You already find the gradient of a straight line from two points. On a curve the gradient changes from point to point: \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) evaluated at \(x=a\) gives the gradient of the curve — and of its tangent — at that point. At \(P\) the tangent (red) has the same instantaneous direction as the curve; the normal (green) is the line through \(P\) perpendicular to the tangent. Tangent and normal at \(P(x_1,\ y_1)\) LineGradientEquation Tangent\(m=\dfrac{\mathrm{d}y}{\mathrm{d}x}\) evaluated at \(x=x_1\)\(y-y_1=m(x-x_1)\) Normal\(-\dfrac{1}{m}\quad(m\ne 0)\)\(y-y_1=-\dfrac{1}{m}(x-x_1)\) Method Differentiate to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Substitute the \(x\)-coordinate of the point to find the tangent gradient \(m\) (and the \(y\)-coordinate if it is not given). Write the equation with the point–gradient form; for the normal replace \(m\) by \(-\dfrac{1}{m}\). If a gradient is given instead of a point, set \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to that value and solve for \(x\) to locate the point on the curve. Exam check Differentiate first, then substitute \(x=a\). When both gradients are finite, their product is \(-1\). If the tangent is horizontal, the normal is the vertical line \(x=x_1\). A tangent can cross the curve at its point of contact, as at the origin of \(y=x^3\). Use \(-1/m\) for the normal only when \(m\ne0\).

Tangents and Normals

Secondary 4

TGM Original Questions

You already find the gradient of a straight line from two points. On a curve the gradient changes from point to point: \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) evaluated at \(x=a\) gives the gradient of the curve — and of its tangent — at that point.

At \(P\) the tangent (red) has the same instantaneous direction as the curve; the normal (green) is the line through \(P\) perpendicular to the tangent.

Tangent and normal at \(P(x_1,\ y_1)\)

Line	Gradient	Equation
Tangent	\(m=\dfrac{\mathrm{d}y}{\mathrm{d}x}\) evaluated at \(x=x_1\)	\(y-y_1=m(x-x_1)\)
Normal	\(-\dfrac{1}{m}\quad(m\ne 0)\)	\(y-y_1=-\dfrac{1}{m}(x-x_1)\)

Method

Differentiate to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
Substitute the \(x\)-coordinate of the point to find the tangent gradient \(m\) (and the \(y\)-coordinate if it is not given).
Write the equation with the point–gradient form; for the normal replace \(m\) by \(-\dfrac{1}{m}\).

If a gradient is given instead of a point, set \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to that value and solve for \(x\) to locate the point on the curve.

Exam check

Differentiate first, then substitute \(x=a\). When both gradients are finite, their product is \(-1\). If the tangent is horizontal, the normal is the vertical line \(x=x_1\).

A tangent can cross the curve at its point of contact, as at the origin of \(y=x^3\). Use \(-1/m\) for the normal only when \(m\ne0\).

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