}{rcl}{e}^2& =& 1-\frac{e^2}{3}\\ & \Rightarrow & \frac{4e^2}{3}=1\\ & \Rightarrow & \frac{2e}{\sqrt{3}}=1\\ \Rightarrow e& =& \frac{\sqrt{3}}{2}\end{array}$

Hence, option A is correct.