Calculus · 0606 Topic 14

Tangents & Normals

Teacher Rig, IGCSE Add Math tutor

Written by Teacher Rig

8 years teaching IGCSE Add Math · Updated 12 June 2026

The tangent at a point on a curve has gradient dydx\frac{dy}{dx} evaluated there; the normal is perpendicular to it. One routine covers both, and the only real hazard is answering the wrong one.

The routine

Find the equation of the normal to y=x34x+1y = x^3 - 4x + 1 at the point where x=2x = 2.

  1. Point: y(2)=88+1=1(2,1)y(2) = 8 - 8 + 1 = 1 \to (2, 1)
  2. Differentiate: dydx=3x24\frac{dy}{dx} = 3x^2 - 4
  3. Substitute: at x=2x = 2, gradient of tangent m=8m = 8
  4. Tangent or normal? Normal asked \to gradient == 18-\frac{1}{8} (negative reciprocal, stated)
  5. Build the line: y1=18(x2)y - 1 = -\frac{1}{8}(x - 2) \to 8y+x=108y + x = 10 (point-gradient form)

Five visible steps, each a marking point. The examiner-report perennial: students execute steps 1–3 perfectly, then build the tangent when the normal was asked (or vice versa). Underline the word in the question, and write step 4 explicitly even when the tangent is wanted (“tangent gradient =8= 8”), the stated decision is cheap insurance.

The variants

  • “Find where the tangent is parallel to y=5x2y = 5x - 2”: set dydx=5\frac{dy}{dx} = 5, solve for xx, gradient condition, parallel means equal gradients
  • “The tangent at PP passes through the origin”: build the tangent with unknown contact point, impose the through-origin condition, harder, but the same five steps with algebra carried symbolically
  • “Where does the normal meet the curve again?”: build the normal, then solve simultaneously with the curve
  • With product/quotient/chain functions: step 2 invokes the differentiation rules, the routine is unchanged

Tangents to circles need no calculus, radius \perp tangent does it (circle tangents); reaching for dydx\frac{dy}{dx} there is legal but slow.

Common mistakes

  • Tangent built for normal (and vice versa), the topic’s defining error
  • The yy-coordinate never computed (line built through (2,0)(2, 0) by reflex)
  • Gradient function used as the gradient (unsubstituted 3x243x^2 - 4 in the line equation)
  • Final form ignored when the question specified one
  • The 1m-\frac{1}{m} flip applied to the normal of a normal (double-flipping back to the tangent)

Full topic context: Calculus notes · the exam drill: differentiation technique.

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