Chapter 10.4, Problem 42AYU

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 42AYU

$\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{16}=1$

Explanation of Solution

Given:

Center at $\left(-3,1\right)$; focus at $\left(-3,6\right)$; vertex at $\left(-3,4\right)$.

Formula used:

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

Calculation:

The center of the hyperbola is $\left(h,k\right)=\left(-3,1\right)$; $h=-3,k=1$.

Focus: $\left(-3,6\right)=\left(h,k \pm c\right)$; $c=5$.

Vertex: $\left(-3,4\right)=\left(h,k \pm a\right)$; $a=3$.

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 $b$;

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

$b²=5²-3²=16;b=\text{ 4}$

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

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

$\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{16}=1$

Asymptotes: $\left(y-1\right)=\pm 3/4\left(x+3\right)$.

To determine

c.

To graph: The equation $\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{16}=1$.

Answer to Problem 42AYU

Explanation of Solution

Given:

Center at $\left(-3,1\right)$; focus at $\left(-3,6\right)$; vertex at $\left(-3,4\right)$.

Graph:

The graph of the equation of $\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{16}=1$ is plotted.

Using the given information Center at $\left(-3,1\right)$; focus at $\left(-3,6\right)$; vertex at $\left(-3,4\right)$.

We find $a=3$, $b=\text{ 4}$. We have formed the equation and plotted the graph for $\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{16}=1$.

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