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!
Best Answer
The sum of the distances to the foci is $6\sqrt{3}$. Via the "distance formula," this translates to $$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}.$$ Now let us grind. Rewrite as $$\sqrt{x^2+(y-3)^2}=6\sqrt{3}-\sqrt{x^2+(y+3)^2}.$$ Square both sides. We arrive at $$x^2+(y-3)^2=108 -12\sqrt{3}\sqrt{x^2+(y+3)^2} +x^2+(y+3)^2.$$ There is some cancellation. After the cancellation, we can divide through by $12$ and arrive at $$\sqrt{3}\sqrt{x^2+(y+3)^2}=9+y.$$ Square both sides again. Again, there is some cancellation, and we arrive at $$3x^2+2y^2=54.$$