Proof of the canonical equation

We derive the canonical equation directly from the definition of locus, translating the condition on the sum of the distances into coordinates.

Proof — Canonical equation

From the definition of locus, setting $P(x;y)$: $\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a.$ The two distances are both $\ge 0$, but the sum of two radicals cannot be squared “directly”: one of them must first be isolated. $\sqrt{(x+c)^2 + y^2} = 2a - \sqrt{(x-c)^2 + y^2}.$ We square both sides (the right-hand side must be $\ge 0$, that is $2a \ge \sqrt{(x-c)^2+y^2}$, which is true by the definition of the ellipse): $(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 + y^2.$ Expanding and cancelling the common terms: $4cx - 4a^2 = -4a\sqrt{(x-c)^2+y^2} \iff a^2 - cx = a\sqrt{(x-c)^2+y^2}.$ We square again (both sides $\ge 0$ inside the region of definition): $a^4 - 2a^2 cx + c^2 x^2 = a^2\bigl[(x-c)^2 + y^2\bigr] = a^2 x^2 - 2a^2 cx + a^2 c^2 + a^2 y^2.$ Simplifying and rearranging: $a^4 - a^2 c^2 = (a^2 - c^2)x^2 + a^2 y^2.$ Setting $b^2 = a^2 - c^2$ (positive because $a > c$) and dividing everything by $a^2 b^2$ we arrive at the canonical form. $\blacksquare$

Warning — Double squaring

The only real difficulty in the derivation is that one has to square twice, because after the first squaring a radical still appears. Once the remaining radical is isolated and the second squaring is done, the computations simplify magically.

Links

Topics: Ellipse
Concepts: Ellipse · Canonical equation · Locus
Methods: Canonical ellipse
Skills: Proving · Analytic geometry · Solving equations