TheoremProved

Focal difference theorem for the hyperbola

Statement

If $P(x,y)$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ 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?

Just as the ellipse is built from a constant focal sum, the hyperbola is built from a constant focal difference. This theorem checks that the algebraic equation with a minus sign between the squared terms produces that constant-difference behavior, and pins the constant down to $2a$ .

Proof sketch

Let P(x,y) be a point on the right branch (x >= a), and set r1 = PF1, r2 = PF2. As before, r1^2 - r2^2 = (x+c)^2 - (x-c)^2 = 4cx, so (r1-r2)(r1+r2) = 4cx.

On the hyperbola the difference r1-r2 stays constant in sign and, by symmetry of the two branches, equal in size to some constant d. Writing r1+r2 = 4cx/d and combining with r1-r2 = d gives r1 = d/2 + 2cx/d.

Substitute this into r1^2 = (x+c)^2+y^2 and use y^2 = b^2(x^2/a^2-1) from the hyperbola's equation. Expanding and collecting the x^2 coefficients, the equation is satisfied for all x when d = 2a and c^2 = a^2+b^2.

Hence r1-r2 = 2a on the right branch, and by the mirror-symmetric argument r2-r1 = 2a on the left branch, so in all cases |PF1-PF2| = 2a.

Topics that use this theorem

The three conic sections

Step-by-step proofs

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