Equations of Tangents and Normals

Q: What form should I write tangent/normal equations in?

Edexcel accepts any correct equivalent form, but or are preferred unless the question explicitly specifies a format.

Q: What do I do if the tangent gradient is zero?

If , the tangent is horizontal (form ) and the normal is vertical (form ). Do not attempt to calculate as this will lead to division by zero.

Edexcel International GCSE Further Pure Mathematics· 9F· 15 min read

1. Gradient of the Tangent to a Curve★★☆☆☆⏱ 3 min

📘 Definition

Gradient of Tangent

$m_{t} = f^{'} (a)$

The gradient of the tangent to curve $y = f (x)$ at point $x = a$ equals the value of the derivative of $f (x)$ evaluated at $x = a$ .

To find the tangent gradient, first differentiate $f (x)$ using the appropriate differentiation rule, then substitute the x-coordinate of the given point into the derivative. If you are only given the x-coordinate, calculate the corresponding y-coordinate by substituting into the original function $f (x)$ .

📐 Worked Example

Find the gradient of the tangent to the curve $y = 3 x^{2} + 2 x - 1$ at the point where $x = 1$ .

1
Step 1: Differentiate the function using the power rule:
$f'(x) = 6x + 2$
2
Step 2: Substitute $x = 1$ into the derivative:
$f'(1) = 6(1) + 2 = 8$
3
Step 3: The gradient of the tangent at $x = 1$ is 8.

Exam tip:

Always use the original function, not the derivative, to calculate the y-coordinate of the point of contact.

2. Equation of the Tangent★★★☆☆⏱ 4 min

Once you have the tangent gradient $m_{t}$ and the coordinates $(x_{1}, y_{1})$ of the point of contact, use the straight line equation $y - y_{1} = m_{t} (x - x_{1})$ to find the equation of the tangent. Rearrange to the required form if specified in the question.

📐 Worked Example

Find the equation of the tangent to the curve $y = (2 x - 1)^{3}$ at the point $(1, 1)$ . Give your answer in the form $y = m x + c$ .

1
Step 1: Differentiate using the chain rule:
$\frac{dy}{dx} = 6(2x - 1)^2$
2
Step 2: Evaluate the derivative at $x = 1$ :
$\frac{dy}{dx}\bigg|_{x=1} = 6(2(1)-1)^2 = 6$
3
Step 3: Substitute into the straight line equation:
$y - 1 = 6(x - 1)$
4
Step 4: Rearrange to $y = m x + c$ form:
$y = 6x - 5$

Exam tip:

Double-check that the point of contact lies on both the curve and your final tangent equation to catch rearrangement errors.

3. Gradient of the Normal★★★☆☆⏱ 3 min

📘 Definition

Gradient of Normal

$m_{n} = - \frac{1}{m _{t}}$ (for $m_{t} \neq = 0$ )

The normal is perpendicular to the tangent at the point of contact, so its gradient is the negative reciprocal of the tangent gradient.

If the tangent gradient is 0 (horizontal tangent), the normal is vertical, with an undefined gradient and equation of the form $x = x_{1}$ . If the tangent gradient is undefined (vertical tangent), the normal is horizontal, with equation $y = y_{1}$ .

📐 Worked Example

Find the gradient of the normal to the curve $y = \frac{x ^{2} + 1}{x}$ at the point where $x = 1$ .

1
Step 1: Simplify the function and differentiate:
$f(x) = x + x^{-1} \implies f'(x) = 1 - \frac{1}{x^2}$
2
Step 2: Evaluate the derivative at $x = 1$ :
$f'(1) = 1 - 1 = 0$
3
Step 3: The tangent gradient is 0, so the normal is vertical (undefined gradient), with equation $x = 1$ .

Exam tip:

Never try to calculate $- 1/ m_{t}$ when $m_{t} = 0$ , as this leads to division by zero. Recognize the horizontal/vertical line case instead.

4. Equation of the Normal★★★★☆⏱ 4 min

✓ Calculator OK

Once you have the normal gradient $m_{n}$ and the point of contact $(x_{1}, y_{1})$ , substitute into the same straight line formula $y - y_{1} = m_{n} (x - x_{1})$ used for tangents, then rearrange to the required form.

📐 Worked Example

Find the equation of the normal to the curve $y = x sin x$ at the point where $x = \frac{π}{2}$ . Give your answer in the form $a x + b y + c = 0$ .

1
Step 1: Calculate the y-coordinate of the point:
$y = \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$
2
Step 2: Differentiate using the product rule:
$f'(x) = \sin x + x \cos x$
3
Step 3: Evaluate the derivative at $x = \frac{π}{2}$ :
$f'\left(\frac{\pi}{2}\right) = 1 + 0 = 1$
4
Step 4: Calculate the normal gradient:
$m_n = -\frac{1}{1} = -1$
5
Step 5: Substitute into the straight line equation:
$y - \frac{\pi}{2} = -1\left(x - \frac{\pi}{2}\right)$
6
Step 6: Rearrange to $a x + b y + c = 0$ form:
$x + y - \pi = 0$

5. Common Pitfalls

Wrong move:

Using the derivative function to calculate the y-coordinate of the point of contact

Why:

The derivative gives gradient values, not the y-value of the original curve at a given x

Correct move:

Always substitute the x-coordinate into the original $f (x)$ function to get the correct $y_{1}$ value

Wrong move:

Using the positive reciprocal of the tangent gradient for the normal

Why:

Perpendicular lines have gradients whose product is -1, so the negative reciprocal is required

Correct move:

Calculate normal gradient as $m_{n} = - 1/ m_{t}$ for non-zero $m_{t}$

Wrong move:

Attempting to calculate $- 1/ m_{t}$ when the tangent gradient is 0

Why:

Division by zero is undefined, leading to invalid answers

Correct move:

If $m_{t} = 0$ , tangent is horizontal ( $y = y_{1}$ ) and normal is vertical ( $x = x_{1}$ )

Wrong move:

Rearranging the straight line equation incorrectly with sign errors

Why:

Simple arithmetic errors cost easy marks in structured exam questions

Correct move:

After rearranging, substitute the point $(x_{1}, y_{1})$ back into your final equation to verify it is correct

Wrong move:

Differentiating the function incorrectly using the wrong rule

Why:

Tangent/normal questions rely entirely on accurate differentiation to get the correct gradient

Correct move:

Double-check your derivative against the correct rule (power/product/quotient/chain) before substituting values

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Tangent gradient at $(x_{1}, y_{1})$	$m_{t} = f^{'} (x_{1})$	Evaluate derivative at $x_{1}$
Tangent equation	$y - y_{1} = m_{t} (x - x_{1})$	Rearrange to required form
Normal gradient (non-zero $m_{t}$ )	$m_{n} = - 1/ m_{t}$	Negative reciprocal of tangent gradient
Normal equation (non-zero $m_{t}$ )	$y - y_{1} = m_{n} (x - x_{1})$	Use same line formula as tangent
Horizontal tangent ( $m_{t} = 0$ )	Tangent: $y = y_{1}$ , Normal: $x = x_{1}$	No division by zero needed
Vertical tangent ( $m_{t}$ undefined)	Tangent: $x = x_{1}$ , Normal: $y = y_{1}$	For points where $f^{'} (x)$ is undefined

7. Frequently Asked

What form should I write tangent/normal equations in?

Edexcel accepts any correct equivalent form, but $y = m x + c$ or $a x + b y + c = 0$ are preferred unless the question explicitly specifies a format.

What do I do if the tangent gradient is zero?

If $f^{'} (x) = 0$ , the tangent is horizontal (form $y = k$ ) and the normal is vertical (form $x = k$ ). Do not attempt to calculate $- 1/ m_{t}$ as this will lead to division by zero.

Going deeper

What's Next

Now that you can calculate equations of tangents and normals, you are ready to apply your calculus skills to more advanced topics in the Edexcel IGCSE Further Pure Math syllabus. Tangent and normal equations are often combined with stationary point problems, where you will identify maximum, minimum and inflection points on curves, and calculate tangents at these key positions. You will also encounter these skills in coordinate geometry problems that ask you to find the area of shapes formed by tangents, normals and the coordinate axes, or the intersection points of multiple tangents to a single curve. Practicing mixed questions that combine differentiation, coordinate geometry and algebra will help you secure full marks on this high-frequency exam topic.