A circle has the same centre as an ellipse & passes through the foci F₁ & F₂ 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 PF₁F₂ is 30, then the distance between the foci is :

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.

Given area of $∆P$$F_1{F}_2$=$P{F}_1.P{F}_2$=30
⇒$P{F}_1.P{F}_2$=60 ⋯(1)
Also, we know $P{F}_1+P{F}_2$=2a (major axis)
⇒$P{F}_1+P{F}_2$=17 ⋯(2)
$$\begin{gathered} Now {\left(F_1{F}_2\right)}^2={\left(P{F}_1\right)}^2+{\left(P{F}_2\right)}^2 \\ ={(P{F}_1+P{F}_2)}^2-2P{F}_{1}P{F}_2 \\ =289-120 \\ =169 \\ \Rightarrow F_1{F}_2=13 \end{gathered}$$