Chapter 10.4, Problem 60AYU

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.

$2y^2-x^2+2x+8y+3=0$

To determine

a.

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

Answer to Problem 60AYU

center: $\left(1,\mathrm{-2}\right)$

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

Vertices: $\left(1,-2\pm \sqrt{2}\right)$,

Foci: $\left(1,-2\pm \sqrt{6}\right)$,

Asymptotes: $y+2=\pm \sqrt{2}/2\left(x-1\right)$

Explanation of Solution

Given:

$2y²-x²+8y+2x+3=0$

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:

$2y²-x²+8y+2x+3=0$ can be rewritten as

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

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

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

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 $\left(h,k\right)=\left(1,\mathrm{-2}\right)$; $h=1,k=\mathrm{-2}$.

$a=\surd 2,b=2$

Vertices: $\left(h,k\pm a\right)=\left(1,- 2\pm \sqrt{2}\right)$

$c²=\surd 2²+4=\surd 6$

Foci: $\left(h,k\pm c\right)=\left(1,-2\pm \sqrt{6}\right)$

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

Asymptotes: $y+2=\pm \sqrt{2}/2\left(x-1\right)$.

To determine

b.

To graph: The equation $2y²-x²+8y+2x+3=0$.

Answer to Problem 60AYU

Explanation of Solution

Given:

$2y²-x²+8y+2x+3=0$

Graph:

The graph of the equation of $2y²-x²+8y+2x+3=0$ is plotted.

Using the given equation $2y²-x²+8y+2x+3=0$ we find the Center at $\left(1,\mathrm{-2}\right)$; Vertices: $\left(1,-2\pm \sqrt{2}\right)$, Foci: $\left(1,-2\pm \sqrt{6}\right)$, Asymptotes: $y+2=\pm \frac{\sqrt{2}}{2}\left(x-1\right)$.

Transverse axis is parallel to $y\text{-axis}$. We find, $a=\surd 2,b=2$. With the information available we plot the graph of $2y²-x²+8y+2x+3=0$ and verify the same using the graphing tool.