Chapter 10.4, Problem 62AYU

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.

$x^2-3y^2+8x-6y+4=0$

To determine

a.

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

Answer to Problem 62AYU

Center: $\left(-4,-1\right)$.

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

Foci: $\left(-4\pm 2\sqrt{3},-1\right)$.

Vertices: $\left(-7,-1\right),\left(-1,-1\right)$.

Asymptotes: $y+1=\pm \frac{\sqrt{3}}{3}\left(x+4\right)$.

Explanation of Solution

Given:

$x² -3y²+8x-6y+4=0$

Formula used:

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

Calculation:

$x² -3y²+8x-6y+4=0$

$x²+8x-3y²-6y+4=0$

${\left(x+4\right)}^2-3{\left(y+1\right)}^2+4=16-3$

$\frac{{\left(x+4\right)}^2}{9}-\frac{3{\left(y+1\right)}^2}{9}=1$

The above equation is of the form $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$;

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

The center of the hyperbola is $\left(h,k\right)=\left(-4,-1\right)$; $h=-4, k=-1$.

$a=3, b=\sqrt{3}$

Vertices: $\left(h\pm a,k\right)=\left(-4\pm 3,-1\right)=\left(-7,-1\right),\left(-1,-1\right)$.

$c²=3²+\sqrt{3^2}=\sqrt{12}=2\sqrt{3}$

Foci: $\left(h\pm c,k\right)=\left(-4\pm 2\sqrt{3},-1\right)$.

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

Asymptotes: $y+1=\pm \frac{\sqrt{3}}{3}\left(x+4\right)$.

To determine

b.

To graph: The equation $x² -3y²+8x-6y+4=0$.

Answer to Problem 62AYU

Explanation of Solution

Given:

$x²-3y²+8x-6y+4=0$

Calculation:

The graph of the equation of $x²-3y²+8x-6y+4=0$ is plotted.

Using the given equation $x²-3y²+8x-6y+4=0$. We find the Center at $\left(-4,-1\right)$; Foci: $\left(-4\pm 2\sqrt{3},-1\right)$, Vertices: $\left(-1,-1\right)$, $\left(-7,-1\right)$, Asymptotes: $y+1=\pm \frac{\sqrt{3}}{3}\left(x+4\right)$.

Transverse axis is parallel to $x\text{-axis}$. We find, $a=3,b=\sqrt{3}$. With the information available we plot the graph of $x²-3y²+8x-6y+4=0$ and verify the same using the graphing tool.