Question

If the lines joining the foci of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$, where $a>b$ and an extremity of its minor axis are inclined at an angle of $60°$, then the eccentricity of the ellipse is:

• A. $-\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{5}}{2}$
• D. $\frac{\sqrt{7}}{3}$
• E. $\sqrt{3}$

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option

Step 1: Evaluate the measure of unknown angles

It is given that $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$ is an ellipse

It is given that the coordinates of $P$ will be $(0,b)$ and $\angle SPS\text{'}=60°$.

For $△POS$ and $△POS\text{'}$,

$\to$$OP$ is the common side

$\to OS=OS\text{'}$ [equal distance]

$\to$$\angle POS=\angle POS\text{'}$ [right angle]

So, by $SAS$ property, $△POS\cong △POS\text{'}$

Thus, we can conclude that $\angle OPS=\angle OPS\text{'}$ which is also half of $\angle SPS\text{'}$

$$\begin{gathered} \Rightarrow \angle OPS=\frac{1}{2}\angle SPS\text{'} \\ \Rightarrow \angle OPS=30º \end{gathered}$$

Step 2: Find the eccentricity of the ellipse

According to the given information, $a>b$, so, $S(ae,0)$ and $S\text{'}(-ae,0)$.

In $△POS$,

$$\begin{gathered} \Rightarrow \tan 30°=\frac{OS}{OP}[\because \tan \mathrm{\theta }=\frac{\mathrm{Perpendicular}}{\mathrm{Base}}] \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{ae}{b} \\ \Rightarrow \frac{b}{a}=\sqrt{3}e \end{gathered}$$

We know that eccentricity, $e^2=1-\frac{b^2}{a^2}$, thus,

$$\begin{gathered} \Rightarrow e^2=1-3e^2\left[\because \frac{b}{a}=\sqrt{3}e\right] \\ \Rightarrow 4e^2=1 \\ \Rightarrow e=\frac{1}{2} \end{gathered}$$

Therefore, option (B) i.e.$\frac{1}{2}$ is the correct answer.