Let an ellipse E:x^2a^2+y^2b^2=1, a^2b^2, pass through (32,1) and have eccentricity 1√3.

Question

Let an ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a^2>b^2$, pass through $\left(\dfrac32,1\right)$ and have eccentricity $\dfrac1{\sqrt3}$. If a circle, centred at focus $F(\alpha,0)$, $\alpha>0$, of $E$ and radius $\dfrac2{\sqrt3}$, intersects $E$ at two points $P$ and $Q$, then $PQ^2$ is equal to

Accepted Answer

$e^2=\dfrac13=1-\dfrac{b^2}{a^2}\Rightarrow a^2=\dfrac32b^2$. Through $\left(\dfrac32,1\right)$: $\dfrac{9/4}{a^2}+\dfrac1{b^2}=1$; solving simultaneously gives $b^2=2,\ a^2=3$.

$$E:\frac{x^2}3+\frac{y^2}2=1,\qquad \text{focus }F=(1,0).$$

Circle: $(x-1)^2+y^2=\dfrac43$. Solving with the ellipse: $x=1$, $y=\pm\dfrac2{\sqrt3}$.

$$PQ=\frac4{\sqrt3}\ \Longrightarrow\ PQ^2=\frac{16}3.$$

Solutions · 1