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${A circle has the same center as an ellipse and passes through the foci F}_{1} {, and F}_{2} of the ellipse, such that$
$the two curves intersect at four points. Let P be any one of their points of intersection. If the major axis of the ellipse is$
$17 and the area of triangle P F_{1} F_{2} is 30, then the distance between the foci is$

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Detailed Solution

$Let the ellipse be \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . Then the circle is x^{2} + y^{2} = a^{2} e^{2}$

$Radius of circle = a e$

$One of the points of intersection of the circle and the ellipse is$

$(\frac{a}{e} \sqrt{2 e^{2} - 1}, \frac{a}{e} (1 - e^{2}))$

$Now, area of Δ P F_{1} F_{2} is$

$\begin{array}{l} \frac{1}{2} |\frac{a}{e} (1 - e^{2}) (2 a e)| = 30 (Given) \\ or a^{2} (1 - e^{2}) = 30 \\ or a^{2} e^{2} = a^{2} - 30 = {(\frac{17}{2})}^{2} - 30 = \frac{169}{4} \\ or 2 a e = 13 \end{array}$

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