Question

A circle and an ellipse have centres at $(0,0)$ and the circle passes through foci $F_1$ and $F_2$ of the ellipse, such that the two curves intersect at four points. Let $P$ be any one of their points of intersection. If the length of major axis of the ellipse is $17$ and area of the triangle $P{F}_1{F}_2$ is $30$, then distance between foci is

• A. $11$
• B. $12$
• C. $13$
• D. $15$

Answer 1

The correct option is C $13$


Assume horizontal ellipse.
Radius of circle $=ae$
$C:x^2+y^2=a^2{e}^2$
$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

So, $\text{Area}\left(P{F}_1{F}_2\right)=\frac{1}{2}\left|\begin{array}{ccc}ae& 0& 1\\ -ae& 0& 1\\ x& y& 1\end{array}\right|=\pm 30$

Expand about $2$nd column
$-y(ae+ae)=\pm 60$
$\Rightarrow y=\frac{\mp 30}{ae}$

$C:x^2=a^2{e}^2-y^2$
$E:x^2=a^2(1-\frac{y^2}{b^2})$

$a^2(1-\frac{y^2}{b^2})=a^2{e}^2-y^2$

$\Rightarrow a^2(1-\frac{900}{a^2{e}^2{b}^2})=a^2{e}^2-\frac{900}{a^2{e}^2}$

$\Rightarrow a^2(1-e^2)=\frac{900}{e^2{b}^2}-\frac{900}{a^2{e}^2}$

$\Rightarrow a^2(1-e^2)=\frac{900}{e^2}[\frac{1}{a^2(1-e^2)}-\frac{1}{a^2}]$

$\Rightarrow \left(a^2\right(1-e^2){)}^2=900$
$\Rightarrow 1-e^2=\frac{30}{a^2}$

$\Rightarrow e^2=1-\frac{30}{289}\times 4$

$\Rightarrow e=\frac{13}{17}$
$\therefore$ Distance between focii is $2ae=13$