G3PureCore

Tangents and Normals

Q: How do I find the y-coordinate if only x is given?

Substitute the given x-value into the ORIGINAL curve equation y=f(x) (not the derivative) to find the matching y-value.

Differentiation gives you the gradient at any single point on a curve — but the equation of the actual tangent line there needs one more step: combine that gradient with the point itself using the straight-line equation you already know.

28 min Video by TLMaths Differentiation

The big picture

This lesson turns the derivative into something concrete: an actual equation of a line. The $tangent$ touches the curve and shares its gradient exactly at that point; the $normal$ is perpendicular to the tangent there. Together they connect calculus directly back to coordinate geometry — every tangent/normal question is really "differentiate to find the gradient, then use $y - y_{1} = m (x - x_{1})$ ", combining two topics you already have.

A-Level Maths G3-02: Introducing Tangents and NormalsWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Find the gradient of a curve at a specific point by substituting into the derivative
Find the equation of the tangent to a curve at a given point
Use the perpendicular gradient rule to find the equation of the normal
Solve problems combining tangents/normals with other curve properties

The gradient at a point

The derivative $\frac{d y}{d x}$ is itself a FUNCTION of $x$ — it gives a different gradient at every point on the curve. To find the gradient at one specific point, substitute that point's $x$ -coordinate into the derivative.

1Differentiate:

\frac{d y}{d x} = 2 x - 3

2Substitute

x = 4

2 (4) - 3 = 5

AnswerGradient

= 5

Tip — The derivative and the gradient AT a point are different things — the derivative is a formula in $x$ ; the gradient at a point is a single number found by substitution.

The equation of a tangent

A tangent touches the curve at exactly one point (locally) and has the SAME gradient as the curve there. Find the gradient (differentiate and substitute), find the $y$ -coordinate of the point (substitute into the original equation if not given), then use $y - y_{1} = m (x - x_{1})$ .

1Gradient at

x = 4

m = 5

(from the previous example).

y

-coordinate:

y = 4^{2} - 3 (4) = 16 - 12 = 4

. So the point is

(4, 4)

3Use

y - y_{1} = m (x - x_{1})

y - 4 = 5 (x - 4)

4Rearrange:

y = 5 x - 20 + 4 = 5 x - 16

Answer

y = 5 x - 16

The equation of a normal

The $normal$ is the line perpendicular to the tangent at the same point. Perpendicular gradients multiply to $- 1$ , so the normal's gradient is the negative reciprocal of the tangent's gradient.

m_{normal} = - \frac{1}{m _{tangent}}

Perpendicular gradients multiply to give −1.

1Tangent gradient at this point is 5 (found above).

2Normal gradient:

- \frac{1}{5}

3Use

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

4Rearrange:

5 y - 20 = - (x - 4) = - x + 4

, so

5 y = - x + 24

Answer

5 y = - x + 24

(or

y = - \frac{1}{5} x + \frac{24}{5}

)

Why this works

A normal is only needed occasionally in real applications (e.g. the direction light reflects, or the shortest path to a curve) — but every question follows the identical two-step recipe: find the tangent gradient, then negative-reciprocal it.

Think like an examiner

Always find BOTH the gradient (from the derivative) AND the $y$ -coordinate of the point (from the original equation) before writing the line equation — a common lost mark is skipping the $y$ -coordinate when it is not given directly.
For a normal, take the negative reciprocal of the TANGENT gradient, not of the original function value.
Leave the final equation in the form requested (often $a x + b y + c = 0$ ) — check the question's required format.

Common misconceptions

Using the original function value as the gradient in the line equation.

The gradient comes from the DERIVATIVE evaluated at the point, not from

y = f (x)

itself.

Using the tangent's gradient directly for the normal equation.

The normal is perpendicular: its gradient is

- \frac{1}{m _{tangent}}

, the negative reciprocal.

Forgetting to find the

y

-coordinate of the point when only

x

is given.

Substitute the given

x

-value into the ORIGINAL equation

y = f (x)

to get the full point before using

y - y_{1} = m (x - x_{1})

Tangents and normals

y - y_{1} = m (x - x_{1})

Line through a point with gradient m.

m_{normal} = - \frac{1}{m _{tangent}}

Perpendicular gradient rule.

Stretch yourself

The curve $y = x^{3} - 4 x$ has a tangent at the point where $x = 2$ . Find the equation of the normal at this point, in the form $a x + b y + c = 0$ .

Hint — Differentiate to find the tangent gradient at $x = 2$ , take its negative reciprocal for the normal, find the $y$ -coordinate, then build the line equation.

Questions students ask

What is the difference between a tangent and a normal?

The tangent touches the curve and shares its exact gradient at that point. The normal is the line perpendicular to the tangent at the same point — its gradient is the negative reciprocal of the tangent's.

How do I find the y-coordinate if only x is given?

Substitute the given

x

-value into the ORIGINAL curve equation

y = f (x)

(not the derivative) to find the matching

y

-value.

Why is the normal gradient the negative reciprocal?

Two lines are perpendicular exactly when the product of their gradients is

- 1

— this is the same perpendicular-gradient rule from coordinate geometry, applied to the tangent's gradient.

Key takeaways

Find the gradient at a point by substituting its $x$ -value into the derivative.
The tangent shares that gradient; use $y - y_{1} = m (x - x_{1})$ with the point.
The normal is perpendicular: gradient $= - \frac{1}{m _{tangent}}$ .
Always find both the gradient and the full point before writing the line equation.

How this fits the course

Build on

Stationary Points

Leads to

Stationary Points

Test yourself

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