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}} $