Standard equations of a hyperbola

With the centre at the origin and foci (±c, 0), the condition PF₁ − PF₂ = 2a, squared twice, simplifies to x²/a² − y²/(c² − a²) = 1, that is x²/a² − y²/b² = 1. Conversely every point of this curve has a difference of distances equal to 2a.

With the foci on the y-axis the equation is y²/a² − x²/b² = 1. A hyperbola with a = b is called an equilateral hyperbola.

From x²/a² ≥ 1, either x ≤ −a or x ≥ a: no part of x²/a² − y²/b² = 1 lies between the lines x = −a and x = a, so the curve has two separate branches and does not meet the y-axis.

A hyperbola is symmetric about both axes. The foci lie on the transverse axis, which is set by the variable whose square carries the positive sign: x²/9 − y²/16 = 1 has a transverse axis of length 6 along the x-axis, and y²/25 − x²/16 = 1 one of length 10 along the y-axis.

The latus rectum (the chord through a focus perpendicular to the transverse axis) has length 2b²/a, the same expression as for the ellipse.

x²/9 − y²/16 = 1: a = 3, b = 4, c = √(9 + 16) = 5; foci (±5, 0), vertices (±3, 0), e = 5/3, latus rectum 32/3.

y² − 16x² = 16 becomes y²/16 − x²/1 = 1: a = 4, b = 1, c = √17; foci (0, ±√17), vertices (0, ±4), e = √17/4, latus rectum 1/2.

Foci (0, ±12) and latus rectum 36 give 2b²/a = 36, so b² = 18a; with c² = a² + b², 144 = a² + 18a, so a = 6 (the root −24 is rejected) and b² = 108: y²/36 − x²/108 = 1, or 3y² − x² = 108.

Foci (0, ±3) and vertices (0, ±√11/2) give a² = 11/4 and b² = 9 − 11/4 = 25/4, so the hyperbola is 100y² − 44x² = 275.