Question

The eccentricity of the hyperbola $3x^2-4y^2=-12$ is

• A. $\frac{\sqrt{7}}{3}$
• B. $\frac{\sqrt{7}}{2}$
• C. $\sqrt{\frac{7}{3}}$
• D. $\sqrt{\frac{7}{2}}$

Answer 1

The correct option is B $\frac{\sqrt{7}}{2}$

Given the equation of the hyperbola is $3x^2-4y^2=-12$ or, $\frac{y^2}{3}+\frac{x^2}{4}=1$.......(1).

Comparing (1) with $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ we get, $a^2=4$ and $b^2=3$.

Then the eccentricity of the hyperbola (1) is $\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{3}{4}}=\frac{\sqrt{7}}{2}$.