Tangents and normals

AS · P1 · 14 min

A tangent touches a curve at a point and points the same way as the curve there; a normal crosses the curve at right angles at that point. Finding their equations is the most direct use of the derivative, and it is in almost every Paper 1: sometimes as a 3-mark opener, often as the first part of a longer question that goes on to find where the normal meets the curve again, or the area of a triangle formed with the axes. The method never changes, so these marks should be secure.

Tangent and normal at a point

Definition

The tangent to a curve at a point $P$ is the straight line through $P$ whose gradient equals the gradient of the curve at $P$ .

The normal to a curve at $P$ is the straight line through $P$ perpendicular to the tangent at $P$ .

Both lines pass through the same point, so you need just two things: the coordinates of $P$ , and a gradient.

The gradient of the tangent is the value of $\dfrac{dy}{dx}$ at $P$ .
The gradient of the normal comes from the perpendicular rule from coordinate geometry: two lines with gradients $m_1$ and $m_2$ are perpendicular when $m_1 m_2 = -1$ . So the normal gradient is the negative reciprocal of the tangent gradient.

Key result

At the point $(x_1, y_1)$ on a curve, with $m = \dfrac{dy}{dx}$ evaluated at $x = x_1$ :

\text{tangent: } y - y_1 = m(x - x_1)

\text{normal: } y - y_1 = -\frac{1}{m}(x - x_1)

If $m = 0$ the tangent is horizontal, $y = y_1$ , and the normal is vertical, $x = x_1$ .

y = x^3 - 3x + 1 y = 9x - 15 y = (29 - x)/9 (2, 3)

The curve is $y = x^3 - 3x + 1$ . At $(2, 3)$ the tangent $y = 9x - 15$ runs along the curve, and the normal $x + 9y = 29$ crosses it at right angles. Because the tangent is steep (gradient $9$ ), the normal is very shallow (gradient $-\tfrac{1}{9}$ ). The two lines only look perpendicular when the axes have equal scales, which this graph does not; the gradients are what matter.

Equation of a tangent or normal

Find the $y$ -coordinate of the point if you are only given $x$ (substitute into the curve).
Differentiate to get $\dfrac{dy}{dx}$ .
Substitute the $x$ -coordinate into $\dfrac{dy}{dx}$ to get the tangent gradient $m$ .
For a normal, use $-\dfrac{1}{m}$ .
Write $y - y_1 = (\text{gradient})(x - x_1)$ and rearrange into the form the question asks for.

Variations the exam uses

The basic method is the same, but the information arrives in different orders.

You are given	What to do first
The $x$ -coordinate of the point	Find $y$ from the curve, then the gradient
A point where the curve meets an axis	Set $y = 0$ (or $x = 0$ ) to find the point
The gradient of the tangent, or a line it is parallel to	Solve $\dfrac{dy}{dx} = m$ to find the point(s)
The gradient of the normal	Convert to the tangent gradient, then solve $\dfrac{dy}{dx} = m$
A line that is a tangent, with an unknown constant	Match the gradient to find the point, then substitute for the constant
A point the tangent passes through, not on the curve	Let the point of contact be $(a, f(a))$ and solve for $a$

Two lines are parallel when their gradients are equal. A tangent parallel to $y = 9x$ has gradient $9$ . A tangent parallel to the $x$ -axis has gradient $0$ , which is a stationary point.

After finding the lines, questions often ask for where they meet an axis, where they meet each other, or the area of the triangle they enclose. Those are coordinate-geometry steps: set $x = 0$ or $y = 0$ , solve simultaneous equations, and use $\tfrac{1}{2} \times \text{base} \times \text{height}$ with a base along an axis.

Worked examples

Tangent and normal at a given x

Find the equations of the tangent and the normal to $y = x^3 - 3x + 1$ at the point where $x = 2$ , giving the normal in the form $ax + by = c$ .

Solution

When $x = 2$ : $y = 8 - 6 + 1 = 3$ . The point is $(2, 3)$ .

\frac{dy}{dx} = 3x^2 - 3 \quad\Rightarrow\quad m = 3(4) - 3 = 9

Tangent: $y - 3 = 9(x - 2)$ , so $y = 9x - 15$ .

Normal: gradient $-\tfrac{1}{9}$ , so $y - 3 = -\tfrac{1}{9}(x - 2)$ . Multiply by $9$ : $9y - 27 = -x + 2$ , so

x + 9y = 29

A normal and the area of a triangle

The normal to the curve $y = \dfrac{8}{\sqrt{x}}$ at the point $P$ where $x = 4$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $OAB$ , where $O$ is the origin.

Solution

When $x = 4$ , $y = \dfrac{8}{2} = 4$ , so $P$ is $(4, 4)$ .

$y = 8x^{-\frac{1}{2}}$ , so $\dfrac{dy}{dx} = -4x^{-\frac{3}{2}}$ . At $x = 4$ : $\dfrac{dy}{dx} = -\dfrac{4}{8} = -\dfrac{1}{2}$ .

The normal gradient is $-1 \div \left(-\tfrac{1}{2}\right) = 2$ :

y - 4 = 2(x - 4) \quad\Rightarrow\quad y = 2x - 4

At $A$ , $y = 0$ , so $x = 2$ : $A(2, 0)$ . At $B$ , $x = 0$ , so $y = -4$ : $B(0, -4)$ .

Triangle $OAB$ has a right angle at $O$ , with $OA = 2$ and $OB = 4$ :

\text{area} = \tfrac{1}{2} \times 2 \times 4 = 4

Tangents parallel to a given line

Find the equations of the two tangents to the curve $y = x^3 - 3x^2 + 2$ which are parallel to the line $y = 9x$ .

Solution

Parallel to $y = 9x$ means gradient $9$ :

\begin{aligned} 3x^2 - 6x &= 9 \\ x^2 - 2x - 3 &= 0 \\ (x - 3)(x + 1) &= 0 \end{aligned}

At $x = 3$ : $y = 27 - 27 + 2 = 2$ , so the tangent is $y - 2 = 9(x - 3)$ , i.e. $y = 9x - 25$ .

At $x = -1$ : $y = -1 - 3 + 2 = -2$ , so the tangent is $y + 2 = 9(x + 1)$ , i.e. $y = 9x + 7$ .

y = x^3 - 3x^2 + 2 y = 9x - 25 y = 9x + 7 (3, 2) (-1, -2)

The two parallel tangents from the last example touch the curve at $(-1, -2)$ and $(3, 2)$ .

Where the normal meets the curve again

The point $P(3, 2)$ lies on the curve $y = x^2 - 4x + 5$ . The normal to the curve at $P$ meets the curve again at $Q$ . Find the coordinates of $Q$ .

Solution

$\dfrac{dy}{dx} = 2x - 4$ , which is $2$ at $x = 3$ . The normal gradient is $-\tfrac{1}{2}$ :

y - 2 = -\tfrac{1}{2}(x - 3) \quad\Rightarrow\quad y = -\tfrac{1}{2}x + \tfrac{7}{2}

Solve with the curve:

\begin{aligned} x^2 - 4x + 5 &= -\tfrac{1}{2}x + \tfrac{7}{2} \\ 2x^2 - 8x + 10 &= -x + 7 \\ 2x^2 - 7x + 3 &= 0 \\ (2x - 1)(x - 3) &= 0 \end{aligned}

$x = 3$ is $P$ itself, which is a useful check. So $Q$ has $x = \tfrac{1}{2}$ and

y = \tfrac{1}{4} - 2 + 5 = \tfrac{13}{4}

$Q$ is $\left(\tfrac{1}{2}, \tfrac{13}{4}\right)$ .

y = x^2 - 4x + 5 y = -x/2 + 7/2 (3, 2) (0.5, 3.25)

Tangent, normal and the x-axis

The curve $y = \sqrt{2x + 1}$ passes through the point $P(4, 3)$ . The tangent to the curve at $P$ meets the $x$ -axis at $A$ , and the normal at $P$ meets the $x$ -axis at $B$ . Find the area of triangle $APB$ .

Solution

$y = (2x + 1)^{\frac{1}{2}}$ , so by the chain rule

\frac{dy}{dx} = \tfrac{1}{2}(2x + 1)^{-\frac{1}{2}} \times 2 = \frac{1}{\sqrt{2x + 1}}

At $x = 4$ : $\dfrac{dy}{dx} = \dfrac{1}{3}$ .

Tangent: $y - 3 = \tfrac{1}{3}(x - 4)$ . At $A$ , $y = 0$ : $-9 = x - 4$ , so $x = -5$ and $A$ is $(-5, 0)$ .

Normal: gradient $-3$ , $y - 3 = -3(x - 4)$ . At $B$ , $y = 0$ : $-3 = -3x + 12$ , so $x = 5$ and $B$ is $(5, 0)$ .

The base $AB$ lies along the $x$ -axis with length $5 - (-5) = 10$ , and the height is the $y$ -coordinate of $P$ , which is $3$ :

\text{area} = \tfrac{1}{2} \times 10 \times 3 = 15

y = sqrt(2x + 1) y = (x + 5)/3 y = 15 - 3x fill -5 4 y = (x + 5)/3 fill 4 5 y = 15 - 3x (4, 3)

The shaded triangle $APB$ from the last example, with the tangent on the left and the normal on the right.

Tangents from a point not on the curve

Find the equations of the two tangents to the curve $y = x^2 + 4$ that pass through the origin.

Solution

The origin is not on the curve, so you cannot use it as the point of contact. Let the tangent touch the curve at $(a, a^2 + 4)$ . The gradient there is $2a$ , so the tangent is

y - (a^2 + 4) = 2a(x - a) \quad\Rightarrow\quad y = 2ax - a^2 + 4

It passes through $(0, 0)$ :

0 = -a^2 + 4 \quad\Rightarrow\quad a = 2 \text{ or } a = -2

With $a = 2$ the tangent is $y = 4x$ ; with $a = -2$ it is $y = -4x$ .

Check with the discriminant: $x^2 + 4 = 4x$ gives $x^2 - 4x + 4 = (x - 2)^2 = 0$ , a repeated root, so $y = 4x$ does touch the curve.

Watch out

Using the derivative itself as the gradient. The tangent gradient is a number, $\dfrac{dy}{dx}$ evaluated at the point. Writing $y - 3 = (3x^2 - 3)(x - 2)$ gives a curve, not a line.

Wrong normal gradient. The normal gradient is $-\dfrac{1}{m}$ , not $-m$ and not $\dfrac{1}{m}$ . For $m = -\tfrac{1}{2}$ the normal gradient is $2$ .

Finding $y$ from the derivative. The point's $y$ -coordinate comes from the equation of the curve.

Forgetting the second point. "Tangents parallel to $y = 9x$ " usually has two answers, because $\dfrac{dy}{dx} = 9$ is a quadratic equation.

Treating an external point as the point of contact. If the given point is not on the curve, introduce $(a, f(a))$ as the unknown point of contact.

Exam tip

Typical marks. A 4-mark tangent or normal question usually pays: one for differentiating, one for substituting to get the gradient, one for the perpendicular gradient (normals) and one for the final equation. Show each line.
Form of the answer. If the question asks for $ax + by + c = 0$ or $ax + by = c$ , give integer coefficients. Otherwise $y = mx + c$ or even $y - y_1 = m(x - x_1)$ is accepted, but a fully simplified form is safer.
Exact values. Keep fractions exact, e.g. $y = -\tfrac{1}{2}x + \tfrac{7}{2}$ , rather than $y = -0.5x + 3.5$ if decimals might become long.
"Meets the curve again" always leads to an equation with the original point as one root. Use that to factorise, and as a check on your normal.
Area questions. Sketch the lines. With a base along an axis, the height is a coordinate of the third vertex; there is no need for a distance formula.

Summary

Tangent gradient $m = \dfrac{dy}{dx}$ at the point; normal gradient $-\dfrac{1}{m}$ .
Equation of either line: $y - y_1 = (\text{gradient})(x - x_1)$ .
Always find the point on the curve first, from the curve's equation.
Parallel lines share a gradient; perpendicular gradients multiply to $-1$ .
Given a gradient, solve $\dfrac{dy}{dx} = m$ to find the point or points.
For tangents through an external point, use a general point of contact $(a, f(a))$ .
To find where a normal meets the curve again, solve simultaneously; one root is the original point.

Practice questions

Question

Find the equation of the tangent to $y = 2x^2 - 3x + 1$ at the point where $x = 2$ .
Find the equation of the normal to $y = x^3 - 2x$ at the point $(1, -1)$ .
Find the equation of the normal to $y = \dfrac{6}{x}$ at the point $(2, 3)$ , in the form $ax + by + c = 0$ .
Find the equation of the tangent to $y = (3x - 5)^4$ at the point where $x = 2$ .
The line $y = 2x + c$ is a tangent to the curve $y = x^2 - 4x + 10$ . Find the coordinates of the point of contact and the value of $c$ .
Find the coordinates of the point on $y = \sqrt{x}$ at which the normal has gradient $-4$ , and the equation of this normal.
The tangents to $y = x^2 - 2x + 3$ at the points where $x = 0$ and $x = 3$ meet at $P$ . Find the coordinates of $P$ .
Find the equation of the normal to $y = x^3$ at $(1, 1)$ , and show that it does not meet the curve again.
The tangent to the curve $y = \dfrac{4}{2x - 1}$ at the point where $x = 1$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $OAB$ .
Find the equations of the two tangents to $y = x^2 - 2x + 9$ that pass through the origin, and the coordinates of their points of contact.

Answers

At $x = 2$ , $y = 8 - 6 + 1 = 3$ . $\dfrac{dy}{dx} = 4x - 3 = 5$ . Tangent: $y - 3 = 5(x - 2)$ , i.e. $y = 5x - 7$ .
$\dfrac{dy}{dx} = 3x^2 - 2 = 1$ at $x = 1$ . Normal gradient $-1$ : $y + 1 = -(x - 1)$ , i.e. $y = -x$ .
$\dfrac{dy}{dx} = -6x^{-2} = -\tfrac{3}{2}$ at $x = 2$ . Normal gradient $\tfrac{2}{3}$ : $y - 3 = \tfrac{2}{3}(x - 2)$ , so $3y - 9 = 2x - 4$ , giving $2x - 3y + 5 = 0$ .
At $x = 2$ , $y = 1^4 = 1$ . $\dfrac{dy}{dx} = 4(3x - 5)^3 \times 3 = 12(3x - 5)^3 = 12$ . Tangent: $y - 1 = 12(x - 2)$ , i.e. $y = 12x - 23$ .
The tangent has gradient $2$ : $2x - 4 = 2$ , so $x = 3$ and $y = 9 - 12 + 10 = 7$ . The point of contact is $(3, 7)$ , and $7 = 6 + c$ gives $c = 1$ .
Normal gradient $-4$ means tangent gradient $\tfrac{1}{4}$ . $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} = \dfrac{1}{4}$ , so $\sqrt{x} = 2$ , $x = 4$ . The point is $(4, 2)$ . Normal: $y - 2 = -4(x - 4)$ , i.e. $y = -4x + 18$ .
$\dfrac{dy}{dx} = 2x - 2$ . At $x = 0$ : point $(0, 3)$ , gradient $-2$ , tangent $y = -2x + 3$ . At $x = 3$ : point $(3, 6)$ , gradient $4$ , tangent $y - 6 = 4(x - 3)$ , i.e. $y = 4x - 6$ . Solving $-2x + 3 = 4x - 6$ gives $x = \tfrac{3}{2}$ , $y = 0$ . So $P$ is $\left(\tfrac{3}{2}, 0\right)$ .
$\dfrac{dy}{dx} = 3x^2 = 3$ at $x = 1$ , so the normal has gradient $-\tfrac{1}{3}$ : $y - 1 = -\tfrac{1}{3}(x - 1)$ , i.e. $x + 3y = 4$ . Substituting $y = x^3$ : $x + 3x^3 = 4$ , so $3x^3 + x - 4 = 0$ . Since $x = 1$ is a root, $3x^3 + x - 4 = (x - 1)(3x^2 + 3x + 4)$ . The quadratic has discriminant $9 - 48 = -39 < 0$ , so it has no real roots. The only intersection is $(1, 1)$ , and the normal does not meet the curve again.
At $x = 1$ , $y = 4$ . $y = 4(2x - 1)^{-1}$ , so $\dfrac{dy}{dx} = -8(2x - 1)^{-2} = -8$ at $x = 1$ . Tangent: $y - 4 = -8(x - 1)$ , i.e. $y = 12 - 8x$ . $A$ is $\left(\tfrac{3}{2}, 0\right)$ and $B$ is $(0, 12)$ . Area $= \tfrac{1}{2} \times \tfrac{3}{2} \times 12 = 9$ .
Let the point of contact be $(a, a^2 - 2a + 9)$ , where the gradient is $2a - 2$ . Tangent: $y = (2a - 2)(x - a) + a^2 - 2a + 9$ . Through $(0, 0)$ : $0 = -(2a - 2)a + a^2 - 2a + 9 = -a^2 + 9$ , so $a = \pm 3$ . For $a = 3$ : contact $(3, 12)$ , gradient $4$ , tangent $y = 4x$ . For $a = -3$ : contact $(-3, 24)$ , gradient $-8$ , tangent $y = -8x$ .

Tangent and normal at a point

Variations the exam uses

Worked examples

Practice questions

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