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.

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

Short answer

The center of ellipse is at $(-2,1)$, vertices are $(0,1)and(-4,1)$and the foci are $(-2\pm \sqrt{3},1)$

The required graph is

Step-by-step solution

Step 1. Given Information 

The given equation is$x^2+4x+4y^2-8y+4=0$

Step 2. Calculation

Rewrite the given equation in the general form of equation,

$$\begin{gathered} x^2+4x+4y^2-8y+4=0 \\ {x}^2+4x+4(y^2-2y)+4=0 \\ (x^2+4x+4)+4(y^2-2y)=0 \\ {(x+2)}^2+4(y^2-2y)+4=4 \\ {(x+2)}^2+4(y^2-2y+1)=4 \\ {(x+2)}^2+4{(y-1)}^2=4 \\ \frac{{(x+2)}^2}{4}+\frac{4{(y-1)}^2}{4}=\frac{4}{4} \\ \frac{{(x+2)}^2}{4}+{(y-1)}^2=1 \end{gathered}$$

The major axis is parallel to x-axis.

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

$$\begin{gathered} (h,k)=(-2,1) \\ a=2,b=1 \\ {c}^2=4-1 \\ {c}^2=3 \\ c=\sqrt{3} \end{gathered}$$

The vertices of major axis are at

$(h\pm a,k)=(-2\pm 2,1)i.e.,(0,1)and(-4,1)$

And the foci are at point$(h\pm c,k)=(-2\pm \sqrt{3},1)$

Step 3. Graph 

On plotting all the points, we get the required graph,