The Hyperbola

Part 13, Chapter 13: Conic Sections in the Coordinate Plane

Learning objectives

  • Write the standard equation of a hyperbola
  • Find the asymptotes of a hyperbola
  • Locate the foci and identify the transverse axis
  • Sketch a hyperbola from its equation

If you change "sum of distances" to "difference of distances" in the ellipse definition, you get the hyperbola. The ellipse closes up into a single loop because adding two positive distances bounds the locus. The hyperbola opens out into two separate branches because the difference can grow without bound. Hyperbolas show up wherever something inversely proportional appears, the curves y=k/xy = k/x are hyperbolas in disguise, and they describe the paths of objects gravitationally slingshotting around the Sun fast enough to escape.

The standard equation

A hyperbola centred at the origin with transverse axis along the xx-axis satisfies

dfracx2a2dfracy2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1

The vertices are (pma,0)(\pm a, 0); the transverse axis has length 2a2a and joins the two vertices through the centre. Notice the minus sign, the only algebraic difference between an ellipse and a hyperbola.

Asymptotes

For large x|x| the 11 on the right becomes negligible and the equation tells you y/xtopmb/ay/x \to \pm b/a. So the hyperbola approaches two straight lines:

y=pmdfracbaxy = \pm \dfrac{b}{a} x

These are the asymptotes. A handy construction trick: draw the central rectangle with corners at (pma,pmb)(\pm a, \pm b). The asymptotes are the diagonals of this rectangle, and the hyperbola hugs them as it heads to infinity.

Foci and eccentricity

The foci sit on the transverse axis at distance cc from the centre, where

c=sqrta2+b2c = \sqrt{a^2 + b^2} (note the plus sign)

The defining property: for any point on the hyperbola, the absolute difference of distances to the two foci equals 2a2a. The eccentricity is e=c/a>1e = c/a > 1, always greater than 11, in contrast to the ellipse where e<1e < 1.

Where this shows up
  • Navigation: LORAN (long-range navigation) systems located ships by measuring time differences between radio pulses from two stations; the locus of constant time-difference is a hyperbola with the stations as foci.
  • Particle Physics: Rutherford's gold-foil experiment showed alpha particles deflected on hyperbolic paths in the field of the nucleus, the asymptotes give the scattering angle, and this is how the atomic nucleus was discovered.
  • Economics: Indifference curves in microeconomics often have a hyperbolic shape; the asymptotes encode the limit consumption rates of two substitute goods.

(Choose "hyperbola" mode. Slide aa and bb; the asymptotes tilt as you adjust their ratio. Watch the central rectangle widen and the foci move accordingly.)

Vertical transverse axis

If the equation is y2/a2x2/b2=1y^2/a^2 - x^2/b^2 = 1, the transverse axis is vertical: vertices at (0,pma)(0, \pm a), asymptotes y=pm(a/b)xy = \pm (a/b) x. Which term is positive tells you which axis is transverse.

Try it

  • Find the asymptotes and foci of x2/9y2/16=1x^2/9 - y^2/16 = 1.
  • A hyperbola has vertices (pm4,0)(\pm 4, 0) and asymptotes y=pmtfrac34xy = \pm \tfrac{3}{4} x. Write its equation.
  • What is the eccentricity of x2/4y2/12=1x^2/4 - y^2/12 = 1?

Pause: the rectangular hyperbola xy=1xy = 1 does not look like the standard form, but it is one, rotated by 45°45°. After rotation, it becomes (x)2/2(y)2/2=1(x')^2/2 - (y')^2/2 = 1. The next section explores exactly this kind of axis rotation.

A trap to watch for

The sign in c=sqrta2+b2c = \sqrt{a^2 + b^2} uses addition, opposite to the ellipse where it is subtraction. Beginners constantly write c=sqrta2b2c = \sqrt{a^2 - b^2} for hyperbolas and get nonsense (often imaginary numbers when b>ab > a). Anchor it geometrically: an ellipse is contained inside its bounding rectangle, so its foci must be closer to the centre than the vertices, c<ac < a, forcing subtraction. A hyperbola opens out beyond its vertices, so its foci are farther than the vertices, c>ac > a, forcing addition.

A second trap: foci on the wrong axis. For x2/a2y2/b2=1x^2/a^2 - y^2/b^2 = 1 the transverse axis is horizontal and the foci are at (pmc,0)(\pm c, 0). For y2/a2x2/b2=1y^2/a^2 - x^2/b^2 = 1 they are at (0,pmc)(0, \pm c). Read the sign pattern, not the size of the denominators.

What you now know

You can write the standard hyperbola equation, find the asymptotes by inspection, locate the foci using c=sqrta2+b2c = \sqrt{a^2 + b^2}, compute the eccentricity (always >1> 1), and distinguish horizontal- and vertical-transverse-axis hyperbolas. The next section ties together all four conics and shows how to rotate the axes so that a tilted conic equation simplifies to standard form.

Quick check

Mark section complete →

References

  • Lang, S. (1971). Basic Mathematics. Springer. Chapter 12, §4, the hyperbola as the difference-of-distances locus.
  • Apostol, T. M. (1969). Calculus, Volume 2. Wiley. Chapter 13: conic sections, including the unified focus-directrix-eccentricity view.
  • Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage. Section 10.5: conic-section properties and applications.

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