Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.

Vertices: $(0,+-18)$; asymptotes: $y=+-9x$

a$.\frac{41y^2}{13,122}+\frac{41x^2}{162}=1$

b$.\frac{13,122x^2}{41}-\frac{41y^2}{162}=1$

c$.\frac{41y^2}{13,122}-\frac{41x^2}{162}=1$

d$.\frac{41x^2}{13,122}+\frac{41y^2}{162}=1$

e$.\frac{41x^2}{13,122}-\frac{41y^2}{162}=1$