Let an ellipse


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 :

  1. $\frac{8}{3}$

  2. $\frac{4}{3}$

  3. $\frac{16}{3}$

  4. 3

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


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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