A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 12: Differentiation · Lesson 3

Gradients, tangents and normals

73 marks

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Date

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Quick questions to start.

A1
Needed today
Find an equation of the line through $(- 2, 5)$ with gradient $- \frac{3}{4}$ , in the form $a x + b y + c = 0$ where $a$ , $b$ and $c$ are integers.
[2 marks]
Answer
A2
Needed today
The line $L$ has equation $2 x - 5 y + 3 = 0$ . Find the gradient of a line perpendicular to $L$ .
[2 marks]
Answer
A3
Last lesson
Given that $y = 4 x^{3} - \frac{6}{x ^{2}}$ , find $\frac{d y}{d x}$ .
[2 marks]
Answer
A4
From chapter 2
Solve $2 x^{2} - 3 x - 9 = 0$ .
[2 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

Find an equation of the tangent to the curve $y = x^{3} - 4 x^{2} + 2 x + 7$ at the point $(3, 4)$ .

Hint Differentiate, then put

x = 3

into

\frac{d y}{d x}

to get the gradient. Use

y - y_{1} = m (x - x_{1})

B1
Find an equation of the tangent to the curve $y = 2 x^{3} + x^{2} - 5 x + 1$ at the point where $x = - 1$ .
[3 marks]
Answer $y = \dots\dots$

Example 2

Find an equation of the normal to the curve $y = \frac{x ^{2} + 12}{x}$ at the point where $x = 2$ . Give your answer in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.

Hint Split the fraction:

y = x + 12 x^{- 1}

. The normal is perpendicular to the tangent, so its gradient is

- \frac{1}{m}

B2
Find an equation of the normal to the curve $y = \frac{4}{x ^{2}} + 3 x$ at the point where $x = 2$ . Give your answer in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[4 marks]
Answer

Example 3

Find the coordinates of the points on the curve $y = 2 x^{3} - 3 x^{2} - 10 x + 1$ at which the tangent is parallel to the line $y = 2 x$ .

Hint Parallel lines have equal gradients. Set

\frac{d y}{d x} = 2

and solve.

B3
Find the coordinates of the points on the curve $y = x^{3} - 6 x^{2} + 4 x + 3$ at which the tangent is perpendicular to the line $x - 5 y = 2$ .
[4 marks]
Answer $(\dots\dots, \dots\dots)$ and $(\dots\dots, \dots\dots)$

Example 4

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

Not drawn accurately

Hint Find the normal at

P

, then solve it with the curve. One root of the quadratic is

x = 3

, which helps you factorise.

B4
The point $P (1, 0)$ lies on the curve $y = x^{3} - 2 x^{2} - 3 x + 4$ . The tangent to the curve at $P$ meets the curve again at $Q$ . Find the coordinates of $Q$ .
[5 marks]
Answer $(\dots\dots, \dots\dots)$

Practice

A-Level

C1
The points $A (4, 4)$ and $B (1, - 2)$ lie on the curve $y = x^{2} - 3 x$ . The tangent to the curve at $A$ meets the normal to the curve at $B$ at the point $R$ . Find the coordinates of $R$ .
[5 marks]
Not drawn accurately
Answer $(\dots\dots, \dots\dots)$

C2
The point $P$ on the curve $y = \frac{16}{x}$ has $x$ -coordinate $2$ . The tangent to the curve at $P$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $O A B$ , where $O$ is the origin.
[4 marks]
Not drawn accurately
Answer
C3
Find the coordinates of the point on the curve $y = 6 x$ at which the normal to the curve has gradient $- 2$ .
[3 marks]
Answer $(\dots\dots, \dots\dots)$

Exam-style questions

Edexcel exam style

D1
The curve $C$ has equation $y = x^{3} - 3 x^{2} - 4 x + 7$ . The point $P$ on $C$ has $x$ -coordinate $- 1$ .
(a)
The tangent to $C$ at the point $Q$ is parallel to the tangent to $C$ at $P$ . Find the coordinates of $Q$ .
[4 marks]
Answer $(\dots\dots, \dots\dots)$
(b)
Find an equation of the tangent to $C$ at $P$ , in the form $y = m x + c$ .
[3 marks]
Answer $y = \dots\dots$
(c)
The tangent to $C$ at $P$ meets $C$ again at the point $T$ . Find the coordinates of $T$ . Solutions relying entirely on calculator technology are not acceptable.
[4 marks]
Answer $(\dots\dots, \dots\dots)$
Total for D1: 11 marks

D2
Mia is finding the normal to the curve $y = 2 x^{3} - 3 x + 5$ at the point where $x = 1$ . She writes: "At $x = 1$ , $y = 4$ , so the gradient of the curve is $4$ . The normal has gradient $- 4$ , so the normal is $y - 4 = - 4 (x - 1)$ ."
Mia's answer is wrong.
(a)
Describe the two mistakes Mia has made.
[2 marks]
(b)
Find an equation of the normal to the curve at the point where $x = 1$ , in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[3 marks]
Answer
Total for D2: 5 marks
D3
The curve $C$ has equation $y = x x - 4 x + 9$ , $x > 0$ . The point $P (4, 1)$ lies on $C$ . The tangent to $C$ at $P$ meets the $y$ -axis at $A$ , and the normal to $C$ at $P$ meets the $y$ -axis at $B$ , as shown.
Not drawn accurately
(a)
Find $\frac{d y}{d x}$ .
[2 marks]
Answer
(b)
Show that the tangent to $C$ at $P$ has equation $y = 5 - x$ .
[3 marks]
(c)
Find an equation of the normal to $C$ at $P$ .
[2 marks]
Answer $y = \dots\dots$
(d)
Find the area of triangle $A P B$ .
[3 marks]
Answer
Total for D3: 10 marks
D4
The curve $C$ has equation $y = a x^{3} + b x^{2} - 4$ , where $a$ and $b$ are constants. The point $(2, 8)$ lies on $C$ , and the normal to $C$ at $(2, 8)$ is parallel to the line $x + 4 y = 3$ . Find the value of $a$ and the value of $b$ .
[5 marks]
Answer $a = \dots\dots$ $b = \dots\dots$

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The point $P$ on the curve $y = x^{3}$ has $x$ -coordinate $p$ , where $p \neq = 0$ . The tangent to the curve at $P$ meets the curve again at $Q$ . Find the $x$ -coordinate of $Q$ in terms of $p$ , and show that the gradient of the curve at $Q$ is always the same multiple of the gradient at $P$ . State this multiple.
[6 marks]
Hint 1 · What to try
Write down the tangent at $(p, p^{3})$ , then solve it with $y = x^{3}$ . You already know one root, and it is a repeated root.
Hint 2 · The first line
The tangent is $y = 3 p^{2} x - 2 p^{3}$ , so $x^{3} - 3 p^{2} x + 2 p^{3} = 0$ , and $(x - p)^{2}$ is a factor.
Hint 3 · The full method
$x^{3} - 3 p^{2} x + 2 p^{3} = (x - p)^{2} (x + 2 p)$ , so $Q$ has $x = - 2 p$ . The gradient at $Q$ is $3 (- 2 p)^{2} = 12 p^{2}$ , and at $P$ it is $3 p^{2}$ . So it is always $4$ times as big.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find the equation of a tangent and of a normal to a curve.
I can find the points where a tangent has a given gradient.
I can find where a tangent or normal meets the curve again.
I can find unknown constants from a tangent or normal.

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