Question

A circle has the same centre as an ellipse and 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 such that the area of triangle $P{F}_1{F}_2$ is $30$ sq. unit. If the length of major axis of the ellipse is $17$ unit, then the distance between the foci is unit(s).

Answer 1

As centre is same so $△P{F}_1{F}_2$ will be right angled triangle at $P$ because $F_1{F}_2$ is diameter of the circle.

Area of $△P{F}_1{F}_2=\frac{1}{2}P{F}_1\cdot P{F}_2=30$
$\Rightarrow P{F}_1\cdot P{F}_2=60\cdots \left(1\right)$
Also, we know $P{F}_1+P{F}_2=2a$ (major axis)
$\Rightarrow P{F}_1+P{F}_2=17\cdots \left(2\right)$
Now,
$(F_1{F}_2{)}^2=(P{F}_1{)}^2+(P{F}_2{)}^2\Rightarrow (F_1{F}_2{)}^2=(P{F}_1+P{F}_2{)}^2-2\times P{F}_1\cdot P{F}_2\Rightarrow (F_1{F}_2{)}^2=289-120=169\therefore F_1{F}_2=13\text{units}$