Chapter 10.3, Problem 52AYU

In Problems 43-54, analyze each equation; that is, find the center, foci, and vertices of each ellipse. Graph each equation by hand. Verify your graph using a graphing utility.

$x^2+9y^2+6x-18y+9=0$

To determine

To find: The vertices and foci of the given equation of the ellipse.

Answer to Problem 52AYU

Solution:

Vertices: $\left(-6,1\right)\text{and}(0,1)$.

Foci: $\left(-3+2\sqrt{2},1\right)\text{and}\left(-3-2\sqrt{2},1\right)$.

Explanation of Solution

Given:

$x²+9y²+6x-18y+9=0$

Formula used:

Equation	Center	Major axis	Foci	Vertices
$\begin{array}{l}\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\\ a>b>0\text{and}b²=a²-c²\end{array}$	$\left(h,k\right)$	$\text{parallel}\text{to}x\text{-axis}$	$\left(h+c,k\right)$	$\left(h+a,k\right)$
	$\left(h,k\right)$	$\text{parallel}\text{to}x\text{-axis}$	$\left(h-c,k\right)$	$\left(h-a,k\right)$

Calculation:

Rearranging the given equation $x²+9y²+6x-18y+9=0$.

We get, $\left(x^2+6x+9\right)+9\left(y^2-2y+1\right)+9=9+9$.

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

${\left(x+3\right)}^2+9{\left(y-1\right)}^2=9$ dividing by 9 we get,

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

The equation is of the form $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b²}=1$.

Major axis is parallel to $x\text{-axis}$.

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

$a=3;b=1$

The vertices are $\left(h\pm a,k\right)=\left(-3\pm 3,1\right)$.

Hence, vertices: $\left(-6,1\right)\text{and}\left(0,1\right)$.

To find the value of $c$,

$c²=3²-1²=8;c=2\surd 2$

Foci are $\left(h\pm c,k\right)=\left(-3\pm 2\surd 2,1\right)$.

Hence the foci are $\left(-3+2\sqrt{2},1\right)\text{and}\left(-3-2\sqrt{2},1\right)$.

c.

To graph: The equation $x²+9y²+6x-18y+9=0$.

Solution:

Given:

$x²+9y²+6x-18y+9=0$

Calculation:

The graph of the equation of $x²+9y²+6x-18y+9=0$, is plotted using graphing tool.

Using the given information $x²+9y²+6x-18y+9=0$, we find the points of the equation as shown in the table below.

Using these points given above, we plot the points and draw a graph. We find center $\left(-3,1\right)$, foci $\left(-3-2\sqrt{2},1\right)\text{and}\left(-3+2\sqrt{2},1\right)$ and vertices $\left(-6,1\right)\text{and}\left(0,1\right)$. We see that the major axis is parallel to $x\text{-axis}$. We plot the graph for the equation using the graphing tool to verify it.