Tangents and Normals – A Level Maths Year 1

1 What Tangents and Normals Are

The 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 there. The normal at $P$ is the straight line through $P$ that is perpendicular to the tangent. DfE content area G asks you to use differentiation to find equations of tangents and normals.

The two gradients at x = a

Tangent: $m_T = \mathrm{f}'(a)$ Normal: $m_N = -\dfrac{1}{\mathrm{f}'(a)}$ (since $m_T \times m_N = -1$)

Equation of a line through (x₁, y₁) with gradient m

$y - y_1 = m(x - x_1)$

Find the coordinates of the point: if you are only given $x$, substitute into $y$ (the original equation).
Differentiate and substitute $x$ into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to get the tangent gradient.
For a normal, take the negative reciprocal.
Use $y - y_1 = m(x - x_1)$ and rearrange into the form the question asks for.

Two different substitutions. The $y$-coordinate comes from $y = \mathrm{f}(x)$; the gradient comes from $\mathrm{f}'(x)$. Mixing them up is the most common error on this topic.

2 Finding the Equations

Worked Example 1 — Tangent and normal to a quadratic

Find the equations of the tangent and the normal to $y = x^2 - 3x + 1$ at the point where $x = 4$. Give the normal in the form $ax + by + c = 0$.

①Point: $y = 16 - 12 + 1 = 5$, so $P(4, 5)$.

②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 3$; at $x = 4$ the gradient is $5$.

③Tangent: $y - 5 = 5(x - 4) \Rightarrow y = 5x - 15$.

④Normal gradient $-\dfrac15$: $y - 5 = -\dfrac15(x - 4)$.

⑤Multiply by $5$: $5y - 25 = -x + 4 \Rightarrow x + 5y - 29 = 0$.

Worked Example 2 — A reciprocal curve

Find the tangent and normal to $y = \dfrac{4}{x}$ at $(2, 2)$.

①$y = 4x^{-1}$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4x^{-2} = -\dfrac{4}{x^2}$. At $x = 2$: $-1$.

②Tangent: $y - 2 = -(x - 2) \Rightarrow y = -x + 4$.

③Normal gradient $-\dfrac{1}{-1} = 1$: $y - 2 = x - 2 \Rightarrow y = x$.

The normal passes through the origin — a nice check, since the curve is symmetrical about $y = x$.

Worked Example 3 — A triangle formed by a tangent and normal

The tangent and normal to $y = 2\sqrt x$ at the point where $x = 9$ meet the $x$-axis at $A$ and $B$. Find the area of the triangle formed by $A$, $B$ and the point of contact.

①Point: $y = 2 \times 3 = 6$, so $P(9, 6)$.

②$y = 2x^{1/2}$, $\;\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{-1/2} = \dfrac{1}{\sqrt x}$. At $x = 9$: $\dfrac13$.

③Tangent: $y - 6 = \dfrac13(x - 9) \Rightarrow y = \dfrac13 x + 3$. At $y = 0$: $x = -9$, so $A(-9, 0)$.

④Normal: gradient $-3$: $y - 6 = -3(x - 9) \Rightarrow y = -3x + 33$. At $y = 0$: $x = 11$, so $B(11, 0)$.

⑤Base $AB = 20$, height $= 6$ (the $y$-coordinate of $P$). Area $= \tfrac12 \times 20 \times 6 = 60$.

3 Working Backwards from a Gradient

Sometimes you are told something about the tangent — for example that it is parallel to a given line — and must find the point. Parallel lines have equal gradients, so set $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ equal to that gradient.

Worked Example 4 — Tangents parallel to a line

Find the equations of the tangents to $y = x^3 - 3x^2 + 2$ that are parallel to $y = 9x + 1$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x = 9$

②$3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$

③$x = 3$: $y = 27 - 27 + 2 = 2$. Tangent $y - 2 = 9(x - 3) \Rightarrow y = 9x - 25$.

④$x = -1$: $y = -1 - 3 + 2 = -2$. Tangent $y + 2 = 9(x + 1) \Rightarrow y = 9x + 7$.

Worked Example 5 — Finding a constant so a line is a tangent

The line $y = 2x + k$ is a tangent to $y = x^2 - 4x + 10$. Find $k$.

①The curve must have gradient $2$ at the point of contact: $2x - 4 = 2 \Rightarrow x = 3$.

②Point of contact: $y = 9 - 12 + 10 = 7$, so $(3, 7)$.

③This lies on the line: $7 = 6 + k \Rightarrow k = 1$.

Alternative without calculus: $x^2 - 4x + 10 = 2x + k$ gives $x^2 - 6x + (10 - k) = 0$. A tangent meets the curve once, so the discriminant is zero: $36 - 4(10 - k) = 0 \Rightarrow k = 1$ ✓

Worked Example 6 — Where a normal meets the curve again

The normal to $y = x^2$ at $(1, 1)$ meets the curve again at $Q$. Find $Q$.

①Gradient at $x = 1$ is $2$, so normal gradient is $-\dfrac12$.

②Normal: $y - 1 = -\dfrac12(x - 1) \Rightarrow y = -\dfrac12 x + \dfrac32$.

③Solve with the curve: $x^2 = -\dfrac12 x + \dfrac32 \Rightarrow 2x^2 + x - 3 = 0 \Rightarrow (2x + 3)(x - 1) = 0$.

④$x = 1$ is $P$ itself, so $x = -\dfrac32$, $y = \dfrac94$. $Q\left(-\dfrac32,\, \dfrac94\right)$.

You always know one root already — the original point. Use it to factorise.

4 Horizontal Tangents and Vertical Normals

If $\mathrm{f}'(a) = 0$ the tangent is horizontal, $y = \mathrm{f}(a)$, and the normal is vertical, $x = a$. The formula $-\dfrac{1}{\mathrm{f}'(a)}$ fails here (division by zero), so think geometrically instead.

Worked Example 7 — A zero gradient

Find the tangent and normal to $y = 2\sqrt x + \dfrac{8}{x}$ at $x = 4$.

①$y = 4 + 2 = 6$; $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{-1/2} - 8x^{-2} = \dfrac12 - \dfrac12 = 0$.

②Tangent: $y = 6$. Normal: $x = 4$.

5 Quick Reference

Tangent gradient

$m = \mathrm{f}'(a)$

Normal gradient

$-\dfrac{1}{m}$ (negative reciprocal)

Line equation

$y - y_1 = m(x - x_1)$

The point

$y$-coordinate from $\mathrm{f}(a)$, not $\mathrm{f}'(a)$.

Parallel

Equal gradients: solve $\mathrm{f}'(x) = m$.

Meets again

Solve line and curve; one root is known.

Zero gradient

Tangent $y = \mathrm{f}(a)$; normal $x = a$.

Final form

$ax + by + c = 0$ with integers if asked.

6 Practice Questions

Question 1

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

▶ Show solution

Point $(1, 3)$. $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 2$.

$y - 3 = 2(x - 1) \Rightarrow y = 2x + 1$.

Question 2

Find the equation of the normal to $y = x^3$ at $(1, 1)$, in the form $ax + by + c = 0$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 = 3$, so normal gradient $-\dfrac13$.

$y - 1 = -\dfrac13(x - 1) \Rightarrow 3y - 3 = -x + 1 \Rightarrow x + 3y - 4 = 0$.

Question 3

Find the equation of the tangent to $y = \dfrac{6}{x}$ at the point where $x = 3$, in the form $ax + by + c = 0$.

▶ Show solution

Point $(3, 2)$. $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{6}{x^2} = -\dfrac69 = -\dfrac23$.

$y - 2 = -\dfrac23(x - 3) \Rightarrow 3y - 6 = -2x + 6 \Rightarrow 2x + 3y - 12 = 0$.

Question 4

Find the equations of the tangent and normal to $y = 4\sqrt x - x$ at the point where $x = 4$.

▶ Show solution

$y = 8 - 4 = 4$, so the point is $(4, 4)$.

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-1/2} - 1 = \dfrac{2}{2} - 1 = 0$.

Tangent: $y = 4$ (horizontal). Normal: $x = 4$ (vertical).

Question 5

Find the point on $y = x^2 - 5x$ where the tangent has gradient $3$, and the equation of that tangent.

▶ Show solution

$2x - 5 = 3 \Rightarrow x = 4$, $\;y = 16 - 20 = -4$. Point $(4, -4)$.

$y + 4 = 3(x - 4) \Rightarrow y = 3x - 16$.

Question 6

Find the equations of the tangent and normal to $y = x^3 - 2x^2 + 1$ at the point where $x = 2$.

▶ Show solution

$y = 8 - 8 + 1 = 1$, so $(2, 1)$. $\;\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x = 12 - 8 = 4$.

Tangent: $y - 1 = 4(x - 2) \Rightarrow y = 4x - 7$.

Normal: $y - 1 = -\dfrac14(x - 2) \Rightarrow 4y - 4 = -x + 2 \Rightarrow x + 4y - 6 = 0$.

Question 7

Show that the tangent to $y = x^2 - 4x + 1$ at $(1, -2)$ passes through the origin.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 4 = -2$ at $x = 1$.

$y + 2 = -2(x - 1) \Rightarrow y = -2x$.

At $x = 0$, $y = 0$, so the tangent passes through the origin.

Question 8

The normal to $y = x^2 + 1$ at $(1, 2)$ meets the curve again at $Q$. Find the coordinates of $Q$.

▶ Show solution

Gradient at $x = 1$ is $2$; normal gradient $-\dfrac12$.

Normal: $y - 2 = -\dfrac12(x - 1) \Rightarrow y = -\dfrac12 x + \dfrac52$.

$x^2 + 1 = -\dfrac12 x + \dfrac52 \Rightarrow 2x^2 + x - 3 = 0 \Rightarrow (2x + 3)(x - 1) = 0$.

$x = -\dfrac32$, $\;y = \dfrac94 + 1 = \dfrac{13}{4}$. $Q\left(-\dfrac32,\, \dfrac{13}{4}\right)$.

Question 9

The line $y = -3x + k$ is a tangent to the curve $y = x^2 + x + 5$. Find $k$ and the point of contact.

▶ Show solution

$2x + 1 = -3 \Rightarrow x = -2$, $\;y = 4 - 2 + 5 = 7$. Point $(-2, 7)$.

$7 = -3(-2) + k \Rightarrow k = 1$.

Question 10

The curve $C$ has equation $y = x^3 - 6x^2 + 9x$.

(a) Find the equation of the tangent to $C$ at the origin. (b) This tangent meets $C$ again at the point $R$. Find $R$. (c) Find the equations of the tangent and normal to $C$ at the point $P$ where $x = 2$. (d) The tangent and normal at $P$ cross the $y$-axis at $A$ and $B$. Find the area of triangle $APB$.

▶ Show solution

(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x + 9$; at $x = 0$ it is $9$. Tangent: $y = 9x$.

(b) $x^3 - 6x^2 + 9x = 9x \Rightarrow x^3 - 6x^2 = 0 \Rightarrow x^2(x - 6) = 0$.

$x = 0$ (a repeated root — the point of contact) or $x = 6$, giving $y = 54$. $R(6, 54)$.

(c) $y = 8 - 24 + 18 = 2$, so $P(2, 2)$. Gradient $= 12 - 24 + 9 = -3$.

Tangent: $y - 2 = -3(x - 2) \Rightarrow y = -3x + 8$.

Normal: gradient $\dfrac13$: $y - 2 = \dfrac13(x - 2) \Rightarrow y = \dfrac13 x + \dfrac43$.

(d) $A(0, 8)$ and $B\left(0, \tfrac43\right)$, so $AB = 8 - \tfrac43 = \tfrac{20}{3}$.

The height of the triangle is the horizontal distance from the $y$-axis to $P$, which is $2$.

Area $= \tfrac12 \times \tfrac{20}{3} \times 2 = \tfrac{20}{3}$ square units.

Tangents and Normals · A Level Maths Year 1 · Created with MathJax