If the ellipse $C_{1} : \frac{x^{2}}{a_{1}^{2}} + \frac{y^{2}}{b_{1}^{2}} = 1$ and hyperbola $C_{2} : \frac{x^{2}}{a_{2}^{2}} - \frac{y^{2}}{b_{2}^{2}} = 1$ has the same focus points $S_{1} & S_{2} .$ Point P is one of the point of intersection of $C_{1} & C_{2} .$ and $P S_{1}$ is perpendicular to $P S_{2} .$ If $e_{1}$ is eccentricity of $C_{1} & e_{2}$ of $C_{2,}$ then minimum value of $9 e_{1}^{2} + e_{2}^{2}$ is

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Detailed Solution

$a_{1} e_{1} = a_{2} e_{2}$

$\begin{array}{l} P F_{1} + P F_{2} = 2 a_{1} \\ | P F_{1} + P F_{2} | = 2 a_{2} \\ {(P F_{1} + P F_{2})}^{2} = 4 (a_{1}^{2} + a_{2}^{2}) \Rightarrow 4 a_{1}^{2} e_{1}^{2} = 2 (a_{1}^{2} + a_{2}^{2}) \end{array}$

$\begin{array}{l} P F_{1}^{2} + P F_{2}^{2} = 4 a_{1}^{2} e_{1}^{2} \\ \Rightarrow 2 (2 P F_{1}^{2} + P F_{2}^{2}) = 4 (a_{1}^{2} + a_{2}^{2}) \Rightarrow 4 a_{1}^{2} e_{1}^{2} = 2 (a_{1}^{2} + a_{2}^{2}) \\ \Rightarrow 2 e_{1}^{2} = 1 + {(\frac{a_{2}}{a_{1}})}^{2} \Rightarrow 2 e_{1}^{2} = 1 + {(\frac{e_{1}}{e_{2}})}^{2} \Rightarrow \frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = 2 \\ L e t \frac{1}{e_{1}} = \sqrt{2} \cos θ; \frac{1}{e_{2}} = \sqrt{2} \sin θ \\ E = 9 e_{1}^{2} + e_{2}^{2} = \frac{9}{2} \sec^{2} θ + \frac{1}{2} \cos e c^{2} θ \\ E = \frac{1}{2} (10 + 9 \tan^{2} θ + \cot^{2} θ); E_{\min} = \frac{1}{2} (10 + 6) = 8 \end{array}$