Let an ellipse E: frac{x^{2}}{a^{2}}+frac{y^{ | vedclass

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(D) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$,we have $\frac{3}{2a^{2}} + \

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$\frac{16}{3}$

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(D) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through $\left(\sqrt{\frac{3}{2}}, 1ight)$,we have $\frac{3}{2a^{2}} + \frac{1}{b^{2}} = 1$. Given eccentricity $e = \frac{1}{\sqrt{3}}$,so $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{3}$,which implies $\frac{b^{2}}{a^{2}} = \frac{2}{3}$ or $a^{2} = \frac{3}{2}b^{2}$. Substituting $a^{2}$ in the first equation: $\frac{3}{2(\frac{3}{2}b^{2})} + \frac{1}{b^{2}} = 1$ $\Rightarrow \frac{1}{b^{2}} + \frac{1}{b^{2}} = 1$ $\Rightarrow b^{2} = 2$. Then $a^{2} = \frac{3}{2}(2) = 3$. The ellipse equation is $\frac{x^{2}}{3} + \frac{y^{2}}{2} = 1$. The focus $F(\alpha, 0)$ is given by $\alpha = ae = \sqrt{3} 	imes \frac{1}{\sqrt{3}} = 1$. So $F = (1, 0)$. The circle equation with center $(1, 0)$ and radius $\frac{2}{\sqrt{3}}$ is $(x-1)^{2} + y^{2} = \frac{4}{3}$. From the ellipse equation,$y^{2} = 2(1 - \frac{x^{2}}{3})$. Substitute $y^{2}$ into the circle equation: $(x-1)^{2} + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow x^{2} - 2x + 1 + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow \frac{x^{2}}{3} - 2x + \frac{5}{3} = 0$ $\Rightarrow x^{2} - 6x + 5 = 0$. Solving $(x-5)(x-1) = 0$,we get $x=1$ or $x=5$. Since the circle radius is $\frac{2}{\sqrt{3}} \approx 1.15$,$x=5$ is outside the ellipse. For $x=1$,$y^{2} = 2(1 - \frac{1}{3}) = 2(\frac{2}{3}) = \frac{4}{3}$,so $y = \pm \frac{2}{\sqrt{3}}$. The points are $P(1, \frac{2}{\sqrt{3}})$ and $Q(1, -\frac{2}{\sqrt{3}})$. $PQ^{2} = (1-1)^{2} + (\frac{2}{\sqrt{3}} - (-\frac{2}{\sqrt{3}}))^{2} = (\frac{4}{\sqrt{3}})^{2} = \frac{16}{3}$.

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