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 ${}^{\prime }{P}^{\prime }$ 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. $11$
• B. $12$
• C. $13$
• D. None

Answer 1

Answer: (C)

$13$

$Theellipse\frac{x^2}{2}+\frac{y^2}{2}=1,a>b>0$

$thecircle:x^2+y^2=a^2-b^2$

$Fromthequestionweget2a=17$

$\therefore a=\frac{17}{2}\because \frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow \left|y\right|=\frac{b^2}{\sqrt{a^2-b^2}}$

$wehave,30=\frac{1}{2}\times 2\sqrt{a^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}$

$\Rightarrow b=\sqrt{30}$

$Thus,$

$a^2-b^2=2\left(\sqrt{{\left(\frac{17}{2}\right)}^2-30}\right)=13$

$Hence,theoptionCisthecorrectanswer.$