Given that the equation of asymptotes to the hyperbola be:
$y=\pm\frac{3x}{2}$ and $b=4$
How to find the equation of hyperbola?
I know that asymtotes have the equation $y=\pm\frac{bx}{a}$, comparing and solving we get $a=\frac{8}{3}$
But in the exercise there are two answers given :
$\frac{9x^2}{64}-\frac{y^2}{16}=1$ and $\frac{y^2}{36}-\frac{x^2}{16}=1$.
How are there tw0 answers. Please Help. Thanks.
HINT
If the given asymptotes are for horizontal hyperbola $$\dfrac{b}{a} = \dfrac{3}{2}$$
otherwise for vertical :
$$\dfrac{a}{b} = \dfrac{3}{2}$$