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:

`((y+2)^2)/(36)-((x+5)^2)/(64)=1`

The general equation of the given hyperbola is:

`((y-k)^2)/(a^2)-((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(a)/(b)=\pm(6)/(8)=\pm(3)/(4)`

The equation of the asymptotes are:

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