🎚️ Tangents and Normals

OCR FSMQ Additional Maths · Coordinate Geometry (CG5)

Level 3 · Ages 15–16

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1 Tangent and Normal
CG5: know, understand and use gradient, intercept, tangent and normal in problems involving points that can be defined by equations and inequalities.
The two lines at a point on a curve
Tangent: touches the curve, with the same gradient as the curve there
Normal: perpendicular to the tangent, through the same point
P tangent normal curve At P the tangent matches the curve's gradient; the normal is at right angles to it
The gradient relationship
If the tangent has gradient $m$, the normal has gradient $-\dfrac{1}{m}$
Read which line is wanted. "Find the normal" and "find the tangent" differ by one step — taking the negative reciprocal. Answering the wrong one usually scores nothing.
2 Tangents and Normals to a Circle
The circle shortcut — no calculus needed
The normal at any point on a circle is the radius through that point
So the tangent is perpendicular to the radius
Worked Example 1 — Tangent and normal to a circle

The circle $(x-2)^2 + (y-3)^2 = 25$ passes through $P(6, 6)$. Find the equations of the tangent and the normal at $P$.

①Check $P$ is on the circle: $(6-2)^2 + (6-3)^2 = 16 + 9 = 25$ ✓
②Centre $C(2, 3)$. Gradient of $CP = \dfrac{6-3}{6-2} = \dfrac34$
③Normal (along the radius, gradient $\tfrac34$): $y - 6 = \tfrac34(x-6)$
④$4y - 24 = 3x - 18$, so $4y = 3x + 6$, i.e. $y = \tfrac34 x + \tfrac32$.
⑤Tangent gradient $= -\dfrac43$: $y - 6 = -\tfrac43(x-6)$
⑥$3y - 18 = -4x + 24$, so $4x + 3y = 42$.
The normal to a circle always passes through the centre. Check: does $(2,3)$ satisfy $y = \tfrac34x + \tfrac32$?  $\tfrac34(2) + \tfrac32 = 3$ ✓
3 Tangents and Normals to Other Curves

For a curve that is not a circle, the tangent's gradient comes from differentiation — this is where coordinate geometry and calculus meet.

The method
1. Differentiate to get $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
2. Substitute the $x$-coordinate to get the numerical gradient $m$
3. Tangent: $y - y_1 = m(x - x_1)$. Normal: use $-\tfrac1m$ instead.
Worked Example 2 — The specification's own example

Find the equation of the normal to the curve $y = x^3 - 3x + 2$ at the point $(1, 0)$.

①Check the point: $1 - 3 + 2 = 0$ ✓
②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 3$
③At $x = 1$: $3(1) - 3 = 0$. The tangent is horizontal.
④A normal perpendicular to a horizontal line is vertical.
⑤The normal is $x = 1$.
Do not write $-\tfrac10$. When the tangent gradient is $0$, the normal is the vertical line $x = x_1$. Similarly, if the tangent is vertical the normal is horizontal.
Worked Example 3 — A standard tangent

Find the tangent and the normal to $y = x^2 - 4x + 7$ at the point where $x = 3$.

①$y$-coordinate: $9 - 12 + 7 = 4$, so the point is $(3, 4)$.
②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 4$
③At $x = 3$: $m = 6 - 4 = 2$
④Tangent: $y - 4 = 2(x-3)$, so $y = 2x - 2$.
⑤Normal: gradient $-\tfrac12$, so $y - 4 = -\tfrac12(x-3)$
⑥$2y - 8 = -x + 3$, so $x + 2y = 11$.
Always find the $y$-coordinate first. Being given only $x$ is common, and forgetting to compute $y$ makes the rest of the question impossible.
4 Using the Tangent or Normal Afterwards

Finding the line is usually only part of the question. What follows is standard straight-line work.

Worked Example 4 — Intercepts and a triangle

The tangent to $y = x^2$ at $(2, 4)$ meets the axes at $A$ and $B$. Find the area of triangle $OAB$, where $O$ is the origin.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x$, so at $x = 2$ the gradient is $4$.
②Tangent: $y - 4 = 4(x-2)$, giving $y = 4x - 4$.
③$y$-intercept: $x = 0$ gives $y = -4$, so $A(0, -4)$.
④$x$-intercept: $y = 0$ gives $x = 1$, so $B(1, 0)$.
⑤The triangle has legs along the axes of lengths $4$ and $1$.
⑥Area $= \tfrac12 \times 1 \times 4 = 2$ square units.
Use the lengths of the intercepts, not their signed coordinates. The length here is $4$, even though the intercept is at $-4$.
Worked Example 5 — Where a normal cuts the curve again

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

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x = 2$ at $x = 1$, so the normal's gradient is $-\tfrac12$.
②Normal: $y - 1 = -\tfrac12(x-1)$, so $y = -\tfrac12 x + \tfrac32$.
③Set equal to the curve: $x^2 = -\tfrac12 x + \tfrac32$
④$2x^2 + x - 3 = 0$, so $(2x+3)(x-1) = 0$.
⑤$x = 1$ (the original point) or $x = -\tfrac32$.
⑥At $x = -\tfrac32$: $y = \tfrac94$. The normal meets the curve again at $\left(-\tfrac32, \tfrac94\right)$.
The known point is always a root. That gives you a factor for free — a useful check that your line is right.
Worked Example 6 — A tangent with a given gradient

Find the point on $y = x^2 - 6x + 5$ where the tangent is parallel to $y = 2x$.

①Parallel means the tangent gradient is $2$.
②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 6 = 2$
③$2x = 8$, so $x = 4$.
④$y = 16 - 24 + 5 = -3$. The point is $(4, -3)$.
5 Quick Reference

Tangent

Same gradient as the curve at the point.

Normal

Gradient $-\dfrac1m$.

Circle

The normal is the radius; no calculus needed.

Other curves

Differentiate, then substitute the $x$-value.

Find $y$ first

If only $x$ is given.

Line form

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

$m = 0$

Normal is vertical: $x = x_1$.

Intercepts

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

Meets again

Solve line $=$ curve; the known point is a root.

Given gradient

Set $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ equal to it and solve for $x$.

6 Practice Questions
Question 1

A tangent has gradient $5$. What is the gradient of the normal at that point?

▶ Show solution

$-\dfrac{1}{5}$

Question 2

Find the gradient of the tangent to $y = x^2 + 3x$ at $x = 2$.

▶ Show solution

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

At $x = 2$: $4 + 3 = 7$.

Question 3

Find the equation of the tangent to $y = x^2$ at $(3, 9)$.

▶ Show solution

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

$y - 9 = 6(x - 3)$

$y = 6x - 9$

Question 4

Find the equation of the normal to $y = x^2$ at $(3, 9)$.

▶ Show solution

Tangent gradient $6$, so normal gradient $-\dfrac16$.

$y - 9 = -\tfrac16(x-3)$

$6y - 54 = -x + 3$, so $x + 6y = 57$.

Question 5

Find the tangent to the circle $x^2 + y^2 = 100$ at $(6, 8)$.

▶ Show solution

Centre at the origin; radius gradient $= \dfrac{8}{6} = \dfrac43$.

Tangent gradient $= -\dfrac34$.

$y - 8 = -\tfrac34(x-6)$

$4y - 32 = -3x + 18$, so $3x + 4y = 50$.

Question 6

Find the point on $y = x^2 - 8x$ where the tangent is horizontal.

▶ Show solution

Horizontal means $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.

$2x - 8 = 0$, so $x = 4$.

$y = 16 - 32 = -16$. The point is $(4, -16)$.

Question 7

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

▶ Show solution

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

Normal gradient $-\dfrac13$.

$y - 2 = -\tfrac13(x-1)$

$3y - 6 = -x + 1$, so $x + 3y = 7$.

Question 8

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

▶ Show solution

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

Tangent: $y - 1 = 4(x-2)$, so $y = 4x - 7$.

Setting $y = 0$: $x = \dfrac{7}{4} = 1.75$.

It crosses at $(1.75, 0)$.

Question 9

Find the two points on $y = x^3 - 12x$ where the tangent is parallel to the $x$-axis.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12 = 0$

$x^2 = 4$, so $x = \pm 2$.

At $x = 2$: $y = 8 - 24 = -16$, giving $(2, -16)$.

At $x = -2$: $y = -8 + 24 = 16$, giving $(-2, 16)$.

Question 10

The curve $C$ has equation $y = x^2 - 4x + 5$, and $P$ is the point on $C$ where $x = 1$.

(a) Find the coordinates of $P$.   (b) Find the equations of the tangent and the normal at $P$.   (c) The tangent and the normal cross the $y$-axis at $A$ and $B$. Find the area of triangle $ABP$.   (d) Show that the normal meets $C$ again, and find that point.

▶ Show solution

(a) $y = 1 - 4 + 5 = 2$, so $P(1, 2)$.

(b) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 4$, which at $x = 1$ is $-2$.

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

Normal: gradient $\tfrac12$, so $y - 2 = \tfrac12(x-1)$, giving $y = \tfrac12 x + \tfrac32$.

(c) Tangent meets the $y$-axis at $A(0, 4)$; normal at $B\left(0, \tfrac32\right)$.

$AB = 4 - \tfrac32 = \tfrac52$ — this is the base, lying along the $y$-axis.

The perpendicular height from $P(1,2)$ to the $y$-axis is its $x$-coordinate, $1$.

Area $= \tfrac12 \times \tfrac52 \times 1 = \dfrac{5}{4} = 1.25$ square units.

(d) Set the normal equal to the curve:

$x^2 - 4x + 5 = \tfrac12 x + \tfrac32$

Multiply by $2$: $2x^2 - 8x + 10 = x + 3$

$2x^2 - 9x + 7 = 0$

$\Delta = 81 - 56 = 25 > 0$, so there are two solutions — the normal does meet $C$ again.

$(2x - 7)(x - 1) = 0$, giving $x = 1$ (which is $P$) or $x = \tfrac72$.

At $x = \tfrac72$: $y = \tfrac12\left(\tfrac72\right) + \tfrac32 = \tfrac74 + \tfrac32 = \tfrac{13}{4}$.

The normal meets $C$ again at $\left(\dfrac{7}{2},\; \dfrac{13}{4}\right)$.

Check on the curve: $\left(\tfrac72\right)^2 - 4\left(\tfrac72\right) + 5 = \tfrac{49}{4} - 14 + 5 = \tfrac{49 - 36}{4} = \tfrac{13}{4}$ ✓

Tangents and Normals (CG5) · OCR FSMQ Additional Maths · Created with MathJax