If the eccentricities of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ and the hyperbola $\frac{x^{2}}{64} - \frac{y^{2}}{b^{2}} = 1$ are reciprocals of each other, then $b^{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
$\because$ Eccentricity of ellipse$= \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$
$\because$ Eccentricity of hyperbola$= 2$
$\Rightarrow \ \sqrt{1 + \frac{b^{2}}{64}}\  \Rightarrow \ \ 2$
$\Rightarrow \ 4 = 1 + \frac{b^{2}}{64}\ \  \Rightarrow 192 = b^{2}$

Q. If the eccentricities of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ and the hyperbola $\frac{x^{2}}{64} - \frac{y^{2}}{b^{2}} = 1$ are reciprocals of each other, then $b^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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