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Disclaimer: This title was hard to formulate. Edits welcome.

Problem:

Given foci $$F_1 = (1,0)$$ $$F_2 = (3,0)$$

of a conic section, find the equation for all points $P$ that satisfy $$|PF_1| + |PF_2| = 6$$

My attempt:

I tried going about it algebraically. Letting $$P = (x,y)$$

I formulated

$$|PF_1| = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(x-1)^2 + y^2}$$

and likewise for $|PF_2|$ and then rewriting the above equation, but this turned out to be a mess, and even when I resorted to WolframAlpha, it turned out to be far uglier than I believe is intended.

I expect there is a more elegant solution here, but I'm not able to find it.

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It's not extremely difficult, just take care not to have a square root everywhere: $$ \sqrt{(x-1)^2+y^2}+\sqrt{(x-3)^2+y^2}=6 \\ \sqrt{(x-1)^2+y^2} = 6 - \sqrt{(x-3)^2+y^2} \\ (x-1)^2+y^2 = 36 + (x-3)^2+y^2 - 12\sqrt{(x-3)^2+y^2} \\ x^2-2x+1+y^2-36-(x^2-6x+9)-y^2 = -12\sqrt{(x-3)^2+y^2} \\ -12\sqrt{(x-3)^2+y^2} = 4x-44 \\ 9\left[x^2-6x+9+y^2\right] = (x-11)^2 \\ 8x^2-32x+9y^2=40 \\ 8(x-2)^2+9y^2=72 \\ (x-2)^2/9+y^2/8=1 \\ \left(\frac{x-2}{3}\right)^2+\left(\frac{y}{2\sqrt{2}}\right)^2 = 1 $$ An ellipse with midpoint $(2,0)$ and axes $(3,2\sqrt{2})$ :
enter image description here

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  • $\begingroup$ That was beautiful. But what does it mean that it has axes $(3, 2\sqrt2)$? $\endgroup$ – Alec Jan 19 '15 at 12:38
  • $\begingroup$ At Wikipedia, see Ellipse, elements of an ellipse and especially take notice of major axis and minor axis. $\endgroup$ – Han de Bruijn Jan 19 '15 at 18:05
  • $\begingroup$ Ah, so by axes $(3, 2\sqrt2)$, there refer to the length of the major and minor radius respectively? I can see by plotting it that the major axis is indeed of radius 3, but the minor axis has a radius of 2. $\endgroup$ – Alec Jan 19 '15 at 18:33
  • $\begingroup$ Then my plotting differs from yours: replace the question mark by a number. $\endgroup$ – Han de Bruijn Jan 20 '15 at 10:57

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