Question

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:

• A. 192
• B. 64
• C. 16
• D. 32
• E. 128

Answer 1

Answer: (A)

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