A circle as the same center as an ellipse and passes through the foci $F_{1}$ and $F_{2}$ of the ellipse . Two curves intersects in four points. Let P be any one of their point of intersection, if the major axes of an ellipse is 17 and the area of the triangle $P F_{1} F_{2}$ is 30, then $\bar{F_{1} F_{2}}$ is_

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answer is 13.

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Detailed Solution

$x^{2} + y^{2} = a^{2} e^{2}$ Let ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ And Circle is
Point of intersection of circle and ellipse is $P (\frac{a}{e} \sqrt{2 e^{2} - 1}, \frac{a}{e} (1 - e^{2}))$ where
$b^{2} = a^{2} (1 - e^{2})$ and $F_{1} (a e, 0) and F_{2} (- a e, 0)$ Area of triangle is
$\begin{array}{l} = \frac{1}{2} |\begin{array}{ccc} \frac{a}{e} \sqrt{2 e^{2} - 1} & \frac{a}{e} (1 - e^{2}) & 1 \\ ae & 0 & 1 \\ - ae & 0 & 1 \end{array}| = 30 \Rightarrow a^{2} (1 - e^{2}) = 30 \\ \Rightarrow a^{2} e^{2} = a^{2} - 30 = {(\frac{17}{2})}^{2} - 30 (∵ 2 a = 17, given) \Rightarrow ae = \frac{13}{2} \Rightarrow 2 ae = 13 \end{array}$