# Conjugate Diameters of an Ellipse

> Conjugate diameters of an ellipse in JEE Mathematics: diameter as locus of chord midpoints, slope relation mm' = −b²/a², Apollonius sum and area theorems.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/diameter-conjugate-ellipse
- Exam / course: JEE · Subject: Mathematics
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- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Conjugate Diameters of an Ellipse", PrepElephant, https://prepelephant.com/topics/jee/mathematics/diameter-conjugate-ellipse

## 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.