Tangent and normal to a curve

Differentiation

A-level Maths (9MA0)

ORIGINAL

Rewrite as powers of x, differentiate, find the gradient at the point — the normal's gradient is the negative reciprocal.

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Question

The curve

C

has equation

y = 2x^2 - \dfrac{16}{x}

x \ne 0

. The point

P

C

has

x

-coordinate 2. Find the equation of the tangent to

C

P

, and the equation of the normal to

C

P

in the form

ax + by + c = 0

, where

a

b

and

c

are integers.

Every step worked, with the reasoning.

1
$x = 2$ : $y = 2(2)^2 - \dfrac{16}{2} = 8 - 8 = 0$ , so $P$ is $(2, 0)$ .
Find the $y$ -coordinate of $P$ first — both lines pass through it.
2
$y = 2x^2 - 16x^{-1}$
Write $\frac{16}{x}$ as a power of $x$ so it can be differentiated.
3
$\dfrac{dy}{dx} = 4x + 16x^{-2} = 4x + \dfrac{16}{x^2}$
Multiply by the power and reduce the power by one: $-16 \times (-1)x^{-2} = +16x^{-2}$ .
4
At $x = 2$ : $\dfrac{dy}{dx} = 4(2) + \dfrac{16}{4} = 12$
This is the gradient of the tangent at $P$ .
5
Tangent: $y - 0 = 12(x - 2)$ , so $y = 12x - 24$
Use $y - y_1 = m(x - x_1)$ with $m = 12$ through $(2, 0)$ .
6
Normal gradient $= -\dfrac{1}{12}$ : $y - 0 = -\dfrac{1}{12}(x - 2)$
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of 12.
7
$12y = -x + 2$ , so $x + 12y - 2 = 0$
Multiply by 12 and collect every term on one side to get integer coefficients.

Answer: Tangent $y = 12x - 24$ ; normal $x + 12y - 2 = 0$ .