Question

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

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

Short answer

The equation of the hyperbola is

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

The graph of the hyperbola is

Step-by-step solution

Given data

Vertices are at$\left(-1,-1\right)$and$\left(3,-1\right)$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 axis parallel to $x$-axis is given by

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

where focus is at role="math" localid="1647055818744" $\left(h\pm c,k\right)$, vertices are at $\left(h\pm a,k\right)$and asymptotes are$y-k=\pm \frac{b}{a}\left(x-h\right)$.

Application

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

So, $k=-1$

Given that the vertices are $\left(-1,-1\right)$and $\left(3,-1\right)$

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

Also given that, the one asymptote line is

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

So, $h=1$

$k=-1$

$a=2$

$b=3$

Thus, the equation of the hyperbola is

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

Substituting all values,

role="math" localid="1647057102885" $\frac{{\left(x-1\right)}^2}{2^2}-\frac{{\left(y+1\right)}^2}{3^2}=1$

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

Graph

The hyperbola is,

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

The graph will be