Mathematics 9709 · Pure Mathematics 1

Tangents and normals

A tangent is the straight line that touches a curve at one point and goes the same way as the curve there. The normal is the line at right angles to the tangent at that point.
Use the derivative to find both. After that it is just straight-line work you already know.

Tangent gradient is the derivative at the point, $m=\tfrac{dy}{dx}\big|_{x=a}$ . Normal gradient is its negative reciprocal, $-\tfrac1m$ (turn it upside down, then change the sign). Both lines then use $y-y_1=m(x-x_1)$ .

The point needs a y-coordinate. Put $x=a$ into the curve (not the derivative) to get it. Almost every tangent or normal question checks if you remembered this.

The three-step method

Differentiate to get the gradient at the point.
For the normal, turn the gradient upside down and change its sign.
Get the $y$ -coordinate from the curve. Then put it into $y-y_1=m(x-x_1)$ .

The tangent (gold) goes the same way as the curve; the normal (blue) is at right angles to it.

Worked example

Tangent and normal to $y = 2x^2 + \dfrac{8}{x^2} - 9$ at $x = 2$

Write the fraction as a power, then differentiate: $y=2x^2+8x^{-2}-9\Rightarrow\tfrac{dy}{dx}=4x-16x^{-3}$ .
Find the point and the gradient at $x=2$ : $y=8+2-9=1$ and $m=8-2=6$ , so the point is $(2,1)$ .
Tangent (gradient $6$ ): $y-1=6(x-2)\Rightarrow y=6x-11$ .
Normal (gradient $-\tfrac16$ ): $y-1=-\tfrac16(x-2)\Rightarrow x+6y=8$ .

Answer

\text{tangent } y=6x-11,\quad\text{normal } x+6y=8

What the exam asks

“The normal meets the $x$ -axis at…”: find the normal, set $y=0$ , then solve for $x$ .
“Verify that line $AB$ is the normal”: show its gradient is $-1/m$ , and show it goes through the point. (9709/12 Oct/Nov 2021 Q11; 9709/12 Oct/Nov 2024 Q10)

Worked example

Real question: normal to $y = f(x)$ at $(1, 0)$ , where $f'(x) = 8(2x-3)^{1/3} - 10x^{2/3}$

9709/12 Oct/Nov 2024 Q10(a). They give you $f'(x)$ already, so you do not differentiate. Just put the number in, then turn the gradient upside down.

The gradient of the curve at the point is $f'(1)$ . Use the real cube root $(-1)^{1/3}=-1$ : $f'(1)=8(2-3)^{1/3}-10(1)^{2/3}=8(-1)-10=-18$ .
Normal gradient is the negative reciprocal: $-\dfrac{1}{-18}=\dfrac{1}{18}$ .
Through $(1,0)$ : $y-0=\tfrac{1}{18}(x-1)$ .

Answer

y=\tfrac{1}{18}(x-1)\quad\text{i.e. }18y=x-1

Where the marks go

Common mistake

The normal gradient is

-\tfrac1m

, not

-m

and not

m

. With

m=6

the normal has gradient

-\tfrac16

. (9709/12 Oct/Nov 2021 Q11)

Common mistake

Do not forget the

y

-coordinate. The point on the line is

(a,\,y)

. Put

x=a

into the curve to get

y

, never into the derivative.

Now you try

The point $A$ on the curve $y=2x^2+\dfrac{8}{x^2}-9$ has $x$ -coordinate $2$ . Find the equation of the tangent at $A$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style (from textbook WE 7.8))

$\tfrac{dy}{dx}=4x-16x^{-3}$ , so at $x=2$ : $m=8-2=6$ and $y=8+2-9=1$ .

Tangent through $(2,1)$ with gradient $6$ :

Answer

y=6x-11

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Tangents and normals

A tangent is the straight line that touches a curve at one point and goes the same way as the curve there. The normal is the line at right angles to the tangent at that point.
Use the derivative to find both. After that it is just straight-line work you already know.

The point needs a y-coordinate. Put $x=a$ into the curve (not the derivative) to get it. Almost every tangent or normal question checks if you remembered this.

The three-step method

Differentiate to get the gradient at the point.
For the normal, turn the gradient upside down and change its sign.
Get the $y$ -coordinate from the curve. Then put it into $y-y_1=m(x-x_1)$ .

The tangent (gold) goes the same way as the curve; the normal (blue) is at right angles to it.

Worked example

Tangent and normal to $y = 2x^2 + \dfrac{8}{x^2} - 9$ at $x = 2$

Write the fraction as a power, then differentiate: $y=2x^2+8x^{-2}-9\Rightarrow\tfrac{dy}{dx}=4x-16x^{-3}$ .
Find the point and the gradient at $x=2$ : $y=8+2-9=1$ and $m=8-2=6$ , so the point is $(2,1)$ .
Tangent (gradient $6$ ): $y-1=6(x-2)\Rightarrow y=6x-11$ .
Normal (gradient $-\tfrac16$ ): $y-1=-\tfrac16(x-2)\Rightarrow x+6y=8$ .

Answer

\text{tangent } y=6x-11,\quad\text{normal } x+6y=8

What the exam asks

“The normal meets the $x$ -axis at…”: find the normal, set $y=0$ , then solve for $x$ .
“Verify that line $AB$ is the normal”: show its gradient is $-1/m$ , and show it goes through the point. (9709/12 Oct/Nov 2021 Q11; 9709/12 Oct/Nov 2024 Q10)

Worked example

Real question: normal to $y = f(x)$ at $(1, 0)$ , where $f'(x) = 8(2x-3)^{1/3} - 10x^{2/3}$

9709/12 Oct/Nov 2024 Q10(a). They give you $f'(x)$ already, so you do not differentiate. Just put the number in, then turn the gradient upside down.

The gradient of the curve at the point is $f'(1)$ . Use the real cube root $(-1)^{1/3}=-1$ : $f'(1)=8(2-3)^{1/3}-10(1)^{2/3}=8(-1)-10=-18$ .
Normal gradient is the negative reciprocal: $-\dfrac{1}{-18}=\dfrac{1}{18}$ .
Through $(1,0)$ : $y-0=\tfrac{1}{18}(x-1)$ .

Answer

y=\tfrac{1}{18}(x-1)\quad\text{i.e. }18y=x-1

Where the marks go

Common mistake

The normal gradient is

-\tfrac1m

, not

-m

and not

m

. With

m=6

the normal has gradient

-\tfrac16

. (9709/12 Oct/Nov 2021 Q11)

Common mistake

Do not forget the

y

-coordinate. The point on the line is

(a,\,y)

. Put

x=a

into the curve to get

y

, never into the derivative.

Now you try

The point $A$ on the curve $y=2x^2+\dfrac{8}{x^2}-9$ has $x$ -coordinate $2$ . Find the equation of the tangent at $A$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style (from textbook WE 7.8))

$\tfrac{dy}{dx}=4x-16x^{-3}$ , so at $x=2$ : $m=8-2=6$ and $y=8+2-9=1$ .

Tangent through $(2,1)$ with gradient $6$ :

Answer

y=6x-11