Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

5

Tangents

(^) and
(^) normals
141
normal
tangent
curve
Figure 5.13
If the gradient of the tangent is m 1 , the gradient, m 2 , of the normal is given by
m 2 m
1

=–.^1

This enables you to find the equation of the normal at any specified point on

a curve.

EXAMPLE 5.9 A curve has equation y
x
=−^164 x. The normal to the curve at the point (4, –4)
meets the y axis at the point P. Find the co-ordinates of P.

SOLUTION

You may find it easier to write y=−^16 x 41 xy=− 64 xx−^1

(^12)
as.
Differentiating gives
d
d
y
x=− xx−×
16 −^214 −
2
(^12)
=−^162 −^2
x x
At the point (4, –4), x = 4 and
(^) d
d
y
x

=− −

=− −=−

16

4

2

4

11 2

2

So at the point (4, –4) the gradient of the tangent is −2.

Gradientofnormal

gradientoftangent

= −^1 =^12

The equation of the normal is given by

y − y 1 = m(x − x 1 )

y − (−4) = 12 (x − 4)

y = 12 x − 6

P is the point where the normal meets the y axis and so where x = 0.

Substituting x = 0 into y =^12 x – 6 gives y = –6.

So P is the point (0, −6).