Finding the equation of an ellipse centered at (-1,3) and through the points (1,3) and (-1,4) I've been taking a computer graphics class and as a review we had some questions about some geometric problems. This one I can not seem to figure out though it seems that it should be straight forward.
I know that the standard equation of an ellipse is 
x^2/a^2 + y^2/b^2 = 1

From this we shift to its stated center and the equation becomes
(x+1)^2/a^2 + (y-3)^2/b^2 =1

But from here I am at a loss for where to go from here...

Substitute into your second equation (corrected with $b^2$ in the second denominator) the $x, y$ values for each of the points on the ellipse $(1, 3), (-1, 4)$, and you'll obtain two equations in two unknowns, from which you can solve for $a^2$ and $b^2$.
That is, evaluate $$\dfrac{(x+1)^2}{a^2} + \dfrac{(y-3)^2}{b^2} =1$$
at (1) $(1, 3)$ and again at (2) $(-1, 4)$. That will give you two equations in $a^2, b^2$, for which you can solve simultaneously.
$$\dfrac{(1 + 1)^2}{a^2} + \dfrac{(3-3)^2}{b^2} = 1 \implies \dfrac 4{a^2} = 1 \implies a^2 = 4\tag{1}$$
$$\dfrac{(-1 + 1)^2}{a^2} + \dfrac{(4-3)^2}{b^2} = 1 \implies \dfrac 1{b^2} = 1 \implies b^2 = 1\tag{2}$$
Now, use the general equation with your new-found values for $a^2, b^2$.
