Let an ellipse $\mathrm{E}:\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1, {\mathrm{a}}^2>{\mathrm{b}}^2,$ passes through $\left(\sqrt{\frac{3}{2}},1\right)$ and has eccentricity $\frac{1}{\sqrt{3}}$ If a circle, centered at focus $\mathrm{F}(\mathrm{\alpha }, 0), \mathrm{\alpha } > 0,$ of E and radius $\frac{2}{\sqrt{3}}$ intersects E at two points P and Q, then PQ² is equal to

$$\begin{gathered} \mathrm{e}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right. \\ {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2)={\mathrm{a}}^2\left(1-\frac{1}{3}\right) \\ {\mathrm{b}}^2=\frac{2{\mathrm{a}}^2}{3} \end{gathered}$$

equation  of ellipse passes through $\left(\sqrt{\frac{3}{2}},1\right) is$

$$\begin{gathered} \Rightarrow \frac{3}{2{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{3}{2{\mathrm{a}}^2}+\frac{3}{2{\mathrm{a}}^2}=1 \\ \Rightarrow 2{\mathrm{a}}^2=6 \Rightarrow {\mathrm{a}}^2=3 \\ \Rightarrow {\mathrm{b}}^2=2 \\ \therefore \mathrm{eq} \mathrm{of} \mathrm{an} \mathrm{ellipse} is \frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{2}=1 \\ \mathrm{centre} \mathrm{of} \mathrm{circle}=\mathrm{Focus} \mathrm{of} \mathrm{ellipse}=(1, 0)=(\mathrm{\alpha }, 0) \\ \mathrm{radius}=\frac{2}{\sqrt{3}}=\mathrm{r} \end{gathered}$$

$$\begin{gathered} {Circle is (\mathrm{x}-1)}^2+{\mathrm{y}}^2=\frac{4}{3} \\ \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{2}=1 \mathrm{and} {(\mathrm{x}-1)}^2+{\mathrm{y}}^2=\frac{4}{3} \end{gathered}$$$$\begin{gathered} \frac{{\mathrm{x}}^2}{3}+\frac{{\displaystyle \frac{4}{3}}-{(\mathrm{x}-1)}^2}{2}=1 \\ 2{\mathrm{x}}^2+4-3{(\mathrm{x}-1)}^2=6 2{\mathrm{x}}^2-3{\mathrm{x}}^2-3+6\mathrm{x}=2 \\ {\mathrm{x}}^2-6\mathrm{x}+5=0 \\ (\mathrm{x}-1) (\mathrm{x}-5)=0 \\ \mathrm{x}=1 or 5 \\ {\mathrm{y}}^2=\frac{4}{3}{(\mathrm{x}-1)}^2 \\ If \mathrm{x}=1 then {\mathrm{y}}^2=\frac{4}{3}\Rightarrow \mathrm{y}=\pm \frac{2}{\sqrt{3}} \\ \mathrm{P}\left(1, \frac{2}{\sqrt{3}}\right) \mathrm{Q}\left(1, -\frac{2}{\sqrt{3}}\right) \\ \mathrm{PQ}=\frac{4}{\sqrt{3}} \\ \Rightarrow {\mathrm{PQ}}^2=\frac{16}{3} \end{gathered}$$