HyperbolaHardBloom L3
Question
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 $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse and hyperbola then the correct relation is:
Options
A.$4e_{1}^{2} - e_{2}^{2} = 6$
B.$e_{1}^{2} - 4e_{2}^{2} = 2$
C.$4e_{2}^{2} - e_{1}^{2} = 6$
D.$2e_{1}^{3} - e_{2}^{2} = 4$
Solution
Sol. (c ) $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{B^{2}} = 1$
$$a^{2} + b^{2} = 4R^{2} = a^{2}\left( 2 - e_{1}^{2} \right)$$
$$a^{2} - B^{2} = R^{2} = a^{2}\left( 2 - e_{2}^{2} \right)$$
$$\begin{matrix} 4 & \ = \frac{2 - e_{1}^{2}}{2 - e_{2}^{2}} \\ \Rightarrow 4e_{2}^{2} - e_{1}^{2} & \ = 6 \end{matrix}$$
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