Hyperbola Standard Equations: Forms, Focus and Graph

A hyperbola is the curve of all points whose distances from two fixed foci differ by a constant, equal to 2a. Its standard equation is x²/a² − y²/b² = 1, where the minus sign between the fractions is what separates it from the ellipse. A hyperbola has two separate branches opening left and right, and its eccentricity e is always greater than 1.

The standard form reveals everything at a glance: centre at the origin, vertices at (±a, 0), foci at (±ae, 0), and the asymptotes the branches approach. Hyperbolas describe real shapes — comets swinging past the Sun, cooling towers of power stations — all read off the same equation.

Key Concepts

• Standard form (transverse axis on x): x²/a² − y²/b² = 1, centre (0, 0).
• Transverse axis: joins the vertices (±a, 0); length 2a; the foci lie on it.
• Conjugate axis: length 2b along the y-axis; the hyperbola never crosses it.
• Eccentricity: e = c/a with c² = a² + b² (plus sign — opposite of the ellipse); e > 1 always.
• Foci (±ae, 0); vertices (±a, 0); latus rectum 2b²/a.
• Conjugate hyperbola y²/b² − x²/a² = 1 opens up and down, with vertices (0, ±b) and foci (0, ±be).
• Rectangular hyperbola x² − y² = a² has a = b and e = √2.

Formulas

• Standard: x²/a² − y²/b² = 1 (branches left and right).
• Conjugate: y²/b² − x²/a² = 1 (branches up and down).
• Focal relation: c² = a² + b²; eccentricity: e = c/a > 1.
• Units: a, b, c are lengths; e is dimensionless. The positive term decides the transverse axis; a and b have no size order.

Worked Example

Find e, the foci, vertices and latus rectum of 9x² − 16y² = 144.

1. Standard form. Divide by 144: x²/16 − y²/9 = 1, so a² = 16 (a = 4), b² = 9 (b = 3).
2. Find c. c² = 16 + 9 = 25, so c = 5.
3. Eccentricity. e = 5/4 = 1.25, greater than 1 as required.
4. Vertices and foci. Vertices (±4, 0); foci (±5, 0).
5. Latus rectum. 2·9/4 = 4.5.

Answer: e = 1.25, vertices (±4, 0), foci (±5, 0), latus rectum 4.5.

Common Misconceptions

• "c² = a² − b², like the ellipse." Wrong — the hyperbola adds: c² = a² + b².
• "The hyperbola crosses the y-axis between its branches." Never — it meets only the x-axis at (±a, 0); the 2b segment is the conjugate axis, not part of the curve.
• "a must exceed b." No such rule; the transverse axis is decided by which term is positive.

Real-World Applications

• Comet paths: visitors from outside the solar system swing past the Sun on hyperbolic orbits (e > 1) and never return.
• Cooling towers: the hyperboloid shape gives strength from thin concrete and drives a natural upward draft.
• Radio navigation: the old LORAN system fixed positions from signal time differences — points of constant difference form a hyperbola.
• Sonic booms: a supersonic jet's shock cone meets the ground along a hyperbola.

To recap: x²/a² − y²/b² = 1 gives vertices (±a, 0) and foci (±ae, 0) with c² = a² + b², e > 1, latus rectum 2b²/a. Watch the signs — minus in the equation, plus in c².