Question:
Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse,
$\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)$ and the hyperbola, $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$
respectively satisfying $e_{1} e_{2}=1$. If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to :
Correct Option: , 4
Solution:
Equation of ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$
Then, $e_{1}=\sqrt{1-\frac{b^{2}}{25}}$
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