# How do you write a standard form equation for the hyperbola with Vertices at #(1, -3)# and #(1,1)#; asymptote the line #y + 1 = (3/2)(x - 1)#?

##### 1 Answer

#### Explanation:

The vertices are

The center is the midpoint

Major axis AA' is given by

The asymptotes intersect at

The equation to one asymptote is

So, the equation to the other asymptote has the form

The asymptotes are equally inclined to the major axis x = 1.

The given asymptote has the slope 3/2 and its inclination to x =1 is

So,

And so, the equation to the other asymptote is

Now, the equation to the hyperbola is

This passes through A(1, 1) and #A'(1, -3). So,

C=4.

In the standard form, this is

graph{((y+1)^2/4-(x-1)^2/(10/9)-1)((y+1)^2/4-(x-1)^2/(10/9))=0 [-10, 10, -6, 4]}