Question

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

• A. $3$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{16}{3}$

eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{3}}$
$\Rightarrow 2a^2=3b^2\dots \left(1\right)$
$(\sqrt{\frac{3}{2}},1)$lies on $E:\frac{3}{2a^2}+\frac{1}{b^2}=1$
$\Rightarrow b^2=2$ and $a^2=3$ using equation $\left(1\right)$
$F(ae,0)=(1,0)$
Circle: $(x-1{)}^2+y^2=\frac{4}{3}\dots (2)$
Ellipse: $\frac{x^2}{3}+\frac{y^2}{2}=1\dots \left(3\right)$
Solving $\left(2\right)$ and $\left(3\right)$
$x=1,5$ (rejected)
At $x=1,y=\pm \frac{2}{\sqrt{3}},PQ=\frac{4}{\sqrt{3}}$
$\Rightarrow P{Q}^2=\frac{16}{3}$