Mathematics.

conic sections

Eccentricity of Conics

Analytic Geometry25 minDifficulty4 out of 10

You should know: conic sections

Overview

Eccentricity is a single non-negative real number, e, that measures how much a conic section deviates from being circular. It arises naturally from the unified focus-directrix definition of a conic: the ratio of a point's distance to a fixed focus versus its distance to a fixed directrix line is constant along the curve, and that constant ratio is exactly e. Because e alone determines the conic's type — e = 0 a circle, 0 < e < 1 an ellipse, e = 1 a parabola, e > 1 a hyperbola — it functions as a shape 'dial': turning it from 0 upward morphs a circle into increasingly elongated ellipses, then at e = 1 the curve opens up into a parabola, and beyond that into a hyperbola.

Intuition

Imagine gradually tilting the plane that slices a double cone. When the slice is perfectly perpendicular to the cone's axis, the cross-section is a circle — perfectly 'undistorted,' eccentricity 0. Tilt the plane slightly and the circle stretches into an ellipse; the more you tilt, the more elongated it gets, and e climbs toward 1. At the exact angle where the cutting plane becomes parallel to the cone's slanted side, the curve stops closing on itself entirely — it opens up into a parabola, e = 1. Tilt further still, so the plane also cuts the upper nappe of the double cone, and you get the two separate branches of a hyperbola, e > 1. Eccentricity is precisely a numeric readout of that tilt angle.

Formal Definition

Definition

For a focus F and directrix line ℓ, the eccentricity e is the constant ratio of distances defining the conic; for the ellipse and hyperbola it can also be computed directly from the axis lengths:

e=PFd(P,)for every point P on the conice = \frac{|PF|}{d(P,\ell)} \quad \text{for every point } P \text{ on the conic}
Focus-directrix definition
eellipse=ca=1b2a2,0e<1e_{\text{ellipse}} = \frac{c}{a} = \sqrt{1-\frac{b^2}{a^2}}, \qquad 0 \le e < 1

a = semi-major axis, b = semi-minor axis, c² = a² − b²

ehyperbola=ca=1+b2a2,e>1e_{\text{hyperbola}} = \frac{c}{a} = \sqrt{1+\frac{b^2}{a^2}}, \qquad e > 1

c² = a² + b² for a hyperbola

eparabola=1alwayse_{\text{parabola}} = 1 \quad \text{always}
ecircle=0(degenerate case: foci coincide, directrix recedes to infinity)e_{\text{circle}} = 0 \quad \text{(degenerate case: foci coincide, directrix recedes to infinity)}

Properties

Range by type

e=0circle,  0<e<1ellipse,  e=1parabola,  e>1hyperbolae=0 \Rightarrow \text{circle},\ \ 0<e<1 \Rightarrow \text{ellipse},\ \ e=1 \Rightarrow \text{parabola},\ \ e>1 \Rightarrow \text{hyperbola}

Scale invariance

Eccentricity depends only on shape, not size: scaling a conic by any factor k>0 leaves e unchanged\text{Eccentricity depends only on shape, not size: scaling a conic by any factor } k>0 \text{ leaves } e \text{ unchanged}

Limiting behavior

As e1 an ellipse becomes arbitrarily elongated; as e a hyperbola’s branches become nearly straight lines\text{As } e \to 1^- \text{ an ellipse becomes arbitrarily elongated; as } e \to \infty \text{ a hyperbola's branches become nearly straight lines}

Semi-latus rectum relation

=a(1e2) for an ellipse (half-width of the conic through a focus, perpendicular to the major axis)\ell = a(1-e^2) \text{ for an ellipse (half-width of the conic through a focus, perpendicular to the major axis)}

Applications

Kepler orbits have eccentricity determined by total energy: e=0 circular orbit, 0<e<1 bound elliptical orbit, e=1 parabolic escape trajectory, e>1 hyperbolic (unbound) flyby.

Worked Examples

  1. Identify a² = 25 and b² = 9 (a is under the larger denominator), so a = 5, b = 3.

    a=5,b=3a = 5,\quad b = 3
  2. Compute c using c² = a² − b² for an ellipse.

    c2=259=16    c=4c^2 = 25 - 9 = 16 \implies c = 4
  3. Compute e = c/a.

    e=45=0.8e = \frac{4}{5} = 0.8

Answer: e = 0.8 (a fairly elongated ellipse, since e is close to 1)

Practice Problems

Difficulty 4/10

Find the eccentricity of the ellipse x²/169 + y²/144 = 1.

Difficulty 5/10

A hyperbola has a = 6 and eccentricity e = 5/3. Find b² for its standard equation.

Difficulty 6/10

A comet's orbit is an ellipse with perihelion (closest distance to the sun) 0.5 AU and aphelion (farthest distance) 4.5 AU, with the sun at one focus. Find the orbit's eccentricity.

Common Mistakes

Common Mistake

Using c² = a² − b² for a hyperbola, copying the ellipse formula.

The relation is type-specific: for an ellipse c² = a² − b² (c < a, foci inside the curve), but for a hyperbola c² = a² + b² (c > a, foci outside the vertices), which is exactly why every hyperbola has e = c/a > 1.

Common Mistake

Believing eccentricity can be negative or that a 'more eccentric' ellipse means a smaller ellipse.

Eccentricity is always e ≥ 0, and it measures elongation (shape), not size — a huge ellipse and a tiny ellipse can have the identical eccentricity if their axis ratios b/a match.

Quiz

A conic with eccentricity e = 1 is:
For an ellipse, increasing the eccentricity toward 1 (while keeping a fixed) makes the ellipse:
Eccentricity is best described as a measure of:

Summary

  • Eccentricity e = |PF|/d(P, directrix) is a single dimensionless number classifying a conic's shape.
  • e = 0 is a circle, 0 < e < 1 an ellipse, e = 1 a parabola, e > 1 a hyperbola — a continuous 'shape dial'.
  • For an ellipse e = c/a with c² = a² − b²; for a hyperbola e = c/a with c² = a² + b² (so e > 1 always).
  • Eccentricity depends only on shape, not size, and governs orbital shapes, optical reflector design, and more.

References