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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 ecentricity $\frac{1}{\sqrt{3}} .$ If a circle, centered at focus $\mathrm{F}(\alpha, 0), \alpha>0$, of $\mathrm{E}$ and radius $\frac{2}{\sqrt{3}}$, intersects $\mathrm{E}$ at two points $\mathrm{P}$ and $\mathrm{Q}$, then $\mathrm{PQ}^{2}$ is equal to:
$\frac{8}{3}$
$\frac{4}{3}$
$3$
$\frac{16}{3}$
Solution
$\frac{3}{2 a^{2}}+\frac{1}{b^{2}}=1 \text { and } 1-\frac{b^{2}}{a^{2}}=\frac{1}{3}$
$\Rightarrow a^{2}=3 b^{2}=2$
$\Rightarrow \frac{x^{2}}{3}+\frac{y^{2}}{2}=1…..(i)$
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
$y=\pm \frac{2}{\sqrt{3}}, x=1$
$\Rightarrow P Q^{2}=\left(\frac{4}{\sqrt{3}}\right)^{2}=\frac{16}{3}$
