Tangent and Normal Lines

Finding the equations of tangent and normal lines to curves is a fundamental skill at A-Level. This lesson consolidates the techniques for both Cartesian and parametric curves, as required by the Edexcel 9MA0 specification.

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The Tangent Line

The tangent to a curve at a point P is the straight line that just touches the curve at P. Its gradient equals the gradient of the curve at P, which is found by differentiation.

For y = f(x)

1. Differentiate to find dy/dx.
2. Substitute the x-coordinate of P to find the gradient m.
3. Use the point-gradient form: y − y₁ = m(x − x₁).

Example: Find the tangent to y = x³ − 2x + 1 at x = 1.

dy/dx = 3x² − 2. At x = 1: dy/dx = 3 − 2 = 1. Point: y = 1 − 2 + 1 = 0, so P = (1, 0). Tangent: y − 0 = 1(x − 1), so y = x − 1.

For Parametric Curves

1. Find dy/dx = (dy/dt)/(dx/dt).
2. Substitute the parameter value to find the gradient.
3. Find the point by substituting into x = f(t) and y = g(t).
4. Use point-gradient form.

Example: For x = t², y = 2t, find the tangent at t = 3.

dx/dt = 2t = 6, dy/dt = 2. dy/dx = 2/6 = 1/3. Point: (9, 6). Tangent: y − 6 = (1/3)(x − 9), so y = x/3 + 3.

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The Normal Line

The normal to a curve at a point P is the straight line perpendicular to the tangent at P. If the tangent has gradient m, the normal has gradient −1/m.

Example: Find the normal to y = x² at x = 3.

dy/dx = 2x. At x = 3: gradient = 6. Normal gradient = −1/6. Point: (3, 9). Normal: y − 9 = (−1/6)(x − 3), so y = −x/6 + 9.5 = −x/6 + 19/2.

Or: 6y = −x + 57, i.e. x + 6y − 57 = 0.

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Tangent and Normal to a Circle

At a point P on a circle, the normal passes through the centre (since the radius is perpendicular to the tangent).

Finding the Tangent

1. Find the gradient of the radius from centre C to point P.
2. The tangent gradient is the negative reciprocal.
3. Write the equation using point P.

Example: Find the tangent to (x − 1)² + (y − 2)² = 25 at the point (4, 6).

Gradient of radius CP: (6 − 2)/(4 − 1) = 4/3. Tangent gradient: −3/4. Tangent: y − 6 = (−3/4)(x − 4), so y = (−3/4)x + 9.

Finding the Normal

The normal is simply the line through C and P.

Normal: y − 2 = (4/3)(x − 1), so y = (4/3)x + 2/3.

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Tangent from an External Point

To find the tangent(s) from a point outside a curve:

1. Let the point of tangency be (a, f(a)) on the curve.
2. Write the tangent equation at this general point.
3. Require this tangent to pass through the external point.
4. Solve the resulting equation for a.

Example: Find the equations of the tangents to y = x² from the point (0, −1).

The tangent at (a, a²) has gradient 2a: y − a² = 2a(x − a), so y = 2ax − a².

This must pass through (0, −1): −1 = 2a(0) − a² = −a². a² = 1, so a = 1 or a = −1.

Tangent 1 (a = 1): y = 2x − 1. Tangent 2 (a = −1): y = −2x − 1.

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Where the Tangent/Normal Meets the Curve Again

A common exam question asks you to find where the tangent or normal at a point meets the curve again.

Example: The tangent to y = x³ at x = 1 meets the curve again at what point?

Tangent at (1, 1): dy/dx = 3x² = 3. Equation: y − 1 = 3(x − 1), so y = 3x − 2.

Set equal to the curve: x³ = 3x − 2, so x³ − 3x + 2 = 0.

We know x = 1 is a root (the tangent point), so (x − 1) is a factor. x³ − 3x + 2 = (x − 1)(x² + x − 2) = (x − 1)(x + 2)(x − 1) = (x − 1)²(x + 2).

The tangent meets the curve again at x = −2, y = −8. Point: (−2, −8).

Exam Tip: When the tangent at x = a meets the curve again, x = a will always be a repeated root of the resulting equation (since the tangent touches at that point). Factor out (x − a)² if possible, or (x − a) twice.

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Key Terms

Term	Definition
Tangent	A line that touches a curve at a point with the same gradient as the curve
Normal	A line perpendicular to the tangent at the point of contact
Gradient	The slope of the curve at a given point, found by differentiation
External point	A point not on the curve from which tangent lines are drawn

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Summary

• The tangent at a point has gradient equal to dy/dx (or (dy/dt)/(dx/dt) for parametric curves).
• The normal has gradient −1/(dy/dx), the negative reciprocal.
• For circles, the normal passes through the centre; the tangent is perpendicular to the radius.
• To find tangents from an external point, use a general point of tangency and impose the condition.
• Finding where a tangent meets the curve again involves solving a polynomial equation where the tangent point gives a repeated root.

A-Level Deep Dive: Tangent and Normal Lines

Spec mapping

Edexcel 9MA0-01 specification section 4 — Coordinate geometry in the $(x, y)$(x,y) plane covers the equation of a tangent and a normal to a curve at a given point (refer to the official specification document for exact wording). This is paired explicitly with section 9 — Differentiation, sub-strand 9.1: "interpret $\dfrac{dy}{dx}$dxdy​ as a rate of change and as the gradient of a curve at a point." The tangent at a point is the straight line through the point with gradient equal to the derivative evaluated there; the normal is the perpendicular line, with gradient equal to the negative reciprocal. Tangent–normal questions are examined throughout 9MA0-01 (Pure 1) and 9MA0-02 (Pure 2). Synoptic links extend to section 4 (Circles, where the tangent at a point on a circle is perpendicular to the radius at that point), section 9 sub-strands 9.2–9.5 (chain, product, quotient rule and implicit differentiation, used when the curve is not given as $y = f(x)$y=f(x)), section 7 (Parametric equations and parametric differentiation, where $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​) and 9MA0-03 Mechanics, where the tangent direction at a point on a trajectory is the velocity vector. The Edexcel formula booklet does not list the tangent or normal equation forms — they must be reconstructed from the point–gradient form $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​) and the perpendicularity condition $m_1 m_2 = -1$m1​m2​=−1.

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Worked example with full mark scheme

Question (8 marks): A curve $C$C has equation $y = x^3 - 2x^2 + 1$y=x3−2x2+1.

(a) Find an equation of the tangent to $C$C at the point $P$P where $x = 2$x=2, giving your answer in the form $y = mx + c$y=mx+c. (4)

(b) Find an equation of the normal to $C$C at $P$P, giving your answer in the form $ax + by + c = 0$ax+by+c=0 where $a$a, $b$b and $c$c are integers. (4)

Solution with mark scheme:

(a) Step 1 — find the $y$y-coordinate of $P$P.

At $x = 2$x=2: $y = 2^3 - 2(2^2) + 1 = 8 - 8 + 1 = 1$y=23−2(22)+1=8−8+1=1. So $P = (2, 1)$P=(2,1).

B1 — correct coordinates of $P$P. A surprising number of candidates skip this and end up with the wrong point in their tangent equation. Without $P$P correctly identified, every subsequent A mark is in jeopardy.

Step 2 — differentiate.

$\dfrac{dy}{dx} = 3x^2 - 4x$dxdy​=3x2−4x

M1 — applying the power rule term-by-term. The constant $+1$+1 differentiates to $0$0 and disappears.

Step 3 — evaluate the gradient at $P$P.

At $x = 2$x=2: $\dfrac{dy}{dx} = 3(4) - 4(2) = 12 - 8 = 4$dxdy​=3(4)−4(2)=12−8=4.

M1 — substituting $x = 2$x=2 into $\dfrac{dy}{dx}$dxdy​ to get the numerical gradient. A common error: writing the tangent equation with $\dfrac{dy}{dx}$dxdy​ left as the symbolic expression $3x^2 - 4x$3x2−4x instead of the number $4$4. That is no longer the equation of a straight line and earns no further marks.

Step 4 — write the tangent equation.

Using $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​) with $m = 4$m=4 and $(x_1, y_1) = (2, 1)$(x1​,y1​)=(2,1):

$y - 1 = 4(x - 2) \implies y = 4x - 7$y−1=4(x−2)⟹y=4x−7

A1 — correct tangent equation in the requested form $y = mx + c$y=mx+c, with $m = 4$m=4, $c = -7$c=−7.

(b) Step 1 — gradient of the normal.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $4$4:

$m_{\text{normal}} = -\dfrac{1}{4}$mnormal​=−41​

M1 — using $m_1 m_2 = -1$m1​m2​=−1 to obtain the gradient of the normal as $-\tfrac{1}{4}$−41​.

Step 2 — write the normal equation through $P$P.

$y - 1 = -\dfrac{1}{4}(x - 2)$y−1=−41​(x−2)

M1 — point–gradient form with the correct point $P = (2, 1)$P=(2,1) and the negative-reciprocal gradient.

Step 3 — clear fractions.

Multiply through by $4$4:

$4(y - 1) = -(x - 2) \implies 4y - 4 = -x + 2 \implies x + 4y - 6 = 0$4(y−1)=−(x−2)⟹4y−4=−x+2⟹x+4y−6=0

A1 — correct integer form $x + 4y - 6 = 0$x+4y−6=0.

A1 — final answer presented in the exact requested form $ax + by + c = 0$ax+by+c=0 with integer $a = 1$a=1, $b = 4$b=4, $c = -6$c=−6. Leaving the answer as $y = -\tfrac{1}{4}x + \tfrac{3}{2}$y=−41​x+23​ would lose this final A1 even though the line is correct.

Total: 8 marks (B1 M1 M1 A1 M1 M1 A1 A1).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = e^{x} \sin x$y=exsinx for $0 \leq x \leq \pi$0≤x≤π.

(a) Show that $\dfrac{dy}{dx} = e^{x}(\sin x + \cos x)$dxdy​=ex(sinx+cosx). (2)

(b) Hence find an equation of the tangent to $C$C at the point where $x = \dfrac{\pi}{4}$x=4π​, giving your answer in the form $y = mx + c$y=mx+c, where $m$m and $c$c are exact constants. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying the product rule with $u = e^x$u=ex, $v = \sin x$v=sinx: $\dfrac{dy}{dx} = u'v + uv' = e^x \sin x + e^x \cos x$dxdy​=u′v+uv′=exsinx+excosx.
• A1 (AO1.1b) — factoring out $e^x$ex to give $e^x(\sin x + \cos x)$ex(sinx+cosx) in the printed form.

(b)

• M1 (AO1.1b) — evaluating $y$y at $x = \tfrac{\pi}{4}$x=4π​: $y = e^{\pi/4} \sin(\pi/4) = \dfrac{\sqrt{2}}{2} e^{\pi/4}$y=eπ/4sin(π/4)=22​​eπ/4.
• M1 (AO1.1b) — evaluating the gradient: $\dfrac{dy}{dx} = e^{\pi/4}(\sin(\pi/4) + \cos(\pi/4)) = e^{\pi/4}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\right) = \sqrt{2}\, e^{\pi/4}$dxdy​=eπ/4(sin(π/4)+cos(π/4))=eπ/4(22​​+22​​)=2​eπ/4.
• M1 (AO2.1) — substituting into $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​) with $m = \sqrt{2}\, e^{\pi/4}$m=2​eπ/4 and the point above.
• A1 (AO2.5) — final answer $y = \sqrt{2}\, e^{\pi/4} x + \dfrac{\sqrt{2}}{2} e^{\pi/4}\left(1 - \dfrac{\pi}{2}\right)$y=2​eπ/4x+22​​eπ/4(1−2π​) (or equivalent simplified exact form).

Total: 6 marks split AO1 = 4, AO2 = 2. This is the typical AO profile of a tangent question with non-polynomial $y$y: AO1 dominates because the bulk of the work is procedural (product rule, substitute, write down a line). The AO2 marks pick up the candidate's interpretation — that the tangent passes through the curve at the given point with the gradient just computed — and the discipline of leaving everything in exact form rather than collapsing to decimals.

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Synoptic links

Connects to:

1. Section 9 — Differentiation rules: every tangent or normal beyond the simplest polynomial requires the chain, product or quotient rule. For $y = (3x + 1)^4$y=(3x+1)4 at $x = 1$x=1, the chain rule gives $\dfrac{dy}{dx} = 12(3x + 1)^3$dxdy​=12(3x+1)3, and the tangent gradient at $x = 1$x=1 is $12 \cdot 4^3 = 768$12⋅43=768. Confidence with the differentiation toolkit is the prerequisite for every coordinate-geometry tangent question above grade C.

2. Section 4 — Circles: the tangent to a circle at a point on its circumference is perpendicular to the radius drawn to that point. This gives a non-calculus route to tangent equations for circles — find the radius gradient, take the negative reciprocal, write the line through the point. The same answer drops out of implicit differentiation of $x^2 + y^2 = r^2$x2+y2=r2, but the geometric route is faster.

3. Section 9.5 / 7 — Implicit and parametric differentiation: when the curve is given as $x^2 + xy + y^2 = 7$x2+xy+y2=7 (implicit) or $(x, y) = (t^2, t^3 - 3t)$(x,y)=(t2,t3−3t) (parametric), $\dfrac{dy}{dx}$dxdy​ at a point still gives the tangent gradient. Implicit: differentiate both sides with respect to $x$x, treating $y$y as a function of $x$x, and solve for $\dfrac{dy}{dx}$dxdy​. Parametric: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​. The form of the tangent equation is unchanged.

4. 9MA0-03 Mechanics — kinematics in two dimensions: for a particle whose position vector is $\mathbf{r}(t) = (x(t), y(t))$r(t)=(x(t),y(t)), the velocity $\mathbf{v}(t) = \dot{\mathbf{r}}(t)$v(t)=r˙(t) is tangent to the trajectory at time $t$t, and its direction equals the tangent direction $\dfrac{dy}{dx}$dxdy​ at that point. The same calculation underlies "find the equation of the tangent to the path at the moment when $t = 2$t=2".

5. Section 8 — Integration (area between curve and tangent): A common A* extension is to find the area enclosed between a curve and one of its tangents, or between the curve, its tangent and a coordinate axis. This combines tangent-finding (calculus) with definite integration (also calculus) and routinely appears in the second half of Paper 2.

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Mark-scheme literacy

Tangent and normal questions on 9MA0 split AO marks as:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–75%	Differentiating correctly, evaluating the derivative at the point, applying the point–gradient form, producing an answer in the requested format
AO2 (reasoning / interpretation)	20–30%	Recognising that "gradient at a point" means evaluate the derivative, not leave it symbolic; using the negative reciprocal correctly with sign; presenting answers in exact form and the requested integer-coefficient layout
AO3 (problem-solving)	5–15%	Tangents-from-an-external-point, area enclosed by a tangent, second tangents passing through a given point — the cluster of "open" tangent problems that combine ideas

Examiner-rewarded phrasing: "by the chain rule, $\dfrac{dy}{dx} = \ldots$dxdy​=…"; "the gradient of the tangent at $P$P is $\dfrac{dy}{dx}\bigg|_{x = x_1} = \ldots$dxdy​​x=x1​​=…"; "the gradient of the normal is the negative reciprocal: $m_{\perp} = -\dfrac{1}{m}$m⊥​=−m1​"; "in the form $ax + by + c = 0$ax+by+c=0 with $a, b, c$a,b,c integers". Phrases that lose marks: writing "gradient $= \dfrac{y}{x}$=xy​" (fundamentally wrong — that's the gradient of the chord from the origin, not the tangent gradient); leaving the gradient as the symbolic derivative inside the line equation (e.g. $y - 1 = (3x^2 - 4x)(x - 2)$y−1=(3x2−4x)(x−2), which is not a line); using $+\dfrac{1}{m}$+m1​ instead of $-\dfrac{1}{m}$−m1​ for the normal gradient.

A specific Edexcel pattern: questions phrased "in the form $ax + by + c = 0$ax+by+c=0 where $a$a, $b$b and $c$c are integers" demand that you clear fractions and collect terms. Leaving $y = -\tfrac{1}{4}x + \tfrac{3}{2}$y=−41​x+23​ — mathematically the same line — costs the final A1. Read the form constraint precisely.

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Grade-band model answers

3-mark question

Question: Find an equation of the tangent to the curve $y = x^2 - 3x$y=x2−3x at the point where $x = 1$x=1.

Grade C response (~190 words):