Question

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.

$x^2+3y^2-12y+9=0$

Short answer

The center of ellipse is at $(0,2)$, vertices are $(\sqrt{3},2)and(-\sqrt{3},2)$and the foci are $(\sqrt{2},2)and(-\sqrt{2},2)$

The required graph is

Step-by-step solution

Step 1. Given Information 

The given equation is$x^2+3y^2-12y+9=0$

Step 2. Calculation

Rewrite the given equation in the general form of equation of ellipse,

$$\begin{gathered} x^2+3y^2-12y+9=0 \\ {x}^2+3(y^2-4y+3)=0 \\ {x}^2+3(y^2-4y+4)=3 \\ {x}^2+3{(y-2)}^2=3 \\ \frac{x^2}{3}+\frac{3{(y-2)}^2}{3}=\frac{3}{3} \\ \frac{x^2}{3}+{(y-2)}^2=1 \end{gathered}$$

The major axis is parallel to x-axis.

Now, on comparing the obtained equation with the general form of equation, we get,

$$\begin{gathered} (h,k)=(0,2) \\ {a}^2=3 \\ {b}^2=1 \\ {c}^2=a^2-b^2 \\ {c}^2=3-1 \\ {c}^2=2 \\ c=\sqrt{2} \end{gathered}$$

The vertices of major axis are at $(h\pm a,k)=(0\pm \sqrt{3},2)$

And the foci are at point$(h\pm c,k)=(0\pm \sqrt{2},2)$

Step 3. Graph

On plotting all the points on the graph, we get the required graph as follows,