Chapter 10.4, Problem 54AYU

In Problems 49-62, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.

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

To determine

a:

To find: The center, transverse axis, vertices, foci and asymptotes.

Answer to Problem 54AYU

Center: $\left(-2,3\right)$.

Transverse axis: parallel to $y\text{-axis}$.

Vertices: $\left(-2,5\right),\left(-2,1\right)$.

Foci: $\left(-2,3\pm 2\sqrt{2}\right)$.

Asymptotes: $y-3=\pm \left(x+2\right)$.

Explanation of Solution

Given:

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

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:

${\left(y-3\right)}^2-{\left(x+2\right)}^2=4$ can be rewritten as $\frac{{\left(y-3\right)}^2}{4}-\frac{{\left(x+2\right)}^2}{4}=1$ is of the form $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$;

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

The center of the hyperbola is $\left(h,k\right)=\left(-2,3\right)$; $h=-2,k=3$.

$a=2,b=2$

Vertices: $\left(h,k\pm a\right)=\left(-2,3\pm 2\right)=\left(-2,5\right),\left(-2,1\right)$.

$c²=2²+2²=2\surd 2$

Foci: $\left(h,k\pm c\right)=\left(-2,3\pm 2\sqrt{2}\right)$.

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

Asymptotes: $y-3=\pm \left(x+2\right)$.

To determine

b.

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

Answer to Problem 54AYU

Explanation of Solution

Given:

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

Graph:

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

Using the given equation ${\left(y-3\right)}^2-{\left(x+2\right)}^2=4$ we find the Center at $\left(-2,3\right)$; Vertices:$\left(-2,5\right)$,$\left(-2,1\right)$, Foci: $\left(-2,3\pm 2\sqrt{2}\right)$ Asymptotes: $y-3=\pm \left(x+2\right)$.

Transverse axis is parallel to $y\text{-axis}$. We find, $a=2$, $b=2$. With the information available we plot the graph of ${\left(y-3\right)}^2-{\left(x+2\right)}^2=4$ and verify the same using the graphing tool.