Tangents and normals

How to find the equation of the tangent and the normal to a curve at a point: get the gradient from the derivative, the point from the curve, then write a straight line — and why the normal's gradient is the negative reciprocal.

Pure · Differentiation

Tangents and normals

Why it works

A tangent is just a straight line that grazes the curve at one point, matching its direction there. So to pin it down you only need the two things any straight line needs: a point it passes through, and a gradient.

The point lies on the curve: at $x = a$ it is $(a, f(a))$ — substitute $a$

into the original equation to get the height.

The gradient is the steepness of the curve at that point — which is

exactly what the derivative gives:

m = f'(a)

. Differentiate, then substitute.

Then it's the ordinary straight-line equation

y - y_1 = m(x - x_1), \qquad (x_1, y_1) = (a, f(a)), \quad m = f'(a).

The whole task is "find a point and a gradient, then write the line" — the only new idea is that the gradient comes from differentiating.

The normal is the line perpendicular to the curve at that same point — same point, turned through

90°

. Perpendicular gradients multiply to

-1

, so if the tangent has gradient

m

, the normal has gradient

-\dfrac{1}{m}

: the negative reciprocal (flip it over, change the sign). Everything else is identical — same point, new gradient.

Two traps worth naming now, because they cost the most marks:

The gradient is $f'(a)$ , not $f(a)$ . The curve's height and its steepness

are different numbers; using the

y

-value as the gradient is the classic slip.

The normal's gradient is $-\dfrac{1}{m}$ , not $\dfrac{1}{m}$ . Drop the minus sign

and your "normal" is just a wrong tangent.Above, the dashed line is the tangent to

y = x^2 - 3x + 2

(3, 2)

: it touches the curve there and has the curve's gradient,

f'(3) = 3