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Learn9709Paper 1 · Pure Mathematics 1DifferentiationTangents and normals
Differentiation · Subtopic 3 of 5

Tangents and Normals

Using dydx\dfrac{dy}{dx} to find the equation of the tangent and the normal to a curve at a point, points where the tangent has a given gradient, and the follow-on problems examiners love.

Cambridge 9709 Paper 1Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Find the equation of the tangent to a curve at a given point.
  • Find the equation of the normal, using mnormal=−1mtangentm_{\text{normal}}=-\dfrac{1}{m_{\text{tangent}}}.
  • Find points where the tangent is parallel to a given line.
  • Solve problems where a tangent or normal meets an axis, another line or the curve again.
Pause and think

You know a point on a line and its gradient. What formula gives the equation of the line?

Section 2 of 14

Core idea

Start hereWhere does the skier fly off?

A ski jump follows a curve. At the lip, the skier leaves along a straight line, the tangent at that point. The line at right angles to it, the normal, is the direction the ramp pushes. Find the gradient with calculus, and both lines follow.

The big idea

At x=x1x=x_1: tangent gradient m=f′(x1)m=f'(x_1); normal gradient −1m-\dfrac1m. Both pass through (x1,f(x1))(x_1,f(x_1)); write each as y−y1=m(x−x1)y-y_1=m(x-x_1).

1. The method

The tangent and normal sliding along y = x²
−4−224−11234560xyy = x²normaltangent gradient 2x · normal gradient −1/(2x)
At every point the two dashed lines meet at a right angle. Where the tangent is steep, the normal is shallow, and the other way round.
Tangent and normal in four steps

Never use f′(x)f'(x) (a formula) as the gradient: substitute x1x_1 to get a number.

Worked exampley=x3−2xy=x^3-2x at x=2x=2

Find the tangent and normal.

Goal
Two line equations.
Strategy
Point, gradient, line.
  1. y1=8−4=4y_1=8-4=4
    Point (2,4)(2,4).
  2. dydx=3x2−2⇒m=10\frac{dy}{dx}=3x^2-2\Rightarrow m=10
    At x=2x=2.
  3. y−4=10(x−2)⇒y=10x−16y-4=10(x-2)\Rightarrow y=10x-16
    Tangent.
  4. y−4=−110(x−2)⇒x+10y−42=0y-4=-\tfrac1{10}(x-2)\Rightarrow x+10y-42=0
    Normal; multiply by 1010.
AnswerTangent y=10x−16y=10x-16; normal x+10y−42=0x+10y-42=0.
✓ Check(2,4)(2,4) on both: 20−16=420-16=4 ✓; 2+40−42=02+40-42=0 ✓.
Worked exampleA tangent with a given gradient

Find where the tangent to y=x2−4x+7y=x^2-4x+7 is parallel to y=2xy=2x.

Goal
The point and tangent.
Strategy
Parallel ⇒ gradient 22.
  1. 2x−4=2⇒x=32x-4=2\Rightarrow x=3
    Set f′(x)=2f'(x)=2.
  2. y=9−12+7=4y=9-12+7=4
    Point (3,4)(3,4).
  3. y=2x−2y=2x-2
    y−4=2(x−3)y-4=2(x-3).
AnswerAt (3,4)(3,4); tangent y=2x−2y=2x-2.
✓ CheckLine meets curve: x2−6x+9=(x−3)2=0x^2-6x+9=(x-3)^2=0: a repeated root, so it touches ✓.
Worked exampleThe normal meets the curve again

The normal to y=x2y=x^2 at (1,1)(1,1) meets the curve again at QQ. Find QQ.

Goal
Second intersection.
Strategy
Normal equation, then solve simultaneously.
  1. m=2⇒normal gradient −12m=2\Rightarrow\text{normal gradient }-\tfrac12
    dydx=2x\frac{dy}{dx}=2x.
  2. y=−12x+32y=-\tfrac12x+\tfrac32
    Through (1,1)(1,1).
  3. x2=−12x+32⇒2x2+x−3=0⇒(2x+3)(x−1)=0x^2=-\tfrac12x+\tfrac32\Rightarrow2x^2+x-3=0\Rightarrow(2x+3)(x-1)=0
    Equate; ×2\times2; factorise.
AnswerQ=(−32,94)Q=\left(-\frac32,\frac94\right).
✓ Checkx=1x=1 is the known point, a good sign the algebra is right ✓.
Common misconceptionNormal gradient The normal's gradient is −1m-\frac1m, not −m-m.
Pause and think

Area of the triangle formed by the tangent to y=4xy=\frac4x at (2,2)(2,2) and the axes?

Summary

  • Point and numerical gradient first.
  • Tangent mm; normal −1/m-1/m.
Key words
Normal
Line perpendicular to the tangent at the point.
Tangent and normal to y = x² at (1, 1)
−1123−0.50.511.522.50xyy = x²tangent y = 2x − 1normal y = −½x + 3/2(1, 1)
At x=1x=1 the gradient is dydx=2x=2\dfrac{dy}{dx}=2x=2, so the tangent has gradient 22. The normal is perpendicular to it, so its gradient is −12-\dfrac12 (the two gradients multiply to −1-1).
HistoryTangents before calculus. In the 1630s Pierre de Fermat found tangents and maxima of curves with a method he called "adequality", decades before Newton and Leibniz. Newton later said he got his ideas on calculus from "Fermat's way of drawing tangents".
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