Question

“what is the equation of a conic with vertices (-9,3) and (7,3) and foci (-5,3) and (3,3)”

Answer 1

This is an ellipse.

Equation of an ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Let's find the center point first. We can do this by finding out the midpoint between the vertices or the foci.

(-9+7)/2=-1

Center: (-1,3)

Now we need to find a and b (length along x and length along y).

We know "a". It is the distance between the center and the right or left bound.

a=7-(-1)=8

To find out "b", first see that...

b^2=a^2-(distance from center to foci)^2

foci distance=3-(-1)=4

b^2=8^2-4^2

b^2=64-16

b^2=48

b=4√3

answer: $\frac{(x+1)^2}{64}+\frac{(y-3)^2}{48}=1$