The centre of a circle C is at the centre of the ellipse E:x^2a^2+y^2b^2=1,ab.

Question

The centre of a circle C is at the centre of the ellipse $E:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b$. Let C pass through the foci ${F}_{1}$ and ${F}_{2}$ of E such that the circle $C$ and the ellipse $E$ intersect at four points. Let P be one of these four points. If the area of the triangle ${PF}_{1}{F}_{2}$ is 30 and the length of the major axis of $E$ is 17 , then the distance between the foci of $E$ is :

Accepted Answer

$$\begin{aligned}
& x^2+\frac{a^2 y^2}{b^2}=a^2 \
& \Rightarrow y^2\left(1-\frac{a^2}{b^2}\right)=a^2\left(e^2-1\right)=a^2\left(1-\frac{b^2}{a^2}-1\right) \
& =-b^2 \
& \Rightarrow \frac{y^2\left(b^2-a^2\right)}{b^2}=-b^2 \Rightarrow y^2=\frac{b^4}{\left(a^2-b^2\right)} \
& \text { Height }=|y|=\frac{b^2}{\sqrt{a^2-b^2}}
\end{aligned}$$

$$\begin{aligned}
&\begin{aligned}
& \text { Area }=(2 a e) \times \frac{1}{2} \times \frac{b^2}{\sqrt{a^2-b^2}}=30 \
& =\frac{a b^2 e}{a \sqrt{1-\frac{b^2}{a^2}}}=b^2, a=\frac{17}{2}
\end{aligned}\
\&\text { Distance between foci }=2 a e\
\&=17 \sqrt{1-\frac{b^2}{a^2}}=17 \sqrt{1-\frac{30 \times 4}{289}}=13
\end{aligned}$$

Solutions · 1