Question

Let X-axis be the transverse axis and Y-axis be the conjugate axis of a hyperbola \( H \). Let the eccentricity of \( H \) be the reciprocal of the eccentricity of the ellipse \[ \frac{x^2}{4} + \frac{y^2}{2} = 1 \] If \( (5, 4) \) lies on \( H \), then the length of the transverse axis is:

• A. \( 2\sqrt{2} \)
• B. \( 4 \)
• C. \( 6 \)
• D. \( 10 \)

Hint:

Hyperbola’s transverse axis is \( 2a \). Use point-substitution to solve for unknowns in the equation.

Answer 1

The Correct Option is C

Eccentricity of ellipse \( e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{4}} = \frac{1}{\sqrt{2}} \) ⇒ Eccentricity of hyperbola \( e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2} \Rightarrow e_H = \sqrt{2} = \frac{1}{e_{\text{ellipse}}} \) Let hyperbola be \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) Use point (5, 4) to satisfy the equation and solve for \( a \). You get: \[ \text{Transverse axis} = 2a = \boxed{6} \]