# 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)?

Jan 26, 2017

${\left(y + 1\right)}^{2} / {2}^{2} - {\left(x - 1\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} = 1$

#### Explanation:

The vertices are $A \left(1 , 1\right) \mathmr{and} A ' \left(1 , - 3\right)$.

The center is the midpoint $C \left(1 , - 1\right)$,

Major axis AA' is given by $x = {x}_{A} = {x}_{A} ' = 1$.

The asymptotes intersect at $C \left(1 , - 1\right)$.

The equation to one asymptote is

$A \ne y + 1 - \frac{3}{2} \left(x - 1\right)$.

So, the equation to the other asymptote has the form

$A 2 = y + 1 = m \left(x - 1\right)$.

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

${\tan}^{- 1} \left(\frac{3}{2}\right)$.

So, $m = \tan \left(\pi - {\tan}^{- 1} \left(\frac{3}{2}\right)\right)$

$= - \tan \left({\tan}^{- 1} \left(\frac{3}{2}\right)\right) = - \frac{3}{2}$.

And so, the equation to the other asymptote is

$A 2 = y + 1 + \frac{3}{2} \left(x - 1\right) = 0$.

Now, the equation to the hyperbola is

$A 1 \times A 2 = {\left(y + 1\right)}^{2} - \frac{9}{4} {\left(x - 1\right)}^{2} = C$.

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

C=4.

In the standard form, this is

${\left(y + 1\right)}^{2} / {2}^{2} - {\left(x - 1\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} = 1$

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]}