Conjugate Diameters of an Ellipse
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Direct answer
A diameter of an ellipse is the locus of midpoints of a family of parallel chords — every diameter passes through the centre, and every chord through the centre is itself a diameter. For x²/a² + y²/b² = 1, the midpoints of chords with slope m lie on the line y = −(b²/a²m)x. Two diameters are conjugate when each bisects all chords parallel to the other; their slopes satisfy mm' = −b²/a². The ends of conjugate diameters sit at eccentric angles differing by π/2, and two classical results follow: the squares of conjugate semi-diameters always add to a² + b² (Apollonius I), and the tangents at their four ends enclose a parallelogram of constant area 4ab (Apollonius II). The major and minor axes are the limiting perpendicular conjugate pair.
What you must remember
- Definition: a diameter is the locus of midpoints of parallel chords; a chord through the centre is a diameter, and its conjugate bisects every chord drawn parallel to it.
- Slope relation: the diameter y = m₁x has conjugate y = m₂x iff m₁m₂ = −b²/a² — a negative product, so two finite-slope conjugate diameters always lean opposite ways.
- Chord-midpoint origin: substituting y = mx + c into the ellipse gives midpoint x = −a²mc/(a²m² + b²), y = b²c/(a²m² + b²), so y/x = −b²/(a²m) independent of c — the slope relation in one division.
- Parametric pair: ends at eccentric angles θ and θ + π/2 are P(a cosθ, b sinθ) and D(−a sinθ, b cosθ); then CP² + CD² = a² + b² term by term, and the parallelogram with adjacent sides CP, CD has area ab.
- Apollonius II: the parallelogram formed by the four tangents at the ends of two conjugate diameters has constant area 4ab — the ellipse's replacement for the circumscribed rectangle.
- Equal pair: exactly one pair of conjugate diameters is equal — the equi-conjugate pair along y = ±(b/a)x, each semi-diameter of length √((a² + b²)/2).
- Circle limit: when a = b the relation reads mm' = −1, so every conjugate pair is perpendicular; for a true ellipse only the axes manage perpendicularity, as the m → 0, m' → ∞ case.
Where the slope relation comes from
Write a chord of x²/a² + y²/b² = 1 as y = mx + c and substitute: (a²m² + b²)x² + 2a²mcx + a²(c² − b²) = 0. The roots are the chord's end abscissas, so the midpoint has x = −a²mc/(a²m² + b²) and y = mx + c = b²c/(a²m² + b²). Dividing gives y/x = −b²/(a²m) — no c anywhere, which is precisely the statement that all midpoints of slope-m chords lie on one line through the centre. Conjugacy is symmetric by construction: midpoints of chords parallel to y = m'x lie on y = −(b²/a²m')x, and identifying the two loci yields mm' = −b²/a². The parametric check seals it: with ends at θ and θ + π/2, CP² = a²cos²θ + b²sin²θ and CD² = a²sin²θ + b²cos²θ add to a² + b², while the cross product |CP × CD| works out to ab — both Apollonius results in two lines of algebra.
Where students slip
The sign is the trap: options routinely include m₁m₂ = +b²/a², and habit from the hyperbola — where the product genuinely is +b²/a² — pushes students there. The ellipse's minus is non-negotiable. Second, conjugate is not perpendicular: for finite slopes, perpendicularity would force a = b, so among genuine ellipses only the major–minor axis pair is perpendicular, as the limiting case. Third, Advanced papers love the eccentric-angle wording: the parameters of conjugate ends differ by π/2, but the geometric angles that CP and CD make with the major axis do not — the eccentric angle is measured on the auxiliary circle x² + y² = a², not at the ellipse's centre.
Frequently asked questions
What is the conjugate of the diameter y = mx?
The line y = −(b²/a²m)x; it bisects every chord parallel to y = mx, and the relation holds symmetrically in both directions.
Can two conjugate diameters both have positive slope?
No — mm' = −b²/a² < 0 forces opposite signs for finite-slope pairs; the axes form the special perpendicular case where one slope is zero and the other infinite.
What stays constant across all conjugate pairs?
CP² + CD² = a² + b² for every conjugate pair (Apollonius I), and the tangent parallelogram at their four ends always has area 4ab (Apollonius II).
Is there more than one pair of equal conjugate diameters?
No — the equi-conjugate pair along y = ±(b/a)x, each semi-diameter measuring √((a² + b²)/2), is the only one.
How does the relation change for the hyperbola x²/a² − y²/b² = 1?
The same derivation gives m₁m₂ = +b²/a², and a real conjugate pair requires |m| < b/a — beyond that the line cuts no chords to bisect.