Chapter 10.4, Problem 46AYU

To determine

a & b

To find: An equation of the hyperbola described.

Answer to Problem 46AYU

$\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$

Explanation of Solution

Given:

Focus at $\left(-4,0\right)$; vertices at $\left(-4,-4\right)$; vertex at $\left(-4,2\right)$.

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: $\left(-4,4\right)$ and $\left(-4,2\right)=\left(h,k\pm a\right);h=-4;k=3;a=1$.

Focus: $\left(-4,0\right)=\left(h,k\pm c\right)$; $c=3$.

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

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 $b$;

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

$b²=3²-1²=8;b=2\sqrt{2}$

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

$\frac{{\left(y-3\right)}^2}{1²}-\frac{{\left(x+4\right)}^2}{2\sqrt{2²}}=1$

$\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$

Asymptotes: $\left(y-3\right)=\pm \frac{\sqrt{2}}{4\left(x+4\right)}$.

To determine

c.

To graph: The equation $\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$.

Answer to Problem 46AYU

Explanation of Solution

Given:

Focus at $\left(-4,0\right)$; vertices at $\left(-4,-4\right)$; vertex at $\left(-4,2\right)$.

Graph:

The graph of the equation of $\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$ is plotted.

Using the given information Focus at $\left(-4,0\right)$; vertices at $\left(-4,-4\right)$; vertex at $\left(-4,2\right)$.

We find $a=1$, $b=2\sqrt{2}$. We have formed the equation and plotted the graph for $\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$.

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