Let an ellipse E: x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1, a ^ 2 > b ^ 2 ,...

The ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ passes through the point $\left(\sqrt{\dfrac{3}{2}}, 1\right)$

So, $\dfrac{3}{2{a}^{2}} + \dfrac{1}{{b}^{2}} = 1 . . . \left(1\right)$ and $1 - \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{1}{3} . . .$\left(2\right)$</p><p>On solving equations \left(1\right)$ and \left(2\right)$, we get</p><p>$⇒ {a}^{2} = 3 \text{ and } {b}^{2} = 2$</p><p>$⇒ \dfrac{{x}^{2}}{3} + \dfrac{{y}^{2}}{2} = 1 . . . \left(3\right)$</p><p>We know that, focii of the ellipse$\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$is \left(\pm ae, 0\right)$

Here, $a = \sqrt{3} \text{ and } e = \dfrac{1}{\sqrt{3}}$

$∴$ Its focus is \left(1, 0\right)\left(∵ \alpha > 0\right)$</p><p>Now, equation of circle is</p><p>$( x - 1{)}^{2} + {y}^{2} = \dfrac{4}{3} . . . \left(4\right)$</p><p>Solving \left(3\right)$ and $\left(4\right)$ we get

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

$⇒ P{Q}^{2} = \sqrt{{\left(1 - 1\right)}^{2} + {\left(\dfrac{2}{\sqrt{3}} + \dfrac{2}{\sqrt{3}}\right)}^{2}}$

$= {\left(\dfrac{4}{\sqrt{3}}\right)}^{2} = \dfrac{16}{3}$