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 $\mathrm{F}(\alpha, 0), \alpha>0$, 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 :
Correct Option: , 3
$\frac{3}{2 a^{2}}+\frac{1}{b^{2}}=1$ 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
$y=\pm \frac{2}{\sqrt{3}}, x=1$
$\Rightarrow \mathrm{PQ}^{2}=\left(\frac{4}{\sqrt{3}}\right)^{2}=\frac{16}{3}$
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