Question

Let $ e $ be the eccentricity of the ellipse $ \frac{x^2}{4} + \frac{y^2}{9} = 1 $. If $ \frac{1}{e} $ is the eccentricity of a hyperbola, then the eccentricity of its conjugate hyperbola is:

• A. \( \frac{4}{3} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{4}{5} \)
• D. \( \frac{3}{2} \)

Hint:

The eccentricity of the conjugate hyperbola is calculated using the same formula for eccentricity as that for the ellipse, but with the roles of \( a \) and \( b \) swapped.

Answer 1

The Correct Option is A

The equation of the ellipse is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). The eccentricity \( e \) of an ellipse is given by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] where \( a^2 = 9 \) and \( b^2 = 4 \). Substituting these values: \[ e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \] The eccentricity of the hyperbola conjugate to this ellipse is given by \( \frac{1}{e} \), so: \[ \frac{1}{e} = \frac{3}{\sqrt{5}} \] The eccentricity of the conjugate hyperbola is: \[ \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \] Thus, the eccentricity of the conjugate hyperbola is \( \boxed{\frac{4}{3}} \).