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Mathematics · Ch 15 — Hyperbola

The Conjugate Hyperbola

15.7

The Conjugate Hyperbola

Every hyperbola has a natural partner curve called its conjugate hyperbola: the one obtained by swapping the roles of the transverse and conjugate axes — i.e., the curve whose transverse axis is the original curve's conjugate axis, and vice versa. Algebraically, given

S≡x2a2−y2b2−1=0,S \equiv \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} - 1 = 0,

its conjugate hyperbola is simply the same expression with the sign of the constant flipped:

S′≡x2a2−y2b2+1=0⟺y2b2−x2a2=1.S' \equiv \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} + 1 = 0 \qquad \Longleftrightarrow \qquad \dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1.

Each is the conjugate of the other — the relationship is symmetric, which is why it's called "conjugate" rather than something one-directional like "derived from."

Because S′S' has the y2y^2 term positive, it opens up and down instead of left and right: its transverse axis lies along the yy-axis with length 2b2b, and its conjugate axis lies along the xx-axis with length 2a2a — precisely the swap the name promises. Its own eccentricity is computed the same way as before, just with aa and bb trading places:

e′=a2+b2b2,e' = \sqrt{\dfrac{a^2+b^2}{b^2}},

giving foci at (0,±be′)(0,\pm be') and directrices y=±b/e′y=\pm b/e'.

A neat relationship links the two eccentricities: since e=(a2+b2)/a2e=\sqrt{(a^2+b^2)/a^2} and e′=(a2+b2)/b2e'=\sqrt{(a^2+b^2)/b^2},

1e2+1e′2=a2a2+b2+b2a2+b2=1.\dfrac{1}{e^2} + \dfrac{1}{e'^2} = \dfrac{a^2}{a^2+b^2} + \dfrac{b^2}{a^2+b^2} = 1.

So a hyperbola and its conjugate can never both be "close to a right angle asymptote" or both be "very flat" independently — their eccentricities are locked together by this identity, which is a handy check in problems that give you one eccentricity and ask for the other.

Geometrically, if you sketch S=0S=0 and S′=0S'=0 on the same axes, you get four hyperbola-like branches opening in the four "compass" directions (up, down, left, right) from a shared centre, all four asymptotic to the very same pair of lines found in Section 5.6 — which is exactly what the identity S+S′=2AS+S'=2A was capturing. …