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)

$\frac{3}{2 a^{2}}+\frac{1}{b^{2}}=1$
and $1-\frac{b^{2}}{a^{2}}=\frac{1}{3}$
$\Rightarrow a^{2}=3 b^{2}=3$
$\Rightarrow \frac{x^{2}}{3}+\frac{y^{2}}{2}=1$ ...(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
$y=\pm \frac{2}{\sqrt{3}}, x=1$
$\Rightarrow P Q^{2}=\left(\frac{4}{\sqrt{3}}\right)^{2}=\frac{16}{3}$