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.
On
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 $$9x^2+5y^2 = 45.$$
Hint: Start with $\sqrt{x^2+(y-2)^2}+\sqrt{x^2+(y+2)^2}=6$, and then rewrite this as $\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$.
Then square both sides, isolate the remaining square root, divide by -4, and then square both sides again.
Another way to do this is to first find points $(0,a)$ and $(0,-a)$ on the y-axis which satisfy the given condition, and then find points $(b,0)$ and $(-b,0)$ which also satisfy this condition.
Then you could use the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ for the ellipse, since it has center at the origin and axes which are parallel to the coordinate axes.