Question

Let $S$ and $S^{\prime }$ be foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S^{\prime }BS$ is a right angled triangle with right angle at $B$ and area of $△S^{\prime }BS=8\text{sq. units}$, then the length of a latus rectum of the ellipse (in units) is :

• A. $2$
• B. $2\sqrt{2}$
• C. $4\sqrt{2}$
• D. $4$

Answer 1

The correct option is D $4$

Let equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$
where $S\equiv (ae,0),S^{\prime }\equiv (-ae,0)$ and $B\equiv (0,b)$

$\because \Delta S^{\prime }BS$ is a right angled triangle with right angle at $B$
$\therefore (S^{\prime }B{)}^2+(SB{)}^2=(S{S}^{\prime }{)}^2\Rightarrow a^2{e}^2+b^2+a^2{e}^2+b^2=4a^2{e}^2\Rightarrow b^2=a^2{e}^2\Rightarrow e^2=\frac{b^2}{a^2}\Rightarrow 1-\frac{b^2}{a^2}=\frac{b^2}{a^2}\Rightarrow a^2=2b^2\Rightarrow a=\sqrt{2}b$

Given, area of $△\left(S^{\prime }BS\right)=8\text{sq. units}$
$\Rightarrow \frac{1}{2}\times 2ae\times b=8\Rightarrow \sqrt{2}b\times \frac{b}{\sqrt{2}b}\times b=8\Rightarrow b^2=8\therefore b=2\sqrt{2}\Rightarrow a=4$
​​​​​​​
So, length of latus rectum
​​​​​​​$=\frac{2b^2}{a}=\frac{2\left(8\right)}{4}=4\text{units}$