The eccentricity of the hyperbola (\frac{(x-3)^2}{9}+\frac{4... | Mathematics | SolveeIt | Solveeit

Today, August 24

Question

The eccentricity of the hyperbola $\frac{(x-3)^2}{9}+\frac{4(y-1)^2}{45}=1$ is equal to

A

$\frac{3}{\sqrt{5}}$

B

$\frac{5}{3}$

C

$\frac{5}{\sqrt{3}}$

D

$\frac{5}{2}$

E

$\frac{3}{2}$

Answer

$\frac{3}{2}$

Explanation

Solution

The correct option is (E) : $\frac{3}{2}$