Let an ellipse E: x^ 2 a^ 2 + y^ 2 b^ 2 =1, a^ 2 >b^ 2 , passes t… - Vidyadip

MCQ

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:

A
$\frac{8}{3}$
B
$\frac{4}{3}$
C
$3$
✓
$\frac{16}{3}$

✓

Answer

Correct option: D.

$\frac{16}{3}$

d
$\frac{3}{2 a^{2}}+\frac{1}{b^{2}}=1 \text { and } 1-\frac{b^{2}}{a^{2}}=\frac{1}{3}$

$\Rightarrow a^{2}=3 b^{2}=2$

$\Rightarrow \frac{x^{2}}{3}+\frac{y^{2}}{2}=1.....(i)$

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

$y=\pm \frac{2}{\sqrt{3}}, x=1$

$\Rightarrow P Q^{2}=\left(\frac{4}{\sqrt{3}}\right)^{2}=\frac{16}{3}$

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