Finding the equation of an ellipse given the foci I just wanted someone to check my solutions for this problem:
Find the equation of the ellipse with Foci (2,3) and (-1,1) where the distances from any point on the ellipse to the focus sums to 10. Write your answer in the form $Ax^2+Bxy+Cy^2=D$. Sketch the graph of the ellipse.
Solution:
$d_1$ = distance between point $P(x,y)$ and $(-1,1)$
$d_2$ =  distance between point $P(x,y)$ and $(2,3)$
$d_1 = \sqrt{(x+1)^2+(y-1)^2}$
$d_2 = \sqrt{(x-2)^2+(y-3)^2}$
$d_1+d_2=10$
$d_1=10-d_2$
$\sqrt{(x+1)^2+(y-1)^2} = 10 - \sqrt{(x-2)^2+(y-3)^2}$
$(x+1)^2+(y-1)^2=100-20\sqrt{(x-2)^2+(y-3)^2}+(x-2)^2+(y-3)^2$
$x^2+2x+1+y^2-2y+1=x^2-4x+4+y^2-6y+9+100-20\sqrt{(x-2)^2+(y-3)^2}$
$6x+4y-11=100\sqrt{(x-2)^2+(y-3)^2}$
Squaring both sides:
$36x^2+16y^2+48xy-132x-88y+121=100^2[(x-2)^2+(y-3)^2]$
Is this the correct answer and is it in the correct form? Or would I have to continue expanding terms out and simplifying?
Also, how would I then end up graphing it? I know that the Foci are (2,3) and (-1,1). I also know that $d_1+d_2=10$ so $2a=10$ so $a=5$.

HINT
Suppose you are given an ellipse as follows:
\begin{align*}
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1
\end{align*}
Then we can rotate and translate it in order to obtain the general equation of an ellipse in $\mathbb{R}^{2}$.
Let us start with rotating it first:
\begin{align*}
\frac{(\cos(\alpha)x - \sin(\alpha)y)^{2}}{a^{2}} + \frac{(\sin(\alpha)x + \cos(\alpha)y)^{2}}{b^{2}} = 1
\end{align*}
Now we can translate it with respect to the point $(x_{0},y_{0})$:
\begin{align*}
\frac{(\cos(\alpha)(x - x_{0}) - \sin(\alpha)(y - y_{0}))^{2}}{a^{2}} + \frac{(\sin(\alpha)(x - x_{0}) + \cos(\alpha)(y - y_{0}))^{2}}{b^{2}} = 1
\end{align*}
At the given example, $(x_{0},y_{0}) = (0.5,2)$ and $\tan(\alpha) = 2/3$.
Can you determine $a$ and $b$?
