What is wrong normal gradient?

The normal gradient is the negative reciprocal of the tangent gradient; do not reuse the tangent gradient.

SEAB Original-style practice: Find the equation of the normal to the curve y = x^2 at the point (1, 1).

Gradient of curve: dydx = 2x, so at x = 1 the tangent gradient is 2. Normal gradient is the negative reciprocal: -12. Normal through (1, 1): y - 1 = -12(x - 1), so y = -12x + 32. What markers reward: finding the tangent gradient 2, taking the negative reciprocal for the normal, and forming the equation through the given point.

SingaporeAdditional MathematicsSyllabus dot point

How do we find the equation of the tangent and the normal to a curve at a given point?

Use the derivative to find the gradient of a curve at a point and write the equations of the tangent and normal there

A focused answer to the N(A)-Level Additional Mathematics outcome on tangents and normals. Use the derivative to find the gradient at a point, then write the equation of the tangent and the perpendicular normal.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find the tangent and the normal to a curve at a given point. The tangent is the straight line that just touches the curve there, with the same gradient as the curve; the normal is the line through the same point at right angles to the tangent. The method combines differentiation (to get the gradient) with the straight-line equation work from coordinate geometry.

The answer

The tangent has the gradient of the curve

The gradient of the curve at a point is the value of $\dfrac{dy}{dx}$ there. Because the tangent touches the curve at that point, the tangent gradient equals the derivative evaluated at that point.

Step-by-step for the tangent

Differentiate to find $\dfrac{dy}{dx}$ .
Substitute the $x$ -coordinate of the point to get the gradient $m$ .
If only the $x$ -coordinate is given, find the $y$ -coordinate by substituting into the original curve.
Use the point-gradient form $y - y_1 = m(x - x_1)$ and rearrange.

The normal is perpendicular to the tangent

The normal meets the curve at the same point but at a right angle to the tangent. So its gradient is the negative reciprocal of the tangent gradient:

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}

Then use the same point in the point-gradient form to write the normal's equation.

Putting it together

For both lines you use the same point; only the gradient differs. Find the tangent gradient first, take its negative reciprocal for the normal, then build each line. A quick sketch helps confirm that the tangent lies along the curve and the normal cuts across it.

Worked example

Find the equations of the tangent and normal to $y = x^2 - 4x + 5$ at the point where $x = 3$ .

Step 1: Find the y-coordinate

y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2

So the point is $(3, 2)$ .

Step 2: Differentiate and find the tangent gradient

\frac{dy}{dx} = 2x - 4, \quad \text{at } x = 3:\ 2(3) - 4 = 2

The tangent gradient is $2$ .

Step 3: Write the tangent equation

y - 2 = 2(x - 3) \quad\Rightarrow\quad y = 2x - 4

Step 4: Write the normal equation

Normal gradient $= -\dfrac{1}{2}$ , through $(3, 2)$ :

y - 2 = -\frac{1}{2}(x - 3) \quad\Rightarrow\quad y = -\frac{1}{2}x + \frac{7}{2}

Examples in context

Example 1. Path of a thrown ball. The flight of a ball follows a curve, and the tangent at a point gives the direction of travel at that instant. Tangents describe instantaneous direction, which is why they matter in physics and motion problems.

Example 2. Reflective surfaces. The normal to a curved mirror at a point is the line about which light reflects. Finding the normal to a curve is exactly the calculation used in optics, connecting this skill to applied contexts.

Try this

Q1. Find the gradient of $y = x^2$ at the point $(2, 4)$ . [2 marks]

Cue. $\dfrac{dy}{dx} = 2x$ , so at $x = 2$ the gradient is $4$ .

Q2. Find the equation of the tangent to $y = x^2$ at $(2, 4)$ . [2 marks]

Cue. Gradient $4$ : $y - 4 = 4(x - 2)$ , so $y = 4x - 4$ .

Q3. Find the gradient of the normal to a curve where the tangent gradient is $3$ . [1 mark]

Cue. Negative reciprocal: $-\dfrac{1}{3}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksFind the equation of the tangent to the curve

y = x^2

at the point

(3, 9)

Show worked answer →

Differentiate: $\dfrac{dy}{dx} = 2x$ . At $x = 3$ , the gradient is $2(3) = 6$ .

Tangent through $(3, 9)$ with gradient $6$ : $y - 9 = 6(x - 3)$ , so $y = 6x - 9$ .

What markers reward: differentiating to get the gradient function, evaluating it at $x = 3$ to get $6$ , and using the point-gradient form to give $y = 6x - 9$ .

Original4 marksFind the equation of the normal to the curve

y = x^2

at the point

(1, 1)

Show worked answer →

Gradient of curve: $\dfrac{dy}{dx} = 2x$ , so at $x = 1$ the tangent gradient is $2$ .

Normal gradient is the negative reciprocal: $-\dfrac{1}{2}$ .

Normal through $(1, 1)$ : $y - 1 = -\dfrac{1}{2}(x - 1)$ , so $y = -\dfrac{1}{2}x + \dfrac{3}{2}$ .

What markers reward: finding the tangent gradient $2$ , taking the negative reciprocal for the normal, and forming the equation through the given point.