Question:medium

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: Jun 3, 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: Recall the pair rule for hyperbolas.
Every hyperbola has a partner called its conjugate hyperbola. If one has eccentricity $e_1$ and the other has $e_2$, they obey a neat link.
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \] We will use this to find the partner's eccentricity.

Step 2: Note what we have and want.
We are given $e_1 = \sqrt{3}$. We want $e_2$, the eccentricity of the conjugate. So we put the known value into the rule and solve for the unknown.

Step 3: Square the known eccentricity.
Since $e_1 = \sqrt{3}$, squaring gives $e_1^2 = 3$. So $\frac{1}{e_1^2} = \frac{1}{3}$.

Step 4: Put this into the rule.
Replace $\frac{1}{e_1^2}$ with $\frac{1}{3}$.
\[ \frac{1}{3} + \frac{1}{e_2^2} = 1 \]

Step 5: Solve for $\frac{1}{e_2^2}$.
Move $\frac{1}{3}$ to the other side.
\[ \frac{1}{e_2^2} = 1 - \frac{1}{3} = \frac{2}{3} \]

Step 6: Flip and take the square root.
Flipping gives $e_2^2 = \frac{3}{2}$. Taking the positive square root (eccentricity is always positive) gives the answer.
\[ e_2 = \sqrt{\frac{3}{2}} \]
\[ \boxed{e_2 = \sqrt{\tfrac{3}{2}}} \]
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