Question

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)`?

Answer 1

`(y+1)^2/2^2-(x-1)^2/(4/3)^2=1`

Explanation:

The vertices are `A(1, 1) and A'(1, -3)`.

The center is the midpoint `C(1, -1)`,

Major axis AA' is given by `x =x_A=x_A'=1`.

The asymptotes intersect at `C(1, -1)`.

The equation to one asymptote is

`A!=y+1-3/2(x-1)`.

So, the equation to the other asymptote has the form

`A2=y+1=m(x-1)`.

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

`tan^(-1)(3/2)`.

So, `m = tan(pi-tan^(-1)(3/2))`

`=-tan(tan^(-1)(3/2))=-3/2`.

And so, the equation to the other asymptote is

`A2=y+1+3/2(x-1)=0`.

Now, the equation to the hyperbola is

`A1xxA2=(y+1)^2-9/4(x-1)^2=C`.

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

C=4.

In the standard form, this is

`(y+1)^2/2^2-(x-1)^2/(4/3)^2=1`

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]}