Find the point of contact

The gradient or a point on the tangent is given

In 5 of 20 recent exams
Paper 1 or 2
4 to 7 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 4–7 marks. Typical (5 of 20 exam versions), usually as the middle part of a longer question.

Spot it in the exam

Exam phrases to spot (6)

“the tangent at $P$ is parallel to the line …”
“the tangent at $P$ is perpendicular to …”
“the normal at $P$ is parallel to …”
“the tangent makes an angle of … with the positive x-axis”
“find the point where the gradient is …”
“the tangent passes through the origin”

In disguise (4)

the line is in general form: $x - 2y = 6$ has gradient $\frac{1}{2}$ (a line $ax + by + d = 0$ has gradient $-\frac{a}{b}$ , not $a$ )
"the rate of change of $y$ with respect to $x$ is …", with no context: that is just the gradient
" $y = kx$ is a tangent to …": $y = kx$ passes through $(0, 0)$ , so the origin is a point on the tangent
a line with an unknown "is a tangent to" a cubic, trig or exponential curve: it looks like Two graphs, but substituting gives no quadratic, so there is no $\Delta$

Variants (2)

two curves with equal gradients: $f'(x) = g'(x)$ (for example $f$ and $f^{-1}$ at the same $x$ )
"exactly one tangent has gradient $m$ ": $f'(x) = m$ has exactly one solution. For a cubic, $f'(x) = m$ is a quadratic, so use $\Delta = 0$ , or the vertex of $f'$ (the last Your turn).

Looks like it, but isn't (5)

every $f'(x) = 0$ wording (horizontal tangent, parallel to the x-axis, touches $y = k$ ): Stationary points
"steepest", "the gradient is greatest": the maximum of $f'$ , in Concavity and points of inflexion
a parabola, with nothing known about the point of contact: Two graphs is often quicker (see
the $x$ of the point of contact is given: Tangent or normal at a given point
a line (not a tangent) at a given angle to the positive $x$ -axis, or parallel or perpendicular to another line: $m = \tan\theta$ or the line's own gradient, with no derivative: Equation of a straight line

Worked example

Parallel to a line in general form, one root rejected

Paper 1 (no calculator)

Question

Let $f(x) = \frac{1}{2}x^2 - 3\ln x$ , for $x > 0$ . The tangent to the graph of $f$ at the point $P$ is parallel to the line $x - 2y = 6$ . Find the coordinates of $P$ . [6]

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See it move: Where the tangent is parallel to a line

At $x = 1.00$ : the gradient is $-2$ . That is less than $\frac{1}{2}$ : move right.
$x - 2y = 6$ is $y = \frac{1}{2}x - 3$ : its gradient is $\frac{1}{2}$ , not $1$ . Parallel means the same gradient.
$f'(x) = \frac{1}{2}$ : $x - \frac{3}{x} = \frac{1}{2}$ , so $2x^{2} - x - 6 = 0$ , and $x = 2$ or $x = -\frac{3}{2}$ .
$x = -\frac{3}{2}$ is not in the domain $x > 0$ , so $P$ is $(2, 2 - 3\ln 2)$ , about $(2, -0.079)$ .

Which method?

the gradient $m$
First move
$f'(x) = m$
Then
$y$ from the curve; then the line, or the unknown, if asked
Watch out
there can be several solutions: keep those in the domain
a point on the tangent that is off the curve
First move
the tangent at $(p, f(p))$ ; substitute the point
Then
solve for $p$ ; each value of $p$ gives one tangent
Watch out
never use $f'$ at the given point: it isn't on the curve
a parabola, and a line $y = mx + c$ with one unknown
First move
either of the above, or substitute and use $\Delta = 0$ (Two graphs)
Then
both methods score
Watch out
$\Delta$ works only because the equation is a quadratic
, and $f'(x) = m$ can't be solved by hand
First move
graph $y = f'(x)$ (graph of the derivative) and $y = m$
Then
every intersection in the domain (intersections)
Watch out
radians; write down the equation and the interval

You know	First move	Then	Watch out
the gradient $m$	$f'(x) = m$	$y$ from the curve; then the line, or the unknown, if asked	there can be several solutions: keep those in the domain
a point on the tangent that is off the curve	the tangent at $(p, f(p))$ ; substitute the point	solve for $p$ ; each value of $p$ gives one tangent	never use $f'$ at the given point: it isn't on the curve
a parabola, and a line $y = mx + c$ with one unknown	either of the above, or substitute and use $\Delta = 0$ (Two graphs)	both methods score	$\Delta$ works only because the equation is a quadratic
, and $f'(x) = m$ can't be solved by hand	graph $y = f'(x)$ (graph of the derivative) and $y = m$	every intersection in the domain (intersections)	radians; write down the equation and the interval

Steps

1
Translate the wording
into a gradient or a point:
- parallel to a line with gradient $k$ : $m = k$
- perpendicular to that line, or "the normal is parallel to it": $m = -\frac{1}{k}$
- a line $ax + by + d = 0$ has gradient $-\frac{a}{b}$
- an angle $\theta$ with the positive x-axis: $m = \tan\theta$ (an obtuse angle gives $m < 0$ ); an acute angle $\theta$ with the x-axis where the curve is falling: $m = -\tan\theta$
- the tangent passes through a point that is not on the curve: that point is on the tangent, not the point of contact
Gradient known:
solve $f'(x) = m$ . Point known: write the tangent at $(p, f(p))$ , substitute the point, and solve for $p$ .
Solve
according to the type of equation: on factorise, use logs, or use exact trig values in radians; on use the GDC (intersections). Keep every solution in the domain, and reject the others with a reason.
Take $y$ from the curve.
Then answer what was asked: the point, the tangent, or the unknown in the line.
Check:
the gradient at your point is $m$ (and the tangent passes through the given point).

Common mistakes

Reading the gradient of $ax + by + d = 0$ as $a$ .
It is $-\frac{a}{b}$ : $x - 2y = 6$ has gradient $\frac{1}{2}$ .
"The normal is parallel to $L$ " solved as $f'(x) = m_L$ .
The normal has gradient $m_L$ , so $f'(x) = -\frac{1}{m_L}$ .
Taking $y$ from the line, or from $f'$ .
The point of contact is on the curve: $y = f(x)$ .

More mistakes (5)

Using $f'$ at the given external point
(for example $f'(0)$ when the tangent passes through the origin). That point is not on the curve.
Missing trig solutions
, or finding them in degrees. Use the whole domain, in radians.
Ignoring the sign of $\tan\theta$
an obtuse angle gives a negative gradient.
Leaving a rejected root in the answer
giving $x = -\frac{3}{2}$ and $x = 2$ in Example 4.1 scores , even though $2$ is there.
Using $\Delta = 0$ with a cubic, trig or exponential curve
substituting the line gives no quadratic.

Easy to confuse with

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