Let the major axis of a standard ellipse equals the transverse axis of a standard hyperbola and their Director circles have radius equal to 2R and R respectively. If $e_{1} a n d e_{2}$ are the eccentricities of the Ellipse and hyperbola then the correct relation is

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$e_{1}^{2} - 4 e_{2}^{2} = 2$

$4 e_{1}^{2} - e_{2}^{2} = 6$

$4 e_{2}^{2} - e_{1}^{2} = 6$

$2 e_{1}^{2} - e_{2}^{2} = 4$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ a^{2} + b^{2} = 4 R^{2} = a^{2} (2 - e_{1}^{2}) \\ a^{2} - b^{2} = R^{2} = a^{2} (2 - e_{2}^{2}) \\ 4 = \frac{2 - {e_{1}^{2}}^{}}{2 - e_{2}^{2}} \\ \Rightarrow 4 e_{2}^{2} - e_{1}^{2} = 6 \end{array}$