Normal to rectangular hyperbola xy=c²

Short answer

The normal at t (ct, c/t) on xy = c² has slope t². Substituting the second intersection t₁ gives t³ t₁ = −1.

Prove that the normal at a point t of the rectangular hyperbola xy = c² meets the curve again at a point t₁ such that t³t₁ = −1.

For the rectangular hyperbola xy = c², a point t is (ct, c/t).

Differentiating xy = c²: x dy/dx + y = 0, so dy/dx = −y/x = −(c/t)/(ct) = −1/t².

The normal has slope t² (negative reciprocal).

y − c/t = t²(x − ct)

Let the normal meet the curve again at t₁, the point (ct₁, c/t₁). Substitute:

c/t₁ − c/t = t²(ct₁ − ct)

c(t − t₁)/(t t₁) = c t²(t₁ − t)

Since t ≠ t₁, divide by c(t − t₁):

1/(t t₁) = −t²

t³ t₁ = −1

Hence the normal at t meets the curve again at the point t₁ = −1/t³. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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