Tangent and Normal Problems

Read the gradient from a tangent or normal line written with an unknown constant

In the exam for Edexcel International AS Maths · 6 real past-paper questions below

Read the gradient from the line

You can already find a tangent or a normal from a point. AS questions often turn this round. They give you the line, with an unknown constant such as $k$ , and ask you to find the point or the constant.

The constant only slides the line up or down. The gradient is still there to read. So read it first.

Formula

y = m x + k

$m$ — the gradient. Get $y$ on its own, then read the number in front of $x$ .
$k$ — the unknown constant. It does not change the gradient.

Line	Rearranged	Gradient
$y = k - 5 x$	$y = - 5 x + k$	$- 5$
$3 y = 2 x + k$	$y = \frac{2}{3} x + \frac{k}{3}$	$\frac{2}{3}$
$x + 4 y = k$	$y = - \frac{1}{4} x + \frac{k}{4}$	$- \frac{1}{4}$

Worked example

1
The normal to $y = x^{2} - 6 x + 5$ at P has equation $y = k - \frac{1}{4} x$ . Find P and $k$ .
2
$m_{normal} = - \frac{1}{4}$ — Read it from the line.
3
$m_{tangent} = 4$ — Flip and change the sign. $\frac{d y}{d x}$ gives the tangent, so you need this one.
4
$\frac{d y}{d x} = 2 x - 6 = 4$ — Differentiate and set it equal to the tangent gradient.
5
$x = 5, y = 25 - 30 + 5 = 0$ — Find $y$ from the curve. P is (5, 0).
6
$0 = k - \frac{5}{4} \Rightarrow k = \frac{5}{4}$ — P is on the line too. Put it in to find $k$ .

Example

The same idea finds a constant in $f^{'} (x)$ . Say $f^{'} (x) = 2 x^{2} + p x - 1$ , and the normal at $x = 2$ has gradient $- \frac{1}{5}$ . The tangent gradient is 5, so $f^{'} (2) = 8 + 2 p - 1 = 5$ , and $p = - 1$ .

Rule

Marks: there is often a B1 just for the tangent gradient (here 4), written or used. In a “show that”, write the equation in the constant, like $8 + 2 p - 1 = 5$ , before the answer.

Common mistake

Setting $\frac{d y}{d x}$ equal to the normal's gradient. $\frac{d y}{d x}$ is always the tangent's gradient. Flip first.

Key idea

Read the gradient, find the point, then put the point into the line to find the constant.

Equations from the gradient

Setting $\frac{d y}{d x}$ equal to a gradient gives an equation. With a term like $\frac{4}{x}$ in the curve, that equation hides a quadratic.

Worked example

1
$y = x^{3} + \frac{4}{x}$ , $x > 0$ . The line $y = 11 x + k$ is the tangent at P. Show that the x coordinate of P satisfies $3 x^{4} - 11 x^{2} - 4 = 0$ , then find $k$ .
2
$\frac{d y}{d x} = 3 x^{2} - 4 x^{- 2} = 3 x^{2} - \frac{4}{x ^{2}}$ — Write $\frac{4}{x}$ as $4 x^{- 1}$ , then differentiate.
3
$3 x^{2} - \frac{4}{x ^{2}} = 11$ — The line's gradient is 11.
4
$3 x^{4} - 4 = 11 x^{2}$ — Multiply every term by $x^{2}$ . Show this line.
5
$3 x^{4} - 11 x^{2} - 4 = 0$ — All terms on one side, = 0. This is what you had to show.
6
$(3 x^{2} + 1) (x^{2} - 4) = 0$ — A quadratic in $x^{2}$ . Factorise it like $3 u^{2} - 11 u - 4$ .
7
$x^{2} = 4 \Rightarrow x = 2$ — $3 x^{2} = - 1$ is impossible. $x = - 2$ is ruled out by $x > 0$ .
8
$y = 8 + \frac{4}{2} = 10$ — Put $x = 2$ , not $x^{2} = 4$ , into the curve.
9
$10 = 11 (2) + k \Rightarrow k = - 12$ — P is on the line. Solve for $k$ .

Rule

Marks: “show that” needs the line where you multiply through, and = 0 at the end. With “show all stages of your working”, factorise the quadratic in $x^{2}$ by hand. A root typed from a calculator scores nothing.

Worked example

1
$y = x^{3} - 6 x^{2} + 4 x$ . The tangent at $x = 5$ is parallel to the tangent at $x = a$ , and the two tangents are different. Find $a$ .
2
$\frac{d y}{d x} = 3 x^{2} - 12 x + 4$ — Differentiate.
3
$3 (25) - 60 + 4 = 19$ — The gradient at $x = 5$ . Parallel lines share it.
4
$3 x^{2} - 12 x + 4 = 19 \Rightarrow x^{2} - 4 x - 5 = 0$ — Set $\frac{d y}{d x}$ equal to 19. Divide by 3.
5
$(x - 5) (x + 1) = 0$ — One root is always the point you started from.
6
$a = - 1$ — $x = 5$ gives the same tangent, and they must be different.

Common mistake

Putting $x^{2} = 4$ into the curve as if $x = 4$ . Square root first.

Key idea

Multiply through to clear fractions, spot the quadratic, and say why you throw a root away.

Where the tangent meets again

A tangent to a cubic can touch the curve at one point and cut it somewhere else. To find where, set the curve equal to the line, as for any two graphs.

Memory trick

A tangent touches the curve where it kisses it, and a kiss counts twice. So the touching point is a double root: its bracket comes out squared.

Worked example

1
$y = (x + 3) (x - 1) (2 x - 5)$ . The tangent where the curve cuts the y-axis meets the curve again at P. Find P using algebra.
2
$y = 2 x^{3} - x^{2} - 16 x + 15$ — Expand. You need this form to differentiate.
3
$x = 0, y = 15$ — The y-axis is $x = 0$ . The point is (0, 15).
4
$\frac{d y}{d x} = 6 x^{2} - 2 x - 16 \Rightarrow m = - 16$ — Put $x = 0$ into $\frac{d y}{d x}$ .
5
$y = 15 - 16 x$ — The tangent: gradient $- 16$ , through (0, 15).
6
$2 x^{3} - x^{2} - 16 x + 15 = 15 - 16 x$ — Curve = line.
7
$2 x^{3} - x^{2} = 0$ — The 15s and the $- 16 x$ terms cancel. That happens for a tangent at $x = 0$ .
8
$x^{2} (2 x - 1) = 0$ — $x^{2}$ is the double root at the touching point, $x = 0$ .
9
$x = \frac{1}{2}, y = 15 - 8 = 7$ — The new root. P is $(\frac{1}{2}, 7)$ .

Rule

Marks: method marks for the gradient, the tangent, setting curve = line and factorising, then A1 for the answer. “Using algebra” means you solve by hand, not with a calculator.

Key idea

Railway engineers join a straight track to a curved one along a tangent. The train feels no sudden kink, because at the join the curve and the line have the same direction.

Key idea

Set curve = line. The touching point gives a squared bracket, so the root left over is the new meeting point.

A normal that is also a tangent

The hardest type: the normal at A is also the tangent at B. One line, two jobs.

Worked example

1
$y = \frac{1}{3} x^{3} - \frac{2}{3} x^{2} - x + k$ . The normal at A, where $x = - 1$ , is the tangent at B, where $x > 0$ . Find B's x coordinate.
2
$\frac{d y}{d x} = x^{2} - \frac{4}{3} x - 1$ — The constant $k$ disappears.
3
$1 + \frac{4}{3} - 1 = \frac{4}{3}$ — Gradient at A: put $x = - 1$ in.
4
$m_{normal} = - \frac{3}{4}$ — Flip and change the sign.
5
$x^{2} - \frac{4}{3} x - 1 = - \frac{3}{4}$ — At B, the curve's gradient equals the line's.
6
$12 x^{2} - 16 x - 3 = 0$ — Multiply by 12 to clear fractions.
7
$(2 x - 3) (6 x + 1) = 0$ — Factorise.
8
$x = \frac{3}{2}$ — Reject $- \frac{1}{6}$ : it is negative, and B is to the right of the y-axis.

Worked example

1
The y-intercept of this line is 2. Find $k$ .
2
$y = - \frac{3}{4} x + 2$ — The line: gradient $- \frac{3}{4}$ , intercept 2.
3
$x = - 1 \Rightarrow y = \frac{3}{4} + 2 = \frac{11}{4}$ — A is on the line. Find its y value.
4
$\frac{11}{4} = - \frac{1}{3} - \frac{2}{3} + 1 + k$ — A is on the curve too. Put $(- 1, \frac{11}{4})$ in.
5
$k = \frac{11}{4}$ — The numbers before $k$ add to 0.

Rule

Marks: say why you keep a root, for example “B is right of the y-axis”. The root alone loses the A mark.

Common mistake

Putting the y-intercept into the curve. It belongs to the line.

Key idea

The gradient links A to B. A point on the line links the line to the curve.

Which method? Read the question

The wording tells you which tool to use.

You see	You do
A tangent line with an unknown $k$	Read its gradient; set $\frac{d y}{d x}$ equal to it
A normal line with an unknown $k$	Flip its gradient first, then set $\frac{d y}{d x}$ equal
$\frac{1}{x}$ or $\frac{1}{x ^{2}}$ terms in $\frac{d y}{d x}$	Multiply by $x^{2}$ ; solve a quadratic in $x^{2}$
“Parallel and distinct”	Same gradient; reject the root you started from
“The tangent meets the curve again”	Curve = line; factor out the squared bracket
“The normal at A is the tangent at B”	Gradient at A, flip, set $\frac{d y}{d x}$ equal
“Justify” or a sketch	Choose the root that fits; write why
An intercept and an unknown in the curve	Line first; put a point on it into the curve

Key idea

Every one of these ends the same way: a point that sits on both the curve and the line. Check your answer in both.

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