Question

A circle has the same centre as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse such that the two curves intersect in $4$ points. Let '$P$' be any one of their point of intersection. If the major axis of the ellipse is $17$ and the area of the triangle $P{F}_1{F}_2$ is $30$, then the distance between the foci is

• A. $13$
• B. $10$
• C. $11$
• D. $12$

Answer 1

The correct option is A $13$

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and
The circle be $x^2+y^2=a^2{e}^2$
Radius of circle $=ae$
The point of intersection of the circle and the ellipse is
$(\frac{a}{e}\sqrt{2e^2-1},\frac{a}{e}(1-e^2\left)\right)$
Now, area of
$\Delta P{F}_1{F}_2=\frac{1}{2}\left|\begin{array}{ccc}\frac{a}{e}\sqrt{2e^2-1}& \frac{a}{e}(1-e^2)& 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|\Rightarrow 30=\frac{1}{2}\left|\begin{array}{ccc}\frac{a}{e}\sqrt{2e^2-1}& \frac{a}{e}(1-e^2)& 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|$
$\Rightarrow \left|\frac{a}{e}\right(1-e^2\left)\right(2ae\left)\right|=60$
$\Rightarrow a^2(1-e^2)=30$
$\Rightarrow a^2{e}^2=a^2-30\Rightarrow (ae{)}^2={\left(\frac{17}{2}\right)}^2-30\Rightarrow ae=\sqrt{\frac{169}{4}}=\frac{13}{2}$
$\Rightarrow 2ae=13$