Question

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

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{\frac{3}{2}}$
• D. $2\sqrt{3}$

Answer 1

The correct option is C

$\sqrt{\frac{3}{2}}$

Explanation for correct option

Given that the eccentricity of a hyperbola is $\sqrt{3}$

Let the the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The eccentricity of the hyperbola is $e=\sqrt{1+{\left(\frac{b}{a}\right)}^2}=\sqrt{3}$ ….$\left(i\right)$

The conjugate Hyperbola of the given hyperbola is $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

The eccentricity of the conjugate hyperbola is $e\text{'}=\sqrt{1+{\left(\frac{a}{b}\right)}^2}$

From equation $\left(i\right)$ we have

$$\begin{gathered} \therefore \sqrt{1+{\left(\frac{b}{a}\right)}^2}=\sqrt{3} \\ \Rightarrow 1+{\left(\frac{b}{a}\right)}^2=3 \\ \Rightarrow {\left(\frac{b}{a}\right)}^2=2 \\ \Rightarrow {\left(\frac{a}{b}\right)}^2=\frac{1}{2} \end{gathered}$$

Therefore the eccentricity of the conjugate hyperbola is

$$\begin{gathered} \Rightarrow e\text{'}=\sqrt{1+{\left(\frac{a}{b}\right)}^2} \\ \Rightarrow e\text{'}=\sqrt{1+\frac{1}{2}}\left[\because {\left(\frac{a}{b}\right)}^2=\frac{1}{2}\right] \\ \Rightarrow e\text{'}=\sqrt{\frac{3}{2}} \end{gathered}$$

Hence, Option (C) is correct i.e. $\sqrt{\frac{3}{2}}$