Question

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

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

Answer 1

Answer: (D)

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

Let the hyperbola H: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ have eccentricity $\sqrt{3}=e$
So, its a conjugate hyperbola obtained by changing signs of $a^2$ and $b^2$
$\Rightarrow H^{{}^{\prime }}:\frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$--------------1
and its Eccentricity $e^{{}^{\prime }}=\sqrt{1+\frac{a^2}{b^2}}$----------2
Now, Eccentricity e$=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{3}$ (given)
Squaring both the sides, we get $1+\frac{b^2}{a^2}=3$
$\Rightarrow \frac{b^2}{a^2}=2$
$\Rightarrow \frac{a^2}{b^2}=\frac{1}{2}$
Let's put the value of $\frac{a^2}{b^2}$ in Equation 2,
$\Rightarrow e^{{}^{\prime }}=\sqrt{1+\frac{1}{2}}$
$\Rightarrow e^{{}^{\prime }}=\sqrt{\frac{3}{2}}$
$\Rightarrow e^{{}^{\prime }}=\frac{\sqrt{3}}{\sqrt{2}}$