Chapter 10.4, Problem 48AYU

In Problems 41-48, find an equation for the hyperbola described. Graph the equation by hand.

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

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 48AYU

$\frac{{\left(y+1\right)}^2}{4}-\frac{9{\left(x-1\right)}^2}{16}=1$

Explanation of Solution

Given:

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

Formula used:

Center	Transverse axis	Foci	Vertices	Equation	Asymptotes
$\left(h,k\right)$	Parallel to $y\text{-axis}$	$\left(h,k \pm c\right)$	$\left(h,k\pm a\right)$	$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$; $b^2=c^2-a^2$	$y-k=\pm \frac{a}{b}\left(x-h\right)$

Calculation:

Vertex: $\left(1,-3\right)$ and $\left(1,1\right)=\left(h,k\pm a\right);h=1;k=-1$; $a=2$.

The center of the hyperbola is $\left(h,k\right)=\left(1,-1\right)$.

From the given data we see that, transverse axis is parallel to $y\text{-axis}$.

Asymptote the line $y+1=\frac{3}{2}\left(x-1\right)$; $\frac{a}{b}=\frac{3}{2}$; $b=\frac{4}{3}$.

Hence, the equation of the hyperbola is

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

To find $b$;

$b^2=c^2-a^2$

${\left(\frac{4}{3}\right)}^2=c^2-2^2;c²=\frac{16}{9}+4=52/9;c=2\sqrt{13}/3$

Focus: $\left(h,k\pm c\right)$; $\left(1,-1\pm \frac{2\sqrt{13}}{9}\right)$.

Substituting the values of $a$ and $b$ in the equation of the hyperbola, we get,

$\frac{{\left(y+1\right)}^2}{2²}-\frac{{\left(x-1\right)}^2}{\left(\frac{4}{3}\right)²}=1$

$\frac{{\left(y+1\right)}^2}{16}-\frac{9{\left(x-1\right)}^2}{4}=1$

To determine

c.

To graph: The equation $\frac{{\left(y+1\right)}^2}{16}-\frac{9{\left(x-1\right)}^2}{4}=1$.

Answer to Problem 48AYU

Explanation of Solution

Given:

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

Graph:

The graph of the equation of $\frac{{\left(y+1\right)}^2}{16}-\frac{9{\left(x-1\right)}^2}{4}=1$ is plotted.

Using the given information Vertices $\left(1,-3\right)$ and $\left(1,1\right)$.

Asymptote the line $y+1=\frac{3}{2}\left(x-1\right)$. We find $a=2$, $b=4/3$.

We have formed the equation and plotted the graph for $\frac{{\left(y+1\right)}^2}{4}-\frac{9{\left(x-1\right)}^2}{16}=1$.

The transverse axis is parallel to $y\text{-axis}$.