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A circle is concentric with the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and passes through the Foci $F_{1}$ and $F_{2}$ of the ellipse. The two curves intersect at four points. Let P be any point of intersection. If the major axis of the ellipse is 15 and the area of the triangle $P F_{1} F_{2}$ is 26. Then the value $2 {(a^{2} - b^{2})}^{\frac{1}{2}} - 4$

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Detailed Solution

$P F_{1} + P F_{2} = 15 P F_{1} \times P F_{2} = 52$

${(F_{1} F_{2})}^{2} = {(P F_{1} + P F_{2})}^{2} - 2 P F_{1} \times P F_{2} = 225 - 104 = 121$

$F_{1} F_{2} = 11 2 a e = 11 \Rightarrow a e = \frac{11}{2} \begin{array}{l} b^{2} = a^{2} (1 - e^{2}) \\ b^{2} = a^{2} - a^{2} e^{2} \end{array} a^{2} - b^{2} = a^{2} e^{2} 4 (a^{2} - b^{2}) = 4 {(\frac{11}{2})}^{2} = 121$

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