Question

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

Ellipse; foci at $(-3,0)$and $(3,0)$; vertex at$(4,0)$

Short answer

The Equation of the ellipse is$\frac{x^2}{16}+\frac{y^2}{7}=1$

Step-by-step solution

Step 1. Given Information

Foci at $(-3,0),(3,0)$

Vertex at$(4,0)$

Step 2. Calculation

Since $$\begin{gathered} (h+c,k)=(-3,0) \\ (h-c,k)=(3,0) \end{gathered}$$

and $(h+a,k)=(4,0)$

Hence $h=0,c=-3,k=0\&a=4$

Then $$\begin{gathered} b^2=a^2-c^2 \\ =16-9 \\ =7 \end{gathered}$$

Then the equation is

$\frac{x^2}{16}+\frac{y^2}{7}=1$

Step 3. Graph