Chapter 10.4, Problem 43AYU

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

Center at $(-3,-4)$; focus at $(-3,-8)$; vertex at $(-3,-2)$

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 43AYU

$\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$

Explanation of Solution

Given:

Center at $\left(-3,-4\right)$; focus at $\left(-3,-8\right)$; vertex at $\left(-3,-2\right)$.

Formula used:

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

Calculation:

The center of the hyperbola is $\left(h,k\right)=\left(-3,-4\right)$; $h=-3,k=-4$.

Focus: $\left(-3,-8\right)=\left(h,k \pm c\right)$; $c=4$.

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

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

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 $\text{b}$;

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

$b²=4²-2²=12;b=2\sqrt{3}$

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

$\frac{{\left(y+4\right)}^2}{2²}-\frac{{\left(x+3\right)}^2}{2\sqrt{3²}}=1$

$\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$

Asymptotes: $\left(y+4\right)=\pm \sqrt{3}/3\left(x+3\right)$.

To determine

c.

To graph: The equation $\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$.

Explanation:

Given:

Center at $\left(-3,-4\right)$; focus at $\left(-3,-8\right)$; vertex at $\left(-3,-2\right)$.

Graph:

The graph of the equation of $\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$ is plotted.

Using the given information Center at $\left(-3,-4\right)$; focus at $\left(-3,-8\right)$; vertex at $\left(-3,-2\right)$.

We find $a=2$, $b=2\sqrt{3}$. We have formed the equation and plotted the graph for $\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$.

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

Answer to Problem 43AYU

Explanation of Solution

Given:

Center at $\left(-3,-4\right)$; focus at $\left(-3,-8\right)$; vertex at $\left(-3,-2\right)$.

Graph:

The graph of the equation of $\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$ is plotted.

Using the given information Center at $\left(-3,-4\right)$; focus at $\left(-3,-8\right)$; vertex at $\left(-3,-2\right)$.

We find $a=2$, $b=2\sqrt{3}$. We have formed the equation and plotted the graph for $\frac{{\left(y+4\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{12}=1$.

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