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(B) Let the distance between the foci $F_1$ and $F_2$ be $2c$. Since the circle $C$ is centered at the origin and passes through the foci $(\pm c, 0)$

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let the distance between the foci $F_1$ and $F_2$ be $2c$. Since the circle $C$ is centered at the origin and passes through the foci $(\pm c, 0)$,its radius is $r = c$.Since $P$ lies on the circle $C$,the distance $OP = c$. Also,since $F_1$ and $F_2$ are on the circle,the angle $\angle F_1PF_2 = 90^\circ$ because $F_1F_2$ is the diameter of the circle.Let $PF_1 = x$ and $PF_2 = y$. Since $P$ lies on the ellipse,by the definition of an ellipse,$x + y = 2a = 17$.In the right-angled triangle $PF_1F_2$,the area is $\frac{1}{2} xy = 30$,so $xy = 60$.We know that $(x + y)^2 = x^2 + y^2 + 2xy$. By the Pythagorean theorem in $	riangle PF_1F_2$,$x^2 + y^2 = (F_1F_2)^2 = (2c)^2 = 4c^2$.Thus,$(2a)^2 = 4c^2 + 2xy$,which gives $17^2 = 4c^2 + 2(60)$.$289 = 4c^2 + 120$.$4c^2 = 169$.$2c = \sqrt{169} = 13$.The distance between the foci is $2c = 13$.

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