conic sectionslocus
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.
Animation of the specified locus, for $a = 1/2$:
The solution is found! The locus (u,v) is very complicate in v. Help from Mathematica.
$ v=\left(\frac{a}{b x}(u-x)+\frac{b}{a}\right)\sqrt{a^2-x^2} $ $ x=\frac{u}{2}+\frac{1}{2} \surd \left(u^2-\frac{2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}+\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)+\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t\right)+\frac{1}{2} \surd \left(2 u^2-\frac{4 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}-\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)-\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t+\left(-\frac{16 a^4 u}{a^2-b^2}+8 u^3-\frac{8 u \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{a^2-b^2}\right)/\left(4 \surd \left(u^2-\frac{2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}+\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)+\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t\right)\right)\right) $ $ t=\left(108 a^8 \left(a^2-b^2\right) u^2-108 a^4 u^2 \left(a^2 u-b^2 u\right)^2+72 a^4 \left(a^2-b^2\right) u^2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+36 a^4 u \left(a^2 u-b^2 u\right) \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^3+\surd \left(-4 \left(-12 a^4 \left(a^2-b^2\right) u^2+12 a^4 u \left(a^2 u-b^2 u\right)+\left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^2\right)^3+\left(108 a^8 \left(a^2-b^2\right) u^2-108 a^4 u^2 \left(a^2 u-b^2 u\right)^2+72 a^4 \left(a^2-b^2\right) u^2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+36 a^4 u \left(a^2 u-b^2 u\right) \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^3\right)^2\right)\right)^{1/3} $
Best Answer
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 $$9x^2+5y^2 = 45.$$