Question

A circle has the same centre as an ellipse & passes through the foci $F_1$ & $F_2$ 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 $17$ & 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. $Noneofthese$

Answer 1

The correct option is C $13$

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

the circle $:x^2+y^2=a^2-b^2$

$2a=17\Rightarrow a=\frac{17}{2}$

Since

$\frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow |y|=\frac{b^2}{\sqrt{a^2-b^2}}$

we have

$30=\frac{1}{2}\times 2\sqrt{o^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}\Rightarrow b=\sqrt{30}$

Thus, the answer is

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