Question:

If the eccentricity of a hyperbola is $\sqrt{3}$, then the eccentricity of its conjugate hyperbola is:

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The elegant relationship $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$ is incredibly powerful and appears frequently in entrance exams. Memorize it!
Updated On: May 31, 2026
  • $\sqrt{\frac{3}{2}}$
  • $\sqrt{3}$
  • $\frac{\sqrt{3}}{2}$
  • $\sqrt{2}$
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The Correct Option is A

Solution and Explanation


Step 1: Concept

If $e_1$ and $e_2$ are the eccentricities of a hyperbola and its conjugate hyperbola, they satisfy the reciprocal identity: \[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \]

Step 2: Meaning

We are given $e_1 = \sqrt{3}$ and need to solve for $e_2$.

Step 3: Analysis

Substituting $e_1 = \sqrt{3}$ into the identity: \[ \frac{1}{(\sqrt{3})^2} + \frac{1}{e_2^2} = 1 \implies \frac{1}{3} + \frac{1}{e_2^2} = 1 \] \[ \frac{1}{e_2^2} = 1 - \frac{1}{3} = \frac{2}{3} \] \[ e_2^2 = \frac{3}{2} \implies e_2 = \sqrt{\frac{3}{2}} \]

Step 4: Conclusion

The eccentricity of the conjugate hyperbola is $\sqrt{\frac{3}{2}}$. Final Answer: (A)
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