Find the equation of the ellipse with center (-3,-2), a focus at (-3,1), and a vertex at (-3,3).
What do I do first?
$\displaystyle \frac{(x-h)^{2}}{b^{2}} + \frac{(y-k)^{2}}{a^{2}} = 1 $
This is the general form for the equation of the ellipse When the ellipse is vertical. When it is horizontal simply switch the $\displaystyle a$ and the $\displaystyle b$. The ellipse is centered at the point $\displaystyle (h,k)$, so plug in those values:
$\displaystyle \frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{a^{2}} = 1 $
Can you solve it from there?
I gave you this equation:
$\displaystyle
\frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{a^{2}} = 1
$
And I gave you this hint:
$\displaystyle a$ is the distance from the center to the vertex.
Now all you have to do is plug in values for $\displaystyle a$ and $\displaystyle b$.
Sorry, I just realized that you don't have a point on the ellipse. Instead you will have to use trig to find b.
You have a focus (-3,1). The definition of a focus is that the distance from the two focus' to any point on the ellipse is constant.
Eg. If you have point A and point B on the ellipse, the distance from f1 to A to f2 will be the same as f1 to B to f2. Am I explaining this clearly?
Since your one focus is (-3,1), your other will be (-3, -5). The distance from one focus to the vertex given, to the second focus is 10. Since the two focuses are equidistant from each other, the point b is therefore 5 away from both focuses. Use Pythagoras to solve for b in this manner:
$\displaystyle 5^2 = b^2 + 4^2$