Tangents and Normals by Calculus – OCR FSMQ Additional Maths

1 The Method

CA4: find the equation of a tangent and normal at any point on a curve. This is where differentiation meets straight-line coordinate geometry — the derivative supplies the gradient, and the straight-line formula does the rest.

The five steps

1. If only $x$ is given, find $y$ by substituting into the curve
2. Differentiate to get $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
3. Substitute the $x$-value to get the numerical gradient $m$
4. For a normal, replace $m$ by $-\dfrac1m$
5. Use $y - y_1 = m(x - x_1)$

The gradients

Tangent: $m_{\text{T}} = \left.\dfrac{\mathrm{d}y}{\mathrm{d}x}\right|_{x = x_1}$ · Normal: $m_{\text{N}} = -\dfrac{1}{m_{\text{T}}}$

Do not substitute the $x$-value before differentiating. Differentiate the general expression first, then substitute. Substituting first turns the curve into a number, whose derivative is zero.

2 Finding the Two Lines

Worked Example 1 — Tangent and normal together

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

①$y = 4 + 6 - 1 = 9$, so the point is $(2, 9)$.

②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + 3$

③At $x = 2$: $m_{\text{T}} = 4 + 3 = 7$

④Tangent: $y - 9 = 7(x-2)$, so $y = 7x - 5$.

⑤$m_{\text{N}} = -\dfrac17$

⑥Normal: $y - 9 = -\tfrac17(x-2)$

⑦$7y - 63 = -x + 2$, so $x + 7y = 65$.

Check both lines pass through the point. Tangent: $7(2) - 5 = 9$ ✓ Normal: $2 + 7(9) = 65$ ✓

Worked Example 2 — When the tangent is horizontal

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

①$y = 1 - 3 + 2 = 0$, so the point is $(1, 0)$.

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

③At $x = 1$: $m_{\text{T}} = 3 - 3 = 0$ — the tangent is horizontal.

④Tangent: $y = 0$, the $x$-axis itself.

⑤A line perpendicular to a horizontal line is vertical, so the normal is $x = 1$.

Two habits this example teaches. First, always find the $y$-coordinate and confirm the point lies on the curve. Second, when $m_{\text{T}} = 0$, never write $-\tfrac{1}{0}$ — state instead that the normal is the vertical line $x = x_1$.

Worked Example 3 — A cubic

Find the tangent to $y = x^3 - 4x$ at the point $(2, 0)$.

①Check: $8 - 8 = 0$ ✓

②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4$

③At $x = 2$: $m = 12 - 4 = 8$

④$y - 0 = 8(x - 2)$, so $y = 8x - 16$.

3 What Questions Do Next

Finding the line is rarely the end of the question. These are the standard follow-ons.

Worked Example 4 — Intercepts and an area

The tangent to $y = x^2 + 1$ at $(2, 5)$ meets the axes at $P$ and $Q$. Find the area of the triangle $OPQ$.

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

②Tangent: $y - 5 = 4(x-2)$, so $y = 4x - 3$.

③$y$-intercept: $(0, -3)$. $x$-intercept: $0 = 4x - 3$, so $\left(\tfrac34, 0\right)$.

④The legs along the axes have lengths $3$ and $\tfrac34$.

⑤Area $= \tfrac12 \times 3 \times \tfrac34 = \dfrac{9}{8}$ square units.

Use lengths, not signed coordinates. The intercept is at $-3$ but the side length is $3$.

Worked Example 5 — A tangent with a given gradient

Find the points on $y = x^3 - 6x$ at which the tangent is parallel to $y = 6x + 1$.

①Parallel means the tangent gradient is $6$.

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

③$3x^2 = 12$, so $x^2 = 4$ and $x = \pm 2$.

④At $x = 2$: $y = 8 - 12 = -4$, giving $(2, -4)$.

⑤At $x = -2$: $y = -8 + 12 = 4$, giving $(-2, 4)$.

Do not stop at one answer. Setting the derivative equal to a number usually gives a quadratic, so expect two points.

Worked Example 6 — Where a tangent meets the curve again

Find where the tangent to $y = x^3$ at $(1, 1)$ meets the curve again.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 = 3$ at $x = 1$.

②Tangent: $y - 1 = 3(x-1)$, so $y = 3x - 2$.

③Set equal to the curve: $x^3 = 3x - 2$, so $x^3 - 3x + 2 = 0$.

④$x = 1$ is a root (the point of contact). Dividing by $(x-1)$: $x^2 + x - 2$.

⑤$x^2 + x - 2 = (x+2)(x-1)$, so $x^3 - 3x + 2 = (x-1)^2(x+2)$.

⑥The repeated factor $(x-1)^2$ is the tangency — it is why the tangent touches rather than crosses.

⑦The other root is $x = -2$, where $y = -8$. The tangent meets the curve again at $(-2, -8)$.

A repeated root always signals a tangent. That connects this topic straight back to the discriminant work in Setting Up and Solving Equations.

4 Quick Reference

Step order

Find $y$, differentiate, substitute, then write the line.

Never substitute first

Differentiate the expression, not a number.

Check the point

Confirm it lies on the curve.

Tangent gradient

Straight from the derivative.

Normal gradient

$-\dfrac{1}{m}$.

$m = 0$

Tangent $y = y_1$; normal $x = x_1$.

Line form

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

Intercepts

Set $x = 0$, then $y = 0$.

Given gradient

Set the derivative equal to it — expect two answers.

Meets again

Solve line $=$ curve; tangency gives a repeated root.

5 Practice Questions

Question 1

Find the gradient of the tangent to $y = x^2 + 4x$ at $x = 1$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + 4$

At $x = 1$: $m = 6$

Question 2

Find the equation of the tangent to $y = x^2$ at $(4, 16)$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 8$ at $x = 4$.

$y - 16 = 8(x - 4)$

$y = 8x - 16$

Question 3

Find the equation of the normal to $y = x^2$ at $(4, 16)$.

▶ Show solution

Tangent gradient $8$, so normal gradient $-\dfrac18$.

$y - 16 = -\tfrac18(x-4)$

$8y - 128 = -x + 4$, so $x + 8y = 132$.

Question 4

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

▶ Show solution

$y = 3 - 2 + 1 = 2$, so the point is $(1, 2)$.

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - 2 = 4$ at $x = 1$.

$y - 2 = 4(x-1)$, so $y = 4x - 2$.

Question 5

Find the normal to $y = x^3 + 2$ at $(-1, 1)$.

▶ Show solution

Check: $-1 + 2 = 1$ ✓

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 = 3$ at $x = -1$.

Normal gradient $-\dfrac13$.

$y - 1 = -\tfrac13(x+1)$, so $3y - 3 = -x - 1$ and $x + 3y = 2$.

Question 6

Find the point on $y = x^2 - 4x + 7$ where the tangent is horizontal.

▶ Show solution

$2x - 4 = 0$, so $x = 2$.

$y = 4 - 8 + 7 = 3$. The point is $(2, 3)$.

Question 7

Find the points on $y = x^3 - 9x$ where the tangent is parallel to the $x$-axis.

▶ Show solution

$3x^2 - 9 = 0$, so $x^2 = 3$ and $x = \pm\sqrt3$.

At $x = \sqrt3$: $y = 3\sqrt3 - 9\sqrt3 = -6\sqrt3$

At $x = -\sqrt3$: $y = 6\sqrt3$

Points: $\left(\sqrt3,\, -6\sqrt3\right)$ and $\left(-\sqrt3,\, 6\sqrt3\right)$.

Question 8

The tangent to $y = x^2 - 1$ at $(3, 8)$ crosses the $x$-axis. Find where.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 6$ at $x = 3$.

Tangent: $y - 8 = 6(x-3)$, so $y = 6x - 10$.

Setting $y = 0$: $x = \dfrac{10}{6} = \dfrac{5}{3}$.

It crosses at $\left(\tfrac53, 0\right)$.

Question 9

Find the equation of the tangent to $y = x^2 - 5x$ that is parallel to $y = 3x$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 5 = 3$, so $x = 4$.

Then $y = 16 - 20 = -4$, giving the point $(4, -4)$.

$y + 4 = 3(x-4)$, so $y = 3x - 16$.

Question 10

The curve $C$ has equation $y = x^2 - 2x$, and $P$ is the point on $C$ with $x = 3$.

(a) Find the equations of the tangent and the normal at $P$. (b) The normal meets $C$ again at $Q$. Find the coordinates of $Q$. (c) Find the length $PQ$ in exact form. (d) Show that the tangent at $P$ and the tangent at the point where $x = -1$ meet on the line $x = 1$, and explain the significance of that line.

▶ Show solution

(a) $y = 9 - 6 = 3$, so $P(3, 3)$.

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 = 4$ at $x = 3$.

Tangent: $y - 3 = 4(x-3)$, so $y = 4x - 9$.

Normal: gradient $-\tfrac14$, so $y - 3 = -\tfrac14(x-3)$, giving $4y = -x + 15$, i.e. $x + 4y = 15$.

(b) Substitute $y = x^2 - 2x$ into $x + 4y = 15$:

$x + 4\left(x^2 - 2x\right) = 15$

$4x^2 - 7x - 15 = 0$

$x = 3$ is a root (that is $P$), so factorise: $(x-3)(4x+5) = 0$.

The other root is $x = -\tfrac54$, where $y = \tfrac{25}{16} + \tfrac{5}{2} = \tfrac{25 + 40}{16} = \tfrac{65}{16}$.

$Q\left(-\dfrac{5}{4},\; \dfrac{65}{16}\right)$

(c) $\Delta x = 3 - \left(-\tfrac54\right) = \tfrac{17}{4}$ and $\Delta y = 3 - \tfrac{65}{16} = -\tfrac{17}{16}$

$PQ^2 = \left(\tfrac{17}{4}\right)^2 + \left(\tfrac{17}{16}\right)^2 = \dfrac{289}{16} + \dfrac{289}{256} = \dfrac{4624 + 289}{256} = \dfrac{4913}{256}$

$PQ = \dfrac{\sqrt{4913}}{16} = \dfrac{17\sqrt{17}}{16}$ (since $4913 = 17^3$)

As a decimal: $4.38$ (3 s.f.)

(d) At $x = -1$: $y = 1 + 2 = 3$, and the gradient is $2(-1) - 2 = -4$.

Tangent there: $y - 3 = -4(x+1)$, so $y = -4x - 1$.

Setting the two tangents equal: $4x - 9 = -4x - 1 \;\Rightarrow\; 8x = 8 \;\Rightarrow\; x = 1$ ✓

They meet at $(1, -5)$, which lies on $x = 1$.

The significance: $x = 1$ is the line of symmetry of the parabola (the vertex is at $x = 1$, since $\tfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ there). The two points $x = 3$ and $x = -1$ are equally spaced either side of it, so their tangents are mirror images and must meet on the axis of symmetry.

Tangents and Normals (CA4) · OCR FSMQ Additional Maths · Created with MathJax