Tangents & Normals

This lesson covers finding the equations of tangents and normals to curves — a fundamental application of differentiation in A-Level Mathematics. These ideas connect algebra, coordinate geometry, and calculus, and appear frequently on AQA papers.

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Tangent Lines

A tangent to a curve at a point P is the straight line that touches the curve at P and has the same gradient as the curve at that point.

To find the equation of a tangent to y = f(x) at x = a:

1. Find y = f(a) — the y-coordinate of the point.
2. Find the gradient: m = f'(a).
3. Use the point-gradient form: y − y₁ = m(x − x₁).

Example 1

Find the equation of the tangent to y = x³ − 2x + 1 at the point where x = 2.

Step 1: y = 8 − 4 + 1 = 5. So the point is (2, 5).

Step 2: dy/dx = 3x² − 2. At x = 2: m = 3(4) − 2 = 10.

Step 3:

y − 5 = 10(x − 2)
y − 5 = 10x − 20
y = 10x − 15

The equation of the tangent is y = 10x − 15.

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Normal Lines

A normal to a curve at a point P is the straight line that passes through P and is perpendicular to the tangent at P.

If the tangent has gradient m, the normal has gradient −1/m (provided m ≠ 0).

Perpendicular Gradients

Two lines are perpendicular if and only if the product of their gradients is −1:

m₁ × m₂ = −1

Example 2

Find the equation of the normal to y = x³ − 2x + 1 at (2, 5).

From Example 1, the tangent gradient is m = 10.

The normal gradient is −1/10.

y − 5 = −(1/10)(x − 2)
10(y − 5) = −(x − 2)
10y − 50 = −x + 2
x + 10y = 52

The equation of the normal is x + 10y = 52 (or equivalently y = −x/10 + 26/5).

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Horizontal and Vertical Tangents

• If f'(a) = 0, the tangent is horizontal: y = f(a).
• If the curve has a vertical tangent (e.g. x = y²), the gradient is undefined and the normal is horizontal.

Example 3

Find where the curve y = x³ − 3x has horizontal tangents.

dy/dx = 3x² − 3 = 0
x² = 1
x = ±1

At x = 1: y = 1 − 3 = −2. Tangent: y = −2. At x = −1: y = −1 + 3 = 2. Tangent: y = 2.

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Finding Where a Tangent Meets the Curve Again

A tangent at one point on a curve may intersect the curve at another point. To find this intersection, substitute the tangent equation into the curve equation and solve.

Example 4

The tangent to y = x³ at the point (1, 1) has equation y = 3x − 2 (verify: dy/dx = 3x², at x = 1 gives m = 3; y − 1 = 3(x − 1), so y = 3x − 2).

Find where this tangent meets y = x³ again.

Substitute y = 3x − 2 into y = x³:

x³ = 3x − 2
x³ − 3x + 2 = 0

We know x = 1 is a root (the point of tangency is a repeated root):

x³ − 3x + 2 = (x − 1)²(x + 2) = 0

So x = −2 is the other intersection. At x = −2: y = (−2)³ = −8.

The tangent meets the curve again at (−2, −8).

Exam Tip: At the point of tangency, x = a is always a repeated root of f(x) − (tangent line) = 0. Factor out (x − a)² to find the remaining intersection.

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Tangent and Normal to y = eˣ and y = ln x

Example 5

Find the equation of the tangent to y = eˣ at the point (0, 1).

dy/dx = eˣ. At x = 0: m = e⁰ = 1.

y − 1 = 1(x − 0)
y = x + 1

Example 6

Find the equation of the normal to y = ln x at the point (e, 1).

dy/dx = 1/x. At x = e: m = 1/e.

Normal gradient = −e.

y − 1 = −e(x − e)
y = −ex + e² + 1

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Intersection of Tangent and Normal

Example 7

Find the point where the tangent and normal to y = x² at x = 1 intersect.

At x = 1: y = 1, dy/dx = 2x = 2.

Tangent: y − 1 = 2(x − 1) → y = 2x − 1

Normal: y − 1 = −(1/2)(x − 1) → y = −x/2 + 3/2

Set equal:

2x − 1 = −x/2 + 3/2
4x − 2 = −x + 3
5x = 5
x = 1

They intersect at the original point (1, 1) — which makes sense, since the tangent and normal to a curve at a point both pass through that point.

Example 8: Two Different Points

Find where the tangent to y = x² at x = 2 intersects the normal to y = x² at x = −1.

Tangent at x = 2: Point (2, 4), gradient 4. Equation: y − 4 = 4(x − 2) → y = 4x − 4.

Normal at x = −1: Point (−1, 1), tangent gradient = −2, normal gradient = 1/2. Equation: y − 1 = (1/2)(x + 1) → y = x/2 + 3/2.

Set equal:

4x − 4 = x/2 + 3/2
8x − 8 = x + 3
7x = 11
x = 11/7
y = 4(11/7) − 4 = 44/7 − 28/7 = 16/7

Intersection point: (11/7, 16/7).

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Tangent Lines to Parametric Curves (Extension)

If a curve is defined parametrically by x = f(t), y = g(t), then:

dy/dx = (dy/dt) ÷ (dx/dt)

Example 9

A curve is given by x = t², y = 2t. Find the tangent at t = 3.

dx/dt = 2t, dy/dt = 2
dy/dx = 2/(2t) = 1/t

At t = 3: point is (9, 6), gradient = 1/3.

Tangent: y − 6 = (1/3)(x − 9) → y = x/3 + 3.

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Summary

• The tangent at x = a has gradient f'(a) and passes through (a, f(a)).
• The normal at x = a has gradient −1/f'(a) and passes through (a, f(a)).
• Use the point-gradient form: y − y₁ = m(x − x₁).
• Perpendicular lines satisfy m₁ × m₂ = −1.
• To find where a tangent meets the curve again, solve f(x) = tangent equation; the point of tangency gives a repeated root.
• For parametric curves, dy/dx = (dy/dt) ÷ (dx/dt).

Exam Tip: Always show clearly: (i) the coordinates of the point, (ii) the gradient calculation, (iii) the equation in a standard form. This earns full method marks even if you make a small arithmetic error. If asked for the equation in a particular form (e.g. ax + by + c = 0), make sure you rearrange accordingly.

A-Level Deep Dive: Tangents and Normals

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section G: Differentiation. The relevant content statement covers "use of differentiation to find equations of tangents and normals to curves". This sits inside the broader Section G strand on first principles, the power rule, products, quotients, chain rule, and applications to stationary points and rates of change. Tangents and normals are also examined synoptically in Section H (Integration, where the area between a curve and its tangent appears in mensuration applications), Section D (Coordinate geometry, where the relationship "tangent to a circle is perpendicular to the radius at the point of contact" recurs), and Section J (Numerical methods, where Newton-Raphson uses the tangent at $x_n$xn​ to predict $x_{n+1}$xn+1​). The AQA formula booklet provides differentiation rules but does not state the point-gradient form of a straight line — candidates must memorise $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​).

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Worked example with full mark scheme

Question (8 marks):

The curve $C$C has equation $y = x^3 - 4x^2 + 5x - 2$y=x3−4x2+5x−2.

(a) Find the equation of the tangent to $C$C at the point $P(3, 4)$P(3,4), giving your answer in the form $y = mx + c$y=mx+c. (4)

(b) The normal to $C$C at $P$P meets the $x$x-axis at $Q$Q. Find the exact coordinates of $Q$Q. (4)

Solution with mark scheme:

(a) Step 1 — differentiate.

$\dfrac{dy}{dx} = 3x^2 - 8x + 5$dxdy​=3x2−8x+5

M1 — correct application of the power rule term-by-term. The most common slip is leaving the $-2$−2 constant intact in the derivative; the derivative of a constant is zero.

Step 2 — evaluate the gradient at $x = 3$x=3.

$m_T = 3(3)^2 - 8(3) + 5 = 27 - 24 + 5 = 8$mT​=3(3)2−8(3)+5=27−24+5=8

A1 — correct gradient $m_T = 8$mT​=8 obtained by substituting the $x$x-coordinate of $P$P into $\frac{dy}{dx}$dxdy​.

Step 3 — apply point-gradient form.

$y - 4 = 8(x - 3)$y−4=8(x−3)

M1 — substituting both the gradient and the coordinates of $P$P correctly into $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​).

Step 4 — rearrange to the requested form.

$y = 8x - 24 + 4 = 8x - 20$y=8x−24+4=8x−20

A1 — final answer $y = 8x - 20$y=8x−20 in the requested form $y = mx + c$y=mx+c.

(b) Step 1 — find the normal gradient.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

$m_N = -\dfrac{1}{m_T} = -\dfrac{1}{8}$mN​=−mT​1​=−81​

M1 — correct use of the perpendicularity condition $m_T \cdot m_N = -1$mT​⋅mN​=−1.

Step 2 — equation of the normal through $P(3, 4)$P(3,4).

$y - 4 = -\dfrac{1}{8}(x - 3)$y−4=−81​(x−3)

M1 — substituting the new gradient and the same point $P$P into point-gradient form. A frequent error is to use the foot of the normal rather than the point of contact; the normal is anchored at $P$P.

Step 3 — set $y = 0$y=0 to find where the normal meets the $x$x-axis.

$-4 = -\dfrac{1}{8}(x - 3) \implies 32 = x - 3 \implies x = 35$−4=−81​(x−3)⟹32=x−3⟹x=35

M1 — correct manipulation to isolate $x$x.

A1 — exact coordinates $Q(35, 0)$Q(35,0).

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = \dfrac{4}{x} + x^2$y=x4​+x2 for $x > 0$x>0.

(a) Find $\dfrac{dy}{dx}$dxdy​. (2)

(b) The point $A$A on $C$C has $x$x-coordinate $2$2. Find the equation of the normal to $C$C at $A$A, giving your answer in the form $ax + by + c = 0$ax+by+c=0 where $a$a, $b$b, $c$c are integers. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — rewriting $\dfrac{4}{x}$x4​ as $4x^{-1}$4x−1 before differentiating, and applying the power rule.
• A1 (AO1.1b) — correct derivative $\dfrac{dy}{dx} = -4x^{-2} + 2x$dxdy​=−4x−2+2x or equivalent $2x - \dfrac{4}{x^2}$2x−x24​.

(b)

• B1 (AO1.1b) — $y$y-coordinate of $A$A: $y = \dfrac{4}{2} + 4 = 6$y=24​+4=6, so $A(2, 6)$A(2,6).
• M1 (AO1.1b) — tangent gradient $m_T = -1 + 4 = 3$mT​=−1+4=3, hence normal gradient $m_N = -\dfrac{1}{3}$mN​=−31​.
• M1 (AO1.1b) — applying point-gradient form: $y - 6 = -\dfrac{1}{3}(x - 2)$y−6=−31​(x−2).
• A1 (AO2.5) — rearranging and clearing fractions to integer form: $x + 3y - 20 = 0$x+3y−20=0.

Total: 6 marks split AO1 = 5, AO2 = 1. AQA tangent/normal questions are AO1-dominated; the AO2 mark is reserved for the presentation step that converts the answer into the requested integer form.

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Synoptic links

Connects to:

1. Section D — Coordinate geometry (circles): the tangent to a circle at any point is perpendicular to the radius at that point. The same negative-reciprocal-gradient principle that produces the equation of a normal to a curve produces the equation of a tangent to a circle. Confidence with one transfers directly to the other.

2. Section G — Implicit and parametric differentiation: for curves like $x^2 + y^2 = 25$x2+y2=25 or $x = t^2$x=t2, $y = t^3$y=t3, the gradient at a point still equals $\dfrac{dy}{dx}$dxdy​, but obtained via implicit differentiation ($2x + 2y \frac{dy}{dx} = 0$2x+2ydxdy​=0) or the chain rule ($\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​). Once the gradient is in hand, tangent and normal construction is identical.

3. Section H — Integration (area under a tangent): mensuration questions sometimes ask for the area enclosed between a curve, its tangent at a specified point, and the $x$x- or $y$y-axis. This requires the tangent equation first, then a definite integral.

4. Section O — Mechanics (kinematics): the velocity vector of a particle moving on a curve is tangent to the path; speed is the magnitude of the rate-of-change vector along the curve. The geometric meaning of "derivative as gradient of tangent" is the same as "velocity as instantaneous direction of motion".

5. Section J — Numerical methods (Newton-Raphson): the iteration $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$xn+1​=xn​−f′(xn​)f(xn​)​ is exactly the $x$x-intercept of the tangent to $y = f(x)$y=f(x) at $x = x_n$x=xn​. Every Newton-Raphson question is, in disguise, a tangent question repeated.

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Mark-scheme literacy

Tangent/normal questions on AQA 7357 split AO marks heavily toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	70–80%	Differentiating correctly, evaluating the gradient at a specified $x$x, applying point-gradient form, using the negative-reciprocal rule for normals
AO2 (reasoning / interpretation)	15–25%	Justifying the perpendicularity condition, expressing the answer in the requested form, recognising when a tangent passes through a stated external point
AO3 (problem-solving)	0–10%	Open contexts (tangent from an external point, intersection of two normals) appear in Year 2 or in synoptic problems

Examiner-rewarded phrasing: "the gradient of the tangent equals $\frac{dy}{dx}$dxdy​ evaluated at $x = a$x=a"; "the normal is perpendicular to the tangent, so $m_N = -\frac{1}{m_T}$mN​=−mT​1​"; "in the form $ax + by + c = 0$ax+by+c=0 with $a$a, $b$b, $c$c integers". Phrases that lose marks: writing "gradient = $y/x$y/x" (treating the curve as a straight line through the origin); leaving an answer with fractional coefficients when integer form is requested; using the point-gradient formula with a different point (e.g. an axis-intercept rather than the point of contact).