Getting the equation of an ellipse using the constant and the foci Find the equation of the ellipse with the foci at (0,3) and (0, -3) for which the constant referred to in the definition is $6\sqrt{3}$
So I'm quite confused with this one, I know the answer is $3x^2+2y^2=54$ through trial and errror, but is there any way to solve it without trial and error? 
The constant referred to is the sum of the distances from the foci right? How do I use it? I've been experimenting with how you can get $6 \sqrt3$ from the equation like using distance formula etc. I'm kinda stuck with this one, can someone please explain thanks!

Let $(x,y)$ be a point of ellipse then from definition 
$$\sqrt{(x-0)^2+(y-3)^2}+\sqrt{(x-0)^2+(y+3)^2}=6\sqrt{3}$$
$$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}$$
$$2x^2+2y^2+18+2\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=108$$
$$x^2+y^2+9+\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=54$$
$$(x^2+y^2-45)^2=(x^2+(y+3)^2)(x^2+(y-3)^2)$$
$$(x^2+y^2)^2-90x^2-90y^2+45^2=x^4+x^2(y+3)^2+x^2(y-3)^2+(y^2-9)^2$$
$$2x^2y^2+y^4-90x^2-90y^2+45^2=x^2(2y^2+18)+y^4-18y^2+81$$
$$-90x^2-90y^2+45^2=18x^2-18y^2+81$$
$$108y^2+72x^2=36\cdot54$$
$$3x^2+2y^2=54$$
