91Ó°ÊÓ

The given equation is:

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

$$\begin{gathered} x^2+9y^2+6x-18y+9=0 \\ \left(x^2+6x+9\right)+9\left(y^2-2y+1\right)=9 \\ {\left(x+3\right)}^2+9{\left(y-1\right)}^2=9 \\ \frac{{\left(x+3\right)}^2}{9}\frac{+9{\left(y-1\right)}^2}{9}=\frac{9}{9} \\ \frac{{\left(x+3\right)}^2}{9}\frac{+{\left(y-1\right)}^2}{1}=1 \end{gathered}$$

The resultant equation of an ellipse of the form $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

By comparing these two equations, we can find that the center of the ellipse lies at point $\left(h,k\right)=\left(-3,1\right)$.

We have seen that the greater denominator, 9 is located under the x variable.

Therefore the ellipse has its major axis parallel to the x-axis.

The major axis is $y=1$.

From the equation we have $a^2=9\mathrm{and}{b}^2=1$.

The vertices of an ellipse with major axis parallel to the x-axis and center at $\left(h,k\right)$are $\left(hÂ\pm a,k\right)=\left(-3Â\pm 3,1\right)$orlocalid="1647101481558" $\left(0,1\right)\mathrm{and}\left(-6,1\right)$.

$$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=9-1 \\ {c}^2=8 \\ c=2\sqrt{2} \end{gathered}$$

The foci are given by $\left(hÂ\pm c,k\right)$. Therefore the foci of the ellipse are atlocalid="1647101423993" $\left(-3+2\sqrt{2},1\right)\mathrm{and}\left(-3-2\sqrt{2},1\right)$.

Now graph the ellipse and label the vertices, centre, and foci: