Perpendicular to the tangent: The normal at a point is perpendicular to the tangent there.
Since perpendicular gradients multiply to −1, the normal's gradient is −1/f'(a).
It passes through the same point (a, f(a)).
The normal is perpendicular to the tangent at the point, so its gradient is −1 / (tangent gradient).
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Negative reciprocal: Flip the tangent gradient and change its sign: gradient 2 → normal −½; gradient −3 → normal +⅓.
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Tangent gradient first, then flip it: Method: find f'(a) (tangent gradient), take its negative reciprocal for the normal gradient, find the point (a, f(a)), then use y − y₁ = m(x − x₁).
IB-style question — normal equation
Find the equation of the normal to y = x² at the point where x = 1.
Step by step
- Tangent gradient f'(1) = 2, so normal gradient = −1/2. Point (1, 1).
- Use y − y₁ = m(x − x₁).
Final answer
y = −½x + 3/2.
Flip the TANGENT gradient: Take the negative reciprocal of f'(a) (the tangent gradient) — not of the point or the y-value.
IB-style question — normal meets the curve again
The normal to y = x² at the point (1, 1) meets the curve again at B.
Find the coordinates of B.
Step by step
- Gradient of the curve: f′(x) = 2x, so at x = 1 the tangent gradient is 2 and the NORMAL gradient is −½.
- Normal through (1, 1): y − 1 = −½(x − 1) → y = −½x + 3⁄2. Set equal to the curve.
- Factor; x = 1 is the known point, so take the other root.
Final answer
B = (−3⁄2, 9⁄4).
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Stationary point → vertical normal: At a stationary point the tangent is horizontal (gradient 0), so the normal is vertical: x = a.
(You can't take −1/0, but geometrically a vertical line is perpendicular to a horizontal one.)
IB-style question — normal at a vertex
Find the equations of the tangent and the normal to y = x² − 4x + 5 at its vertex (where x = 2).
Step by step
- f'(x) = 2x − 4, f'(2) = 0 → tangent horizontal. Point (2, 1).
- Normal is perpendicular to a horizontal line → vertical.
Final answer
Tangent y = 1 (horizontal); normal x = 2 (vertical).
Horizontal tangent ↔ vertical normal: When f'(a) = 0: tangent y = f(a) (horizontal), normal x = a (vertical).