Parabola Tangents, Normals & Chords: JEE Main Guide

The parabola is the most frequently tested conic in JEE Main Mathematics, and within parabola questions, tangent and normal problems are the highest-frequency sub-type. Students who know the parametric form of the parabola and can write tangent, normal, and chord equations in their parametric forms solve parabola questions in half the time of those who use the standard form and Cartesian equations. This guide builds that parametric fluency alongside the Cartesian approach.

Standard Parabola and Parametric Representation

Standard parabola y² = 4ax: focus (a,0), directrix x = −a, vertex (0,0), axis y=0. Parametric form: a point on the parabola can be written as (at², 2at) for parameter t ∈ ℝ. Equivalently, (x,y) = (at², 2at). This parametric form is the fastest tool for parabola problems because: tangent at parameter t is ty = x + at²; normal at parameter t is y + tx = 2at + at³. Both equations are completely determined by a single parameter t — much faster than the Cartesian approach.

Other standard parabolas: y² = −4ax (opens left), x² = 4ay (opens up), x² = −4ay (opens down). For a shifted parabola (h,k) as vertex: (y−k)² = 4a(x−h). The transformations of all properties are identical — replace x by (x−h) and y by (y−k) throughout. JEE uses shifted parabolas once every two sessions to test whether students can handle the shifted case. Take a free Conic Sections mock focusing on parabola questions. For the complete conic overview, see our conic sections guide.

Tangent Equations: Three Forms

Tangent to y² = 4ax at point (x₁,y₁): yy₁ = 2a(x+x₁) — the "T=0" form. Tangent at parametric point (at², 2at): ty = x + at². Tangent with slope m: y = mx + a/m (for any slope m ≠ 0). The condition of tangency of y = mx+c to y² = 4ax is c = a/m — extremely useful for "find the value of c if y=mx+c is tangent to..." The point of contact of the tangent y=mx+a/m is (a/m², 2a/m).

Director circle of a parabola: locus of points from which the two tangents to the parabola are perpendicular. For y² = 4ax: tangents from (h,k) have slopes m₁ and m₂ where m₁·m₂ = k²/(4ah)... actually for parabolas the locus of perpendicular tangent intersection is the directrix x = −a. JEE tests: "The tangents at two points on y²=4ax meet at P. Show that if the tangents are perpendicular, P lies on the directrix." — use the product of slopes = −1 and slopes from parametric points condition to derive. For related results on normals and chords, see our parabola complete guide.

Normal Equations and the Normal to the Parabola

Normal at (at², 2at) on y²=4ax: y + tx = 2at + at³. The normal has slope −t (negative reciprocal of tangent slope t). For a normal with slope m: the foot of normal is at parameter t = −m, so the foot is (am², −2am) and the normal equation is y = mx − 2am − am³. This last equation — y = mx − 2am − am³ — is the equation of the normal with slope m, and it is the most frequently used normal equation in JEE. Condition for a line y = mx+c to be normal to y²=4ax: c = −2am − am³.

Three normals from an external point: a point (h,k) is on three normals to y²=4ax if the cubic t³ + (2a−h)/a · t + k/a = 0 has three real roots (t₁, t₂, t₃). The sum of these roots t₁+t₂+t₃ = 0 — this is Vieta's formula for the cubic and is a key JEE one-liner: "The sum of slopes of three normals from a point to a parabola is zero." The locus of intersection of two normals at right angles to each other: y² = a(x−3a). For normal-chord-tangent interplay in problems, see our tangents and normals guide.

Focal Chord, Chord of Contact, and the Chord with Given Midpoint

Focal chord: a chord passing through the focus (a,0). For a focal chord joining (at₁², 2at₁) and (at₂², 2at₂): t₁t₂ = −1 (product of parameters = −1). Semi-latus rectum: the perpendicular through focus has endpoints at t = ±1, i.e., (a, 2a) and (a, −2a), so the semi-latus rectum length = 2a. The harmonic mean of the distances of the two endpoints of a focal chord from the vertex: = semi-latus rectum = 2a. This "harmonic mean = semi-latus rectum" result is a very standard JEE integer question.

Chord of contact from external point (x₁,y₁): yy₁ = 2a(x+x₁) (same T=0 formula as the point-on-parabola tangent, but now (x₁,y₁) is external). Chord with midpoint (h,k): using T = S₁: ky − 2a(x+h)/1 = k² − 4ah... the derivation gives slope of the chord = 2a/k. JEE asks "find the equation of the chord of y²=4ax whose midpoint is (h,k)" — use the slope 2a/k through (h,k) to get: y−k = (2a/k)(x−h). For the full analytical geometry chapter, see our coordinate geometry guide.