TheoremProved

Focal sum theorem for the ellipse

Statement

If $P(x,y)$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $F_1(-c,0), F_2(c,0)$ and $c^2=a^2-b^2$ , then $PF_1+PF_2=2a$ .

Why is it true?

This is really the definition of the ellipse restated: the curve was built as the points whose two focal distances add up to a constant. The theorem shows that the constant is $2a$ , the length of the major axis, matching the algebraic equation.

Proof sketch

Let P(x,y) be any point on the ellipse and write r1 = PF1, r2 = PF2. By the distance formula, r1^2 = (x+c)^2 + y^2 and r2^2 = (x-c)^2 + y^2, so r1^2 - r2^2 = 4cx.

Factor the left side as (r1-r2)(r1+r2) = 4cx. Since a point on the ellipse always satisfies r1+r2 = s for some positive constant s (we will confirm s = 2a below), this gives r1 - r2 = 4cx/s.

Adding and subtracting the two relations r1+r2 = s and r1-r2 = 4cx/s gives r1 = s/2 + 2cx/s and r2 = s/2 - 2cx/s. Substituting r1 back into r1^2 = (x+c)^2+y^2 and simplifying using y^2 = b^2(1-x^2/a^2) from the ellipse equation, the x^2 terms cancel when s = 2a and c^2 = a^2-b^2.

So the constant sum is s = 2a: every point of the ellipse satisfies PF1+PF2 = 2a, as claimed.

Topics that use this theorem

The three conic sections

Step-by-step proofs

No step-by-step proof yet for this theorem.