How Do You Find the Equation of a Normal Line at a Point?

Differentiate the curve to find the tangent gradient, then take its negative reciprocal to obtain the normal gradient. Substitute this gradient and the given point into the point-gradient equation of a line.

The Method

For a curve $y=f(x)$, suppose the normal is required at $x=a$.

1. Find the point on the curve: (P(a,f(a))).
2. Differentiate to obtain (f'(x)).
3. Calculate the tangent gradient: $m_t=f'(a)$.
4. Since the tangent and normal are perpendicular, their gradients satisfy $m_tm_n=-1$. Therefore, $$m_n=-\frac{1}{f'(a)}.$$
5. Use the point-gradient form: $$y-f(a)=m_n(x-a).$$

Worked Example

Find the equation of the normal to $y=x^2+1$ at $x=2$.

First, find the point:

$$y=2^2+1=5,$$

so the normal passes through $(2,5)$.

Differentiate the function:

$$\frac{dy}{dx}=2x.$$

At $x=2$, the tangent gradient is

$$m_t=2(2)=4.$$

Therefore, the normal gradient is

$$m_n=-\frac{1}{4}.$$

Using point-gradient form:

$$y-5=-\frac{1}{4}(x-2).$$

This is a complete equation of the normal; it may also be rearranged into $y=-\frac{1}{4}x+\frac{11}{2}$.

A special case occurs when $f'(a)=0$: the tangent is horizontal, so the normal is the vertical line $x=a$.

Why the Negative Reciprocal Works

At the point, the tangent and normal are perpendicular. For two non-vertical perpendicular lines, the product of their gradients is $-1$, giving $m_n=-\frac{1}{m_t}$. Graphically, the normal crosses the curve at the same point as the tangent but turns through .

If the point is given as coordinates rather than by $x=a$, verify that it lies on the curve before differentiating. For example, $(2,5)$ satisfies $5=2^2+1$. This check prevents using the correct gradient through the wrong point.

Do not assume that the normal gradient is simply the opposite sign of the tangent gradient. Perpendicularity, not reflection, determines the negative reciprocal.

IB Exam Technique

This is SL 5.4: Tangents and normal in Mathematics: Analysis and Approaches. For the command term find, show the derivative, evaluate the tangent gradient, calculate the negative reciprocal, and substitute into a line equation. A common misconception is using $1/m_t$; the normal gradient is the negative reciprocal, $-1/m_t$.