Question

Let an ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a^2>b^2$, passes through $\left(\sqrt{\frac{3}{2}},1\right)$ and has eccentricity $\frac{1}{\sqrt{3}}$. If a circle, centered at focus $F(\alpha ,0),\alpha >0$, of $E$ and radius $\frac{2}{\sqrt{3}}$, intersects $E$ at two points $P$ and $Q$, then $P{Q}^2$ is equal to :

• A. $\frac{8}{3}$
• B. $\frac{4}{3}$
• C. $\frac{16}{3}$
• D. $3$

Answer 1

Answer: (C)

$\frac{3}{2a^2}+\frac{1}{b^2}=1$
and $1-\frac{b^2}{a^2}=\frac{1}{3}$
$\Rightarrow a^2=3b^2=3$
$\Rightarrow \frac{x^2}{3}+\frac{y^2}{2}=1$ ...(i)
Its focus is $(1,0)$
Now, eqn of circle is
$(x-1{)}^2+y^2=\frac{4}{3}$ ...(ii)
Solving (i) and (ii) we get
$y=\pm \frac{2}{\sqrt{3}},x=1$
$\Rightarrow P{Q}^2={\left(\frac{4}{\sqrt{3}}\right)}^2=\frac{16}{3}$