Question

The eccentricity of the conjugate hyperbola of the hyperbola $x^{2} - 3y^{2} = 1$, is

• A. 2
• B. $\frac{2}{\sqrt{3}}$
• C. 4
• D. $x + y = 25$

Answer 1

Answer: (A)

2

Answer 2

The given hyperbola is $\frac{x^{2}}{1} - \frac{y^{2}}{1/3} = 1$. Here $a^{2} = 1$ and $b^{2} = \frac{1}{3}$

Since $b^{2} = a^{2}(e^{2} - 1)$ ⇒ $\frac{1}{3} = 1(e^{2} - 1)$ ⇒ $e^{2} = \frac{4}{3}$ ⇒ $e = \frac{2}{\sqrt{3}}$

If $e^{'}$ is the eccentricity of the conjugate hyperbola, then $\frac{1}{e^{2}} + \frac{1}{e^{'2}}$ = 1 ⇒ $\frac{1}{e^{'2}} = 1 - \frac{1}{e^{2}} = 1 - \frac{3}{4} = \frac{1}{4}$ ⇒ $e^{'} = 2$