If $e_{1}, e_{2}, e_{3}, e_{4}$ are eccentricities of the conics, $xy = c^{2}, x^{2} - y^{2} = c^{2}, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ and $(\sqrt{\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} + \frac{1}{e_{3}^{2}} + \frac{1}{e_{4}^{2}}}) = \sec θ (0 < θ < π / 2)$ , then $θ$ is

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$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{12}$

answer is B.

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Detailed Solution

$\begin{array}{l} e_{1} = \sqrt{2}, e_{2} = \sqrt{2}, e_{3} = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}}, e_{4} = \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} \\ also \frac{1}{e_{3}^{2}} + \frac{1}{e_{4}^{2}} = 1 Given \sqrt{\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} + \frac{1}{e_{3}^{2}} + \frac{1}{e_{4}^{2}}} = \sec θ \\ \Rightarrow \sqrt{\frac{1}{2} + \frac{1}{2} + 1} = \sec θ \Rightarrow \sec (θ) = \sqrt{2} \Rightarrow θ = \frac{π}{4} \end{array}$