<s>[INST] Locus of a point - sum of distance from two points Find the locus of the point $P$ such that the sum of its distances from $(0,2)$ and $(0,-2)$ is $6$. 
What I did:
I tried using the distance formula, but I think that's too much of a task. There has to be an easier way. Please point me in the right direction. I do think that the locus is going to be an ellipse.
 [/INST] That is exactly how you are intended to find the locus:  let $(x,y)$ be a point in the locus $P$.  Then $$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6.$$  Squaring both sides and simplifying, we get $$\begin{align*} 36 &= x^2 + (y-2)^2 + 2 \sqrt{(x^2 + (y-2)^2)(x^2 + (y+2)^2)} + x^2 + (y+2)^2 \\ &= 2x^2 + 2y^2 + 8 + 2\sqrt{(x^2+y^2+4-4y)(x^2+y^2+4+4y)} .\end{align*}$$  Now move terms to the LHS, divide by $2$, and square:  $$(14-x^2-y^2)^2 = (x^2+y^2+4)^2 - (4y)^2$$  Now rearrange and use again the difference of squares factorization:  $$(4y)^2 = (x^2+y^2+4)^2 - (x^2+y^2-14)^2 = 18(2x^2+2y^2-10),$$ from which we obtain $$4y^2 = 9(x^2+y^2-5)$$ or $$9x^2+5y^2 = 45.$$
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