5.4 Tangents and normals

Introduction

A tangent is the straight line that touches a curve at a point and has the same direction as the curve there. Subtopic 5.1 introduced it as the limit of chords, and subtopic 5.3 gave a fast way to find its gradient from the derivative. Turning that gradient into the full equation of a line, and finding the line at right angles to it, the normal, describes the behaviour of a curve at one point by two straight lines that can be written down, drawn and used.

Tangents and normals describe real directions. A car that loses grip on a bend continues along the tangent to its path, and a ski jumper leaves the ramp along the tangent to its profile. In optics, the angles at which light meets and leaves a curved mirror are measured from the normal to the mirror's surface. The methods combine the derivative with the equations of straight lines from subtopic 2.1, Equations of straight lines.

Tangents to a curve

The point of contact and its gradient

Every tangent touches its curve at one particular point. Everything about the tangent is fixed by that point: the line must pass through it, and its gradient must equal the gradient of the curve there, which is the value of the derivative at that $x$.

Two different numbers come from $a$, and each comes from a different function. The $y$-coordinate of the point of contact comes from the original function, $f(a)$: it is the height of the curve. The gradient of the tangent comes from the derivative, $f'(a)$: it is the steepness of the curve. Mixing up these two values is a frequent source of error.

The equation of the tangent

Once the point of contact $(a, f(a))$ and the gradient $m = f'(a)$ are known, the tangent is the straight line through a known point with a known gradient. The point-gradient form of a line from subtopic 2.1 writes it down immediately.

The method has four steps, and each step uses one piece of information.

1. Find the point of contact: substitute $x = a$ into $f(x)$ to get $f(a)$ (skip this if the point is given).

2. Differentiate to find $f'(x)$.

3. Substitute $x = a$ into $f'(x)$ to get the gradient $m = f'(a)$

4. Substitute the point and the gradient into $y - y_1 = m(x - x_1)$, then rearrange into the form the question asks for.

An equivalent fourth step substitutes the point into $y = mx + c$ and solves for $c$. Both give the same line; the point-gradient form avoids a separate equation for $c$.

The notation $\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x = 2}$ means "the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $x = 2$", the same number as $f'(2)$.

Forms of the answer

The same tangent can be written in several equivalent forms, and a question often specifies one. The gradient-intercept form $y = mx + c$ shows the gradient and the $y$-intercept. The general form $ax + by + d = 0$ is usually requested with $a, b, d \in \mathbb{Z}$, so every coefficient must be an integer. Moving every term to one side and multiplying through by the lowest common denominator of any fractions produces it.

Any integer multiple of a correct general form describes the same line, so $3x - 4y = 0$ and $-3x + 4y = 0$ are both correct, as is $6x - 8y = 0$. Writing the version with the smallest integers and a positive coefficient of $x$ is tidy and conventional.

A tangent's equation can also be used to prove a property of the line, such as passing through a given point. The tangent is found exactly as before, and the property is then read from the equation.

Local linearity

A tangent as an approximation

A smooth curve that is magnified more and more around one point looks straighter and straighter, and at a high enough magnification it cannot be told apart from its tangent. Close to the point of contact, the tangent is therefore a good approximation to the curve. Far from the point the curve bends away from the line and the approximation becomes poor.

The same curve and tangent as before, viewed on a window only 0.4 units wide around $x = 2$, show this clearly. On the wide window the curve visibly bends away from the line; on the narrow one the two are almost on top of each other.

The tangent gives a quick estimate of a function's value near the point of contact: substitute the nearby $x$ into the equation of the tangent instead of into the function. This is the same idea as estimating a change by "rate × change in $x$": moving from $x = a$ to $x = a + h$ changes $y$ by about $f'(a) \times h$.

Normals to a curve

The perpendicular line at the point of contact

The normal to a curve at a point is the line through the point of contact at right angles to the tangent. Two lines are perpendicular when the product of their gradients is $-1$, so the gradient of the normal is the negative reciprocal of the tangent's gradient.

The normal passes through the same point of contact as the tangent, so the method is the tangent method with one extra step: after finding $f'(a)$, take its negative reciprocal, then substitute the point and the new gradient into the point-gradient form.

The curve $y = \frac{x^3}{3} - x^2 + 2$ at $(3, 2)$ shows both lines. The tangent $y = 3x - 7$ has gradient 3 and the normal $y = -\frac{1}{3}x + 3$ has gradient $-\frac{1}{3}$; the product of the gradients is $-1$, and on axes with equal scales the two lines meet at a right angle.

Horizontal tangents and vertical normals

When $f'(a) = 0$ the tangent is horizontal, with equation $y = f(a)$. The negative reciprocal $-\frac{1}{0}$ is not defined, so the formula cannot give the normal's gradient. A line perpendicular to a horizontal line is vertical, though, so the normal is the vertical line through the point of contact, $x = a$. Points where the tangent is horizontal are the stationary points of subtopic 5.6.

Normals in optics: the law of reflection

When light meets a mirror, the angle of incidence equals the angle of reflection, and both angles are measured from the normal to the surface at the point where the light strikes, not from the surface itself. For a flat mirror the normal is the same everywhere. For a curved mirror, such as a car's wing mirror or the dish of a reflecting telescope, the normal changes from point to point, and finding it at the point of contact uses exactly the method for the equation of a normal.

The tangent at the point of contact behaves like a small flat mirror placed there, which is local linearity again. The normal, perpendicular to that tangent, is the reference line from which both angles are measured, so a curved mirror's effect on every ray is fixed by the normals along its surface.

A real reflector: the FAST radio telescope

The Five-hundred-meter Aperture Spherical Telescope (FAST) in Guizhou, China, pulls part of its reflecting surface into the shape of a parabola, and the radio waves collected by a circle 300 m across are focused on a receiver in a feed cabin suspended about 140 m above the dish. A parabola $y = \frac{x^2}{4f}$ sends every ray that arrives parallel to its axis through the point $(0, f)$, its focus. Taking $f = 140$ gives a model of the working reflector, $y = \frac{x^2}{560}$ for $-150 \le x \le 150$, with $x$ and $y$ in metres, and the equation of a normal shows why a ray striking the very edge still reaches the receiver.

Dimensions: 300 m illuminated aperture, feed cabin about 140 m above the reflector, Wikipedia, "Five-hundred-meter Aperture Spherical Telescope", accessed 2026. The parabola is a model built from these two figures.

Where a normal meets the curve again

A tangent or normal is a straight line, and a straight line can cross a curve at more than one point. Finding the other intersection means solving the equation of the curve and the equation of the line simultaneously: set the two expressions for $y$ equal. The point of contact is always one solution, which is a useful check.

Using technology

Gradients and tangent lines on the GDC

The GDC supports every step of the tangent method. It finds a numerical derivative at a point (TI-84 Plus CE: MATH → 8:nDeriv(; Casio fx-CG50: MENU → Run-Matrix, then OPTN → CALC → d/dx; or, on a graph, TI-84 Plus CE 2nd TRACE (CALC) → dy/dx), which gives $f'(a)$ directly when the differentiation is awkward or needs checking. Most models also draw the tangent at a chosen point of a graph and display its equation: on a TI-84 Plus CE, 2nd PRGM (DRAW) → 5:Tangent( from the graph screen; on a Casio fx-CG50, the Sketch menu of the graph screen has Tangent; on a TI-Nspire, the Graphs application's geometry tools include a tangent.

The displayed equation is rounded, and it gives no working, so it is a check rather than a method. A tangent found by hand that disagrees with the GDC's tangent usually has the $y$-coordinate taken from the derivative, or a sign error in a negative power.

1. Graph $y = f(x)$ in a window that shows the point of contact.

2. Use the draw-tangent tool at $x = a$ and read the equation $y = mx + c$.

3. Compare $m$ with the hand value of $f'(a)$, and $c$ with the constant in the hand-derived equation.

A "show that" question about a tangent is different, because the result is already stated and the derivation is what is being asked for.

Working backwards from a tangent

Finding the point where the tangent has a given gradient

A common reverse problem gives the gradient of a tangent and asks where on the curve it occurs. The gradient of the tangent at $x$ is $f'(x)$, so the $x$-coordinates are the solutions of the equation $f'(x) = m$. Each solution gives a point of contact; the $y$-coordinate again comes from $f(x)$, not from $f'(x)$.

The equation $f'(x) = m$ may have one solution, several or none, and each solution is a separate tangent with gradient $m$. Tangents with the same gradient are parallel, so counting the solutions of $f'(x) = m$ counts the tangents parallel to a given line. Graphing $y = f'(x)$ together with the horizontal line $y = m$ shows the number of solutions at a glance, and the GDC finds each one.

If the gradient of the normal is given instead, it is converted to the tangent's gradient first, using the negative reciprocal again: a normal gradient of $-\frac{1}{5}$ means a tangent gradient of 5.

Finding a parameter from a tangent

When a curve's equation contains an unknown parameter, a condition on a tangent gives an equation for it. The usual conditions are a gradient at a given $x$ (substitute into the derivative and set equal to the gradient), a tangent parallel to a given line (equal gradients) or a point that the curve passes through (substitute into the function). Two unknowns need two such conditions, giving a pair of simultaneous equations.

Where a tangent meets the axes

A tangent meets the $x$-axis where $y = 0$ and the $y$-axis where $x = 0$. Together with the origin, these two points form a right-angled triangle whose area is half the product of the two intercepts' distances from the origin. The same idea finds where a tangent meets any other line: solve the two equations simultaneously.

Tangents and normals in context

Moving along a tangent

An object moving along a curved path has, at each instant, a direction of motion along the tangent to the path. If the force holding it on the curve disappears, it continues in a straight line along that tangent, as Newton's first law states: a car that hits ice on a bend slides off along the tangent, and a hammer thrower's hammer, when released, flies off along the tangent to its circle. The equation of that tangent predicts where the object goes next, at least until other forces act.

Joining a straight edge smoothly to a curve

Designers of roads, rails, metal sheets and furniture often join a curved section to a straight one. For the join to have no corner, the straight section must continue in the direction of the curve at the join, so it lies along the tangent at the point where they meet. Railway and road engineers take this further and match the curvature as well, but matching the tangent is the first requirement, and it is a direct use of the equation of a tangent.

Exam focus

Tangents and normals appear in short Paper 1 questions worth 4–7 marks and as parts of longer Paper 2 questions, almost always straight after a part that asks for the derivative. The command terms are find (the equation of the tangent or normal, the gradient, the coordinates of a point), write down (a gradient read from a derivative already found, or a vertical normal), show that (a given equation of a tangent, or a parameter value, derived algebraically) and determine or justify (the number of tangents with a given gradient, with a reason). Answers are frequently required in a stated form, especially $ax + by + d = 0$ with integer coefficients, and the final accuracy mark is withheld when fractions remain or another form is given. Follow-through from a wrong derivative is allowed in later parts, provided the method is shown.

In part (d) the GDC solves the equation, but the equation itself must be written down, and the solution $x = 2$ has to be recognised as $P$ and discarded. By hand, multiplying by $2x^2$ gives $6x^4 - 21x^2 - 12 = 0$, so $x^2 = 4$ and $x = \pm 2$.

Questions that ask "how many" tangents have a property need a reason as well as a number.

Practice questions

1. Find the equation of the tangent to $y = x^3 - 5x + 2$ at the point where $x = 2$. [4]

2. Find the equation of the normal to $y = \frac{4}{x}$ at the point $(2, 2)$. [4]

3. The function $f$ is defined by $f(x) = 2x^2 - x + 3$. Find the equation of the normal to the graph of $f$ at $x = 1$

4. The curve $y = x^2 - 6x + 10$ has a horizontal tangent at the point $A$. Find the coordinates of $A$ and write down the equation of the normal at $A$

5. Find the coordinates of the points on the curve $y = x^3 - 3x^2 + 1$ at which the tangent is parallel to the line $y = 9x - 4$

6. The curve $y = kx^2 + \frac{3}{x}$, $x \ne 0$

7. The normal to the curve $y = \frac{x^3}{3} + 2$ at the point $P$ has gradient $-\frac{1}{4}$

8. The tangent to $y = x^2 + 1$ at the point where $x = 2$ meets the $x$-axis at $A$

9. Let $f(x) = x^4 - 2x$. (a) Find the equation of the tangent to the graph of $f$ at $x = 1$. [3] (b) Use the tangent to estimate $f(1.05)$

Summary

• The tangent to $y = f(x)$ at $x = a$ passes through the point of contact $(a, f(a))$ and has gradient $f'(a)$: $y - f(a) = f'(a)(x - a)$.

• The $y$-coordinate comes from the function and the gradient from the derivative; the gradient is a number, never an expression in $x$.

• The normal is perpendicular to the tangent at the point of contact, with gradient $-\frac{1}{f'(a)}$, because perpendicular gradients multiply to $-1$.

• A horizontal tangent $y = f(a)$ has the vertical normal $x = a$.

• Answers in the form $ax + by + d = 0$ with $a, b, d \in \mathbb{Z}$ need every fraction cleared; integer multiples are equivalent.

• Points where the tangent has gradient $m$ solve $f'(x) = m$; the number of solutions is the number of tangents parallel to a line of gradient $m$.

• A parameter is found by turning a condition on a tangent into an equation; "show that" working starts from the given facts.

• Close to the point of contact, the tangent approximates the curve (local linearity); the GDC's tangent tool is a check, not a substitute for working.

Self-check

• To do:

  I can find the point of contact $(a, f(a))$ from the original function.

• To do:

  I can find the gradient of a tangent by substituting $x = a$ into the derivative.

• To do:

  I can write the equation of a tangent using $y - y_1 = m(x - x_1)$.

• To do:

  I can rearrange an equation into $y = mx + c$ or into $ax + by + d = 0$ with integer coefficients.

• To do:

  I can find the gradient of a normal as the negative reciprocal of the tangent's gradient.

• To do:

  I can write down the normal $x = a$ when the tangent is horizontal.

• To do:

  I can find a tangent or normal for a curve containing a term such as $\frac{6}{x}$.

• To do:

  I can find where a tangent or normal meets the curve again or meets the axes.

• To do:

  I can find the points where the tangent has a given gradient by solving $f'(x) = m$.

• To do:

  I can convert a given normal gradient into a tangent gradient before solving.

• To do:

  I can count and justify the number of tangents parallel to a given line.

• To do:

  I can find a parameter from a condition on a tangent, working forwards in a "show that".

• To do:

  I can use a tangent to estimate a nearby value of a function and judge how reliable it is.

• To do:

  I can use the GDC's numerical derivative and draw-tangent tool to check my answers.

where $x$ is the horizontal distance, in metres, from the start of the mound and $y$ is the height in metres. The two curves meet at $x = 4$ and have the same gradient there, so that the track is smooth. The second curve passes through the point $(10,\ 2.2)$.

1. (a)

   Find the height of the track and the gradient of the track when $x = 4$.

   [2]2 marks

   Markscheme (a)

   height $= 0.1(4)^2 = 1.6$ (m) A1

   gradient $= 0.2(4) = 0.8$ A1

   [2 marks]

2. (b)

   Show that $8a + b = 0.8$.

   [2]2 marks

   Markscheme (b)

   $\frac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b$

3. (c)

   Write down two further equations in $a$, $b$ and $c$.

   [2]2 marks

   Markscheme (c)

   $16a + 4b + c = 1.6$

4. (d)

   Find the values of $a$, $b$ and $c$.

   [2]2 marks

   Markscheme (d)

   attempt to solve the system using technology (M1)

   $a = -0.117$, $b = 1.73$

5. (e)

   Find the maximum height of the mound.

   [2]2 marks

   Markscheme (e)

   $2ax + b = 0$ OR maximum seen on a graph (M1)

   $x = 7.43$, height $= 2.97$

6. (f)

   After $x = 10$ the track continues in a straight line which is a tangent to the second curve at $x = 10$. Find the horizontal distance from the start of the mound to the point where this straight section reaches the ground ($y = 0$).

   [4]4 marks

   Markscheme (f)

   gradient $= 2(-0.11666\ldots)(10) + 1.7333\ldots$

The point of contact is $(2, 3)$. Differentiate, then substitute $x = 2$ for the gradient.

Substitute the point and the gradient into the point-gradient form and rearrange.

As a check, the line passes through $(2, 3)$: $4(2) - 5 = 3$.

(b) Point of contact: $f(-2) = 3(4) - \frac{8}{-2} = 12 + 4 = 16$, so the point is $(-2, 16)$.

Gradient: $f'(-2) = 6(-2) + \frac{8}{4} = -12 + 2 = -10$.

Equation of the tangent:

The brackets matter: $x - (-2) = x + 2$. All three coefficients, 10, 1 and 4, are integers, as required.

Gradient: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{2} - x^{-2}$, so at $x = 2$ the gradient is $1 - \frac{1}{4} = \frac{3}{4}$.

Equation of the tangent:

The equation has no constant term, so when $x = 0$, $y = 0$: the tangent passes through the origin. In integer form the tangent is $3x - 4y = 0$.

In a "show that" question the conclusion has to be stated. Here that means writing the equation and saying why it shows the line passes through $(0, 0)$, rather than only quoting a GDC result.

(c) At $x = 2.5$ the tangent gives $4(2.5) - 5 = 5$, while the curve has $y = 15.625 - 12.5 + 3 = 6.125$, an error of about 18%. The point $x = 2.5$ is too far from the point of contact: the curve bends away from the tangent, so the linear approximation is only reliable close to $x = 2$.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of 3.

Substitute the point $(3, 2)$ and this gradient into the point-gradient form, then clear the fraction by multiplying every term by 3.

Check: $3 + 3(2) - 9 = 0$, so the point $(3, 2)$ lies on the line.

The gradient of the tangent at $x = 2$ is 0, so the tangent is horizontal.

(b) The point of contact is $(2, f(2))$, with $f(2) = 4 - 6 + 2 = 0$. The horizontal tangent through $(2, 0)$ is $y = 0$, which is the $x$-axis itself. The normal is perpendicular to it, so it is the vertical line $x = 2$.

(b) The normal's gradient is the negative reciprocal, $-\frac{28}{15} = -1.87$ (3 s.f.).

(c) The tangent at $E$ makes an angle $\tan^{-1}\left(\frac{15}{28}\right)$ with the horizontal. The normal is perpendicular to the tangent and the ray is perpendicular to the horizontal, so the angle between the ray and the normal is the same angle.

(d) The reflected ray makes an angle $2i = 56.36\ldots^\circ$ with the vertical and heads towards the axis, which is 150 m away horizontally. In the right-angled triangle formed by the ray, the vertical through $E$ and the horizontal, the ray rises

so it crosses $x = 0$ at a height of $40.18 + 99.82 = 140$ m, the height of the receiver.

The same calculation at any other point of the parabola also ends at $(0, 140)$, which is why one receiver collects the waves from the whole 300 m circle.

(b) At an intersection the two $y$-values are equal.

The solution $x = 1$ is the point $A$, as expected. The other is $x = -1.5$, with $y = (-1.5)^2 = 2.25$, so $B$ is $(-1.5, 2.25)$. With the GDC, graph $y = x^2$ and $y = -0.5x + 1.5$ and find both intersections.

(b) A tangent parallel to $L$ has gradient $-4$, so its point of contact satisfies

Graphing $y = 2x - \frac{2}{x^2}$ and $y = -4$ on the GDC (or solving the equivalent cubic $x^3 + 2x^2 - 1 = 0$, found by multiplying by $x^2$ and dividing by 2) gives three solutions: $x = -1.62$, $x = -1$ and $x = 0.618$. One of them, $x = -1$, is the point $P$ itself, so there are two other tangents parallel to $L$.

(c) Substitute each $x$ into $f(x)$, not $f'(x)$.

The points of contact are $(0.618, 3.62)$ and $(-1.62, 1.38)$. By hand, the cubic factorises as $(x + 1)(x^2 + x - 1) = 0$, giving $x = \frac{-1 \pm \sqrt{5}}{2}$.

Substitute $x = 1$ and set the result equal to 9.

The working starts from the given information and arrives at $k = 3$ without using that value; substituting $k = 3$ to check that the gradient is 9 would not show the result.

(b) With $k = 3$, the point of contact has $y = 1 + 3 - 5 = -1$. The tangent passes through $(1, -1)$ with gradient 9.

(b) On the $x$-axis $y = 0$, so $x = 4$ and $A$ is $(4, 0)$. On the $y$-axis $x = 0$, so $B$ is $(0, 12)$. The area of the right-angled triangle is $\frac{1}{2} \times 4 \times 12 = 24$ square units.

(c) At $x = a$ the point of contact is $\left(a, \frac{12}{a}\right)$ and the gradient is $-\frac{12}{a^2}$.

Setting $y = 0$ gives $\frac{12}{a^2}x = \frac{24}{a}$, so $x = 2a$; setting $x = 0$ gives $y = \frac{24}{a}$. The area is

which does not depend on $a$, so every tangent to $C$ forms a triangle of area 24 with the axes. The working uses a general point, so it proves the result for all $a > 0$, not just the one checked in part (b).

(b) $\frac{\mathrm{d}y}{\mathrm{d}x} = 0.007x - 0.6$, so at $x = 60$ the gradient is $0.42 - 0.6 = -0.18$.

(c) The gradient of a line is the tangent of its angle of inclination (subtopic 2.1). The gradient $-0.18$ means the line falls, so the skier's direction is below the horizontal by

The skier leaves the ramp heading about $10.2^\circ$ below the horizontal. The model ignores the skier's own push at take-off and air resistance, so this is only the initial direction.

The ramp profile is hypothetical.

Award A1 for $3x^2 - 4$ and A1 for $-6x^{-2}$. Award at most (M1)A1A0 if any extra term is seen.

(b) $f'(2) = 3(2)^2 - 4 - \frac{6}{2^2}$ M1

$= 12 - 4 - 1.5$ A1

$= 6.5$ AG

(c) $m_{\text{normal}} = -\frac{1}{6.5} = -\frac{2}{13}$ M1

$y - 3 = -\frac{2}{13}(x - 2)$ M1

$13y - 39 = -2x + 4$ (A1)

$2x + 13y - 43 = 0$ A1

Accept any integer multiple. Do not award the final A1 if the answer is not in the required form.

(d) $3x^2 - 4 - \frac{6}{x^2} = 6.5$ (M1)

$x = -2$ (from the GDC; $x = 2$ is $P$) A1

$f(-2) = -8 + 8 - 3 = -3$, so $Q(-2, -3)$ A1

(e) $y + 3 = 6.5(x + 2)$ (M1)

$y = 6.5x + 10$ A1

Total [14]

$2a(4) + b = 0.8$ A1

$8a + b = 0.8$ AG

[2 marks]

$100a + 10b + c = 2.2$ A1

[2 marks]

[2 marks]

[2 marks]

$= -0.6$ (A1)

$y - 2.2 = -0.6(x - 10)$ (M1)

$0 = 2.2 - 0.6(x - 10)$

$x = 13.7$ (m) $\quad(13.6666\ldots)$ A1

Note: Follow through from part (d). Accept $\frac{41}{3}$.

[4 marks]

[2]2 marks

[3]3 marks