<s>[INST] Find the equation of the locus of a point which has its sum of distance from (0,3) and (0,-3) equal to 8. 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.
 [/INST] 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!
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