Question

I need this answered, it’s from my prep guide I will include a pic of the answer options

Answer 1

Given the graph of the hyperbola:

$\frac{(y+2)^2}{36}-\frac{(x+5)^2}{64}=1$

The general equation of the given hyperbola is:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

Where (h, k) is the center of the hyperbola

So, by comparing the equations:

Center = (h, k) = (-5, -2)

The hyperbola opens up and down

Since a =

$a=\sqrt[]{36}=6$

The coordinates of the vertices are: (h, k + a ) and (h, k - a)

h = -5, k = -2, a = 6

So, the coordinates are:

$(-5,-8),(-5,4)$

The slopes of the asymptotes are:

$\pm\frac{a}{b}=\pm\frac{6}{8}=\pm\frac{3}{4}$

The equation of the asymptotes are:

$\begin{gathered} y-k=\pm\frac{a}{b}(x-h) \\ y+2=\pm\frac{3}{4}(x+5) \end{gathered}$