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Tangents and Normals

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How to find tangent and normal equations, use parallel and perpendicular gradients, and solve intersection problems involving a curve and a line.

Tangents and normals

A tangent to a curve at a point is the straight line with the same gradient as the curve at that point. A normal is the straight line through that point at right angles to the tangent.

Differentiation gives the tangent gradient. The normal uses the perpendicular gradient. Both lines pass through the same point on the curve.

A tangent can meet the curve again elsewhere, and at some points it can cross the curve. What defines it is its gradient at the point of contact.

Explore the tangent and normal

The curve below is y=x^2-2, with gradient function \frac{dy}{dx}=2x. Move P along it to compare the two lines.

Try it: at x=1, the tangent gradient is 2 and the normal gradient is -\frac12. At x=-1, their signs swap. Move to x=0 to see a horizontal tangent and a vertical normal.

The axes use the same scale, so the right angle between the lines is shown accurately.

The method

For a point P(x_1,y_1) on a curve:

  • Find y_1 from the original equation if it has not been given.
  • Differentiate, then substitute x_1 to find the tangent gradient m_{\mathrm{t}}.
  • For a normal, use m_{\mathrm{n}}=-\frac{1}{m_{\mathrm{t}}}, provided m_{\mathrm{t}}\ne0.
  • Put the point and the required gradient into y-y_1=m(x-x_1).
  • Rearrange into the form the question asks for.

The original equation gives a point. The derivative gives a gradient. You need both to find the line.

Worked example 1

Find the equation of the tangent to y=x^2+x-4 at the point where x=2. Give your answer in the form y=mx+c

  1. Your turn. Substitute x=2 into y=x^2+x-4. What is the y-coordinate?

    y=2^2+2-4=2, so the point is (2,2)

  2. Differentiate: \frac{dy}{dx}=2x+1. At x=2, the tangent gradient is 5
  3. Use y-y_1=m(x-x_1): y-2=5(x-2)
  4. Expand and rearrange: y-2=5x-10
  5. Answer: y=5x-8. Check that (2,2) lies on the line: 5(2)-8=2

The normal gradient

For two perpendicular lines with finite gradients, the gradients multiply to -1:

m_{\mathrm{t}}m_{\mathrm{n}}=-1

Take the negative reciprocal: change the sign and turn the fraction upside down. If the tangent gradient is 5, the normal gradient is -\frac15. If it is -\frac23, the normal gradient is \frac32

Changing the sign alone is not enough. The normal to a tangent of gradient 5 has gradient -\frac15, not -5

Worked example 2

Find the equation of the normal to y=x^2+x-4 at (2,2), in the form ax+by=c. Then find where this normal meets the x-axis.

  1. As in worked example 1, \frac{dy}{dx}=2x+1, so the tangent gradient at x=2 is 5

  2. Your turn. What is the normal gradient when the tangent gradient is 5?

    The normal gradient is -\frac15, because 5\times(-\frac15)=-1

  3. The normal also passes through (2,2): y-2=-\frac15(x-2)
  4. Multiply by 5: 5y-10=-x+2. Rearrange to give x+5y=12

  5. Your turn. On the x-axis, y=0. What is the x-coordinate where x+5y=12 meets this axis?

    Set y=0, giving x=12

  6. Answer: the normal is x+5y=12, and it meets the x-axis at (12,0)

When the curve contains a fraction

Rewrite a power of x in the denominator as a negative power before differentiating. Keep gradients as exact fractions until the line equation is complete.

Worked example 3

Find the equation of the normal to y=\frac6x+x at the point where x=2. Give your answer in the form y=mx+c

  1. Find the point using the original equation: y=\frac62+2=5, so P=(2,5)
  2. Write y=6x^{-1}+x, then differentiate: \frac{dy}{dx}=-6x^{-2}+1
  3. At x=2, the tangent gradient is -\frac64+1=-\frac12

  4. Your turn. The tangent gradient is -\frac12. What is the normal gradient?

    The normal gradient is 2, since (-\frac12)\times2=-1

  5. The normal is y-5=2(x-2)
  6. Answer: y=2x+1. Check: when x=2, this gives y=5

When the direction of the tangent is given

Parallel lines have equal gradients. If a tangent is parallel to a given line, set the derivative equal to that line’s gradient and solve for x. There may be more than one point.

If a normal is parallel to a line of non-zero gradient m, first find the tangent gradient -\frac1m, then set the derivative equal to that.

Worked example 4

Find the equations of both tangents to y=x^3-6x^2+5x+2 which are parallel to y=-4x+7

  1. The given line has gradient -4. Differentiate the curve: \frac{dy}{dx}=3x^2-12x+5
  2. Set the derivative equal to -4: 3x^2-12x+5=-4. This gives x^2-4x+3=0

  3. Your turn. Solve x^2-4x+3=0. Enter both values of x, separated by a comma.

    Give every value, separated by commas

    (x-1)(x-3)=0, so x=1 or x=3

  4. Use the original curve to find the points. At x=1, y=2. At x=3, y=-10
  5. Through (1,2): y-2=-4(x-1), giving y=-4x+6
  6. Through (3,-10): y+10=-4(x-3), giving y=-4x+2
  7. Answer: y=-4x+6 and y=-4x+2. Both have the required gradient -4

Horizontal tangents and vertical normals

If \frac{dy}{dx}=0, the tangent is horizontal. Through (x_1,y_1), its equation is y=y_1

The normal is then vertical, with equation x=x_1. A vertical line has no finite gradient, so do not try to calculate -\frac10

Worked example 5

For the curve y=2x^2+4x-1, find the point where the tangent is horizontal. Find the tangent and normal there.

  1. Differentiate: \frac{dy}{dx}=4x+4. A horizontal tangent requires 4x+4=0

  2. Your turn. Solve 4x+4=0

    x=-1

  3. Use the original curve: y=2(-1)^2+4(-1)-1=-3. The point is (-1,-3)
  4. The horizontal tangent has constant y-coordinate -3. The vertical normal has constant x-coordinate -1
  5. Answer: the point is (-1,-3), the tangent is y=-3, and the normal is x=-1

Where a tangent meets the curve again

First find the line equation. Then solve it simultaneously with the curve: equate the two expressions for y, solve for x, and find the corresponding y-values.

One solution is the point of contact you already know. Keep track of which solution is the other intersection. For a polynomial curve, the known tangent point gives a repeated factor in the intersection equation.

Worked example 6

The tangent to y=x^3-2x+3 at P(1,2) meets the curve again at Q. Find the coordinates of Q

  1. Differentiate: \frac{dy}{dx}=3x^2-2. At x=1, the gradient is 1
  2. The tangent is y-2=x-1, or y=x+1
  3. At an intersection, x^3-2x+3=x+1, so x^3-3x+2=0
  4. The known solution is x=1. Factorise: x^3-3x+2=(x-1)(x^2+x-2)
  5. Factorise again: (x-1)^2(x+2)=0. The repeated root x=1 is P; the other root is x=-2

  6. Your turn. Use y=x+1 to find the coordinates of Q when x=-2

    Write your answer as (x, y)

    y=-2+1=-1, so Q=(-2,-1)

  7. Answer: Q=(-2,-1). Check in the curve: (-2)^3-2(-2)+3=-1

Common mistakes

  • Using the tangent gradient for a normal. Take the negative reciprocal.
  • Finding the point’s height from the derivative. Use the original curve equation.
  • Writing y=mx automatically. Most tangents and normals do not pass through the origin.
  • Losing a sign in y-y_1=m(x-x_1). At (-1,-3), this becomes y+3=m(x+1).
  • Setting x=0 for an x-axis intercept. On the x-axis, it is y=0.
  • Dividing by zero for a horizontal tangent. Its normal is a vertical line.
  • Keeping only one point when a gradient equation has two solutions.

Before you move on

You should be able to find a tangent or normal from a given point or x-coordinate, use a specified direction to find the point of contact, handle a horizontal tangent, and solve for intercepts or further intersections.

For each line, check two things: does it pass through the required point, and does it have the required gradient?

Now try it: Tangents and Normals practice questions

More on this topic: Tangents and Normals worksheet with full solutions

Unofficial revision notes written by Teach Me Maths. Not produced or endorsed by AQA.