I know the answer but due to various expressions under root, I'm unable to reach there.
Is there something which I'm missing to make the solution easier? By the way the answer given is $$\frac{x^2}{16} + \frac{y^2}{7}=1$$ which isn't equal to the one I'm getting, in fact, it is nowhere close to it.
My answer: (I just used the one plane distance formula here)$$30y^2+64^2+12(64)y=16^2•(x^2+y^2)+16^2(9+6y)$$ I see no point in expanding this.
On
The center of an ellipse is always the midpoint of the foci, so this ellipse must be centered at the origin.
In addition, the line on which the foci fall is always the major axis, while the perpendicular line is the $y$-axis. In this case, these are the $y$ and $x$ axis respectively.
So, we now know that the ellipse has the equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ Where $a$ is half the minor axis, i.e. the $x$-intercept and $b$ is half the major axis, i.e. the $y$-intercept.
We can now figure out the equation for the ellipse by noting that both $(a,0)$ and $(0,b)$ must be on the ellipse. In particular, we have $$ \sqrt{a^2 + 3^2} + \sqrt{a^2 + 3^2} = 8\\ (b-3) + (b + 3) = 8 $$
We start from $$\sqrt{x^2+(y+3)^2}+\sqrt{x^2+(y-3)^2}=8.$$ Bring $\sqrt{x^2+(y-3)^2}$ to the other side and square. We get $$x^2+(y+3)^2=64-16\sqrt{x^2+(y-3)^2}+x^2+(y-3)^2.$$ A little algebra brings us to $$12y-64=-16\sqrt{x^2+(y-3)^2}.$$ Minor cancellation gives $$4\sqrt{x^2+(y-3)^2}=16-3y.$$ Square. We get some cancellation and arrive at $$16x^2+7y^2=16^2-(16)(9)=(16)(7).$$ Finally, divide both sides by $(16)(7)$.
Remark: You may find it easier to deal with the slightly more general equation $$\sqrt{x^2+(y+p)^2}+\sqrt{x^2+(y-p)^2}=k.$$ The steps are exactly the same, but we don't get the distraction of numbers. I certainly find it easier!