Question

Find an equation for the hyperbola described. Graph the equation by hand.

Vertices at $\left(1,-3\right)$and $\left(1,1\right)$; asymptote the line$y+1=\frac{3}{2}\left(x-1\right)$

Short answer

The hyperbola is

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

The graph of the hyperbola is

Step-by-step solution

Given data

Vertices are at $\left(1,-3\right)$and $\left(1,1\right)$and the asymptote line is $y+1=\frac{3}{2}\left(x-1\right)$.

Concept used

The hyperbola with center $\left(h,k\right)$and transverse line parallel to role="math" localid="1647059465861" $y$-axis is given by

$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$; $b^2=c^2-a^2$

where focus are at $\left(h,k\pm c\right)$, vertices are at $\left(h,k\pm a\right)$and the asymptotes are$y-k=\pm \frac{a}{b}\left(x-h\right)$

Application

Since the vertices are lie on the line $x=1$, the transverse axis is parallel to the $y$-axis.

So, $h=1$

Given that, vertices are $\left(1,-3\right),\left(1,1\right)$and one asymptote is $y+1=\frac{3}{2}\left(x-1\right)$.

Since, in general form of asymptotes are $y-k=\pm \frac{a}{b}\left(x-h\right)$

So, $h=1$

$k=-1$

$a=3$

$b=2$

Thus, the equation of hyperbola is

$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Substituting all values,

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

So, role="math" localid="1647060214960" $\frac{{\left(y+1\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{4}=1$

Graph

The hyperbola is

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

The graph will be