The normal to a curve at a point is the straight line through that point that is perpendicular to the tangent. Its gradient is the negative reciprocal of the tangent gradient: if the tangent has gradient m, the normal has gradient −1/m.
You build on finding a tangent equation, so the point and the tangent gradient come first. This lesson belongs to tangents, normals and rates.
Why the negative reciprocal?
Two perpendicular lines with gradients m₁ and m₂ satisfy m₁ × m₂ = −1. A steep line rising to the right (m = 5) is crossed at a right angle by a shallow line falling to the right (−1/5).
A gradient of 2 becomes −1/2. A gradient of −3/4 becomes 4/3. Each time, flip the fraction and change the sign.
What is the method, step by step?
- Find the point by substituting x into the curve.
- Find the tangent gradient by substituting x into dy/dx.
- Convert to the normal gradient: −1 divided by the tangent gradient.
- Write the line using y − y₁ = m(x − x₁).
- Check that the product of the two gradients is −1 and that the point fits.
Worked example
Find the equation of the normal to y = 2x² − 3x + 1 at the point where x = 2. Give your answer in the form ax + by + c = 0 with integer coefficients.
Step 1, the point: y = 2(4) − 6 + 1 = 3. The point is (2, 3).
Step 2, tangent gradient: dy/dx = 4x − 3, so at x = 2 the gradient is 5.
Step 3, normal gradient: −1/5.
Step 4, the line: y − 3 = −(1/5)(x − 2). Multiply by 5: 5y − 15 = −x + 2.
Step 5, rearrange: x + 5y − 17 = 0.
Check: 5 × (−1/5) = −1. ✓ The point (2, 3): 2 + 15 − 17 = 0. ✓
The mistake to watch for
The usual slip is to do half of the conversion.
Mistaken working: tangent gradient 5, so the normal gradient is −5 (sign changed, not flipped). Or: the normal gradient is 1/5 (flipped, sign unchanged).
Neither line is perpendicular. The product of the gradients would be −25 or +1, not −1.
The correction is to write “m × m = −1” beside your working and test it before you go on. If the product is not −1, one of the two steps is missing.
Check yourself
Try these, then open each answer.
1. Find the normal to y = x² + 1 at x = 1, in the form ax + by + c = 0.
Show answer
y = 1 + 1 = 2, so the point is (1, 2). dy/dx = 2x, so the tangent gradient is 2 and the normal gradient is −1/2.
y − 2 = −(1/2)(x − 1). Multiply by 2: 2y − 4 = −x + 1, so x + 2y − 5 = 0. Check: 1 + 4 − 5 = 0. ✓
2. Find the normal to y = 8/x at the point where x = 2, in the form y = mx + c.
Show answer
y = 8/2 = 4, so the point is (2, 4). Write y = 8x⁻¹, so dy/dx = −8/x² and the tangent gradient is −8/4 = −2. The normal gradient is 1/2.
y − 4 = (1/2)(x − 2), so y = x/2 + 3. Check: 2/2 + 3 = 4. ✓
3. A tangent has gradient −3/4. What is the gradient of the normal at the same point?
Show answer
Flip −3/4 to get −4/3, then change the sign: 4/3. Check: (−3/4) × (4/3) = −1. ✓
Where this leads next
With tangents and normals secured, move on to using a derivative as a rate of change, then try the mixed practice set. The non-calculator working trainer is useful for practising reciprocal fractions, and the quadratic structure explorer shows a parabola with its tangent and normal.
If you understand the rule but still slip on the sign under time pressure, it helps to have someone watch your working live. That is what our teachers do in online one-to-one Additional Mathematics tuition.