Tangent and Normal to a Parabola: Formulas and Examples

The tangent to the parabola y² = 4ax at the point (x₁, y₁) is T = 0, i.e. yy₁ = 2a(x + x₁). The normal there is the perpendicular line, conveniently written with the parameter t (where the point is (at², 2at)) as y = −tx + 2at + at³. A tangent of slope m to y² = 4ax is y = mx + a/m — a compact form that solves "tangent with given slope" problems in one step.

These formulas matter because the tangent's point of contact and slope appear in reflection problems (a ray from the focus reflects parallel to the axis) and in optimisation (shortest normal from a point to the curve). The parametric forms turn tangent/normal questions into single-variable algebra in t, which is far cleaner than eliminating between line and curve equations.

Key Concepts

• Parametric point: every point of y² = 4ax is (at², 2at) for some real t; note y₁ = 2at₁, so t₁ = y₁/(2a).
• Tangent at t: ty = x + at². Normal at t: y = −tx + 2at + at³.
• Tangent in slope form: y = mx + a/m touches y² = 4ax for every non-zero m; the contact point is (a/m², 2a/m).
• Chord of contact: from an external point, T = 0 again gives the chord joining the two contact points, exactly as for circles.

Formulas

• Tangent at (x₁, y₁): yy₁ = 2a(x + x₁). Tangent at parameter t: ty = x + at².
• Normal at t: y = −tx + 2at + at³.
• Slope-m tangent: y = mx + a/m, valid for m ≠ 0; contact point (a/m², 2a/m).
• Units: a, x, t-related terms carry consistent length units; m is dimensionless.
• Assumption: (x₁, y₁) lies on the parabola for the tangent formula; m ≠ 0 for the slope form (vertical tangents need separate handling).

Worked Example

For y² = 8x, find the tangent and normal at the point (2, 4).

1. Identify a: 4a = 8 gives a = 2. Check the point: 4² = 16 and 8·2 = 16, so (2, 4) is on the parabola.
2. Tangent by T = 0: y·4 = 2·2·(x + 2), i.e. 4y = 4x + 8, so y = x + 2.
3. Parametric check: t = y/(2a) = 4/4 = 1; tangent ty = x + at² gives y = x + 2. Matches.
4. Normal: slope of tangent is 1, so normal slope is −1: y − 4 = −(x − 2), i.e. x + y = 6.
5. Formula check: y = −tx + 2at + at³ with t = 1, a = 2 gives y = −x + 4 + 2 = −x + 6. Matches.

Common Misconceptions

• "The tangent formula needs the point to be anywhere." yy₁ = 2a(x + x₁) is a tangent only when (x₁, y₁) is on the parabola; from outside it gives the chord of contact.
• "y = mx + a/m works for m = 0." It fails for m = 0 (division by zero) — the horizontal "tangent" y = 0 is the axis, not a tangent.
• "The normal is hard to remember." In parameter form it is a single line: y = −tx + 2at + at³.

Real-World Applications

• Reflector design: the tangent at each point of a parabolic mirror determines the local surface angle that reflects rays to the focus.
• Projectile safety: tangent lines to a projectile's parabolic path bound the danger zone at each instant.
• CNC machining: cutting a parabolic profile needs the tangent direction at each point to orient the tool.
• Bridge cables: hangers attach along the normal direction of the parabolic main cable for even load transfer.

In short: tangent yy₁ = 2a(x + x₁) (or ty = x + at²), normal y = −tx + 2at + at³, and slope form y = mx + a/m cover every tangent/normal question on y² = 4ax. Confirm the point is on the curve first, then verify the normal is perpendicular to the tangent.