# The eccentricity of the hyperbola $9y^2 -4x^2=36$ is

$\begin{array}{1 1}(A)\;\frac{5}{2} \\(B)\;\frac{\sqrt 7}{2} \\(C)\;\sqrt \frac{13}{2} \\(D)\;2 \end{array}$

$\large\frac{y^2}{4}-\frac{x^2}{9}$$=1 here the coefficient y^2 is +ve and that of x^2 is -ve and hence it represents a hyperbola whose transverse axis is vertical i.e a^2=4 b^2=9 b^2=a^2(e^2-1) \large\frac{9}{4}$$+1=e^2$
$e= \large\frac{\sqrt {13}}{2}$
Hence C is the correct answer.