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
Jan 26, 2017

Answer:

#(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]}