Question

In Problems 11–18, find an equation of the conic described. Graph the equation by hand.

Ellipse; foci at $(-4,2)$and $(-4,8)$; vertex at$(-4,10)$

Short answer

$\frac{{(x+4)}^2}{16}+\frac{{(y-5)}^2}{25}=1$

Step-by-step solution

Step 1. Given Information 

Foci at $(-4,2),(-4,8)$

Vertex at$(-4,10)$

Step 2. Calculation 

Since $$\begin{gathered} (h,k+c)=(-4,2) \\ (h,k-c)=(-4,8) \end{gathered}$$$$\begin{gathered} (h,k+c)=(-4,2) \\ (h,k-c)=(-4,8) \end{gathered}$$

and $(h,k+a)=(-4,10)$

Then we get $h=-4,k=5,c=-3\&a=5$

Then $$\begin{gathered} b^2=a^2-c^2 \\ =25-9 \\ =16 \end{gathered}$$

Then the equation of the ellipse is

$\frac{{(x+4)}^2}{16}+\frac{{(y-5)}^2}{25}=1$

Step 3. Graph