Lines Meeting Curves: the Discriminant

Equate a line and a curve (including k/x and 1/x² curves), clear fractions, rearrange to ax²+bx+c=0

In the exam for Edexcel International AS Maths · 8 real past-paper questions below

Clear the fractions first

You can already put a line into a quadratic curve and test $b^{2} - 4 a c$ . AS questions use curves like $y = \frac{k}{x}$ or $y = \frac{1}{x - 2}$ , or two curves at once. The test is the same. First you must turn the equation into a quadratic.

Formula

curve = line \Rightarrow \times denominator \Rightarrow a x^{2} + b x + c = 0

$curve$ — the curve's right-hand side, fraction and all.
$line$ — the line's right-hand side.
$denominator$ — multiply every term by it, like $x$ or $(x - 2)$ , so no fraction is left.

Worked example

1
The line $y = k x + 3$ is a tangent to the curve $y = 5 - \frac{2}{x}$ . Find $k$ .
2
$k x + 3 = 5 - \frac{2}{x}$ — At a meeting point the two y values are equal.
3
$k x^{2} + 3 x = 5 x - 2$ — Multiply every term by $x$ .
4
$k x^{2} - 2 x + 2 = 0$ — Everything on one side. Now $a = k$ , $b = - 2$ , $c = 2$ .
5
$(- 2)^{2} - 4 (k) (2) = 0$ — Tangent: the discriminant is 0.
6
$4 - 8 k = 0 \Rightarrow k = \frac{1}{2}$ — Solve for $k$ .

Two curves work the same way. Set them equal and collect everything on one side.

Worked example

1
The curves $y = x^{2} + k x + 3$ and $y = - x^{2} + 3 x + k$ meet at exactly one point. Find the possible values of $k$ .
2
$2 x^{2} + (k - 3) x + 3 - k = 0$ — Equate and collect. The constant is the whole of $3 - k$ .
3
$(k - 3)^{2} - 4 (2) (3 - k) = 0$ — One meeting point: discriminant 0.
4
$k^{2} + 2 k - 15 = 0$ — $k^{2} - 6 k + 9 - 24 + 8 k$ .
5
$(k + 5) (k - 3) = 0 \Rightarrow k = - 5 or 3$ — Factorise.
6
$k = 3 : 2 x^{2} = 0, point (0, 3)$ — Put $k$ back to find the point. With $k = - 5$ it is $(2, - 3)$ .

Example

A line parallel to $y = 2 x$ touches $y = x^{2} - 4 x + 1$ . Write it as $y = 2 x + c$ : the gradient is fixed, only $c$ is unknown. Then $x^{2} - 6 x + 1 - c = 0$ , and $36 - 4 (1 - c) = 0$ gives $c = - 8$ . The line is $y = 2 x - 8$ .

Common mistake

Multiplying only the fraction by $x$ . Every term on both sides must be multiplied, or the quadratic is wrong and the marks after it go too.

Key idea

Mark schemes usually give a method mark for forming the quadratic and an accuracy mark for getting it right. Show that line before the discriminant.

Surd boundaries

When the line can cut, miss or meet the curve, you get an inequality in $k$ . In AS questions the critical values are often surds. You must find them by algebra, not by calculator.

Worked example

1
The line $y = k x + 1$ meets the curve $y = \frac{2}{x + 1}$ at least once. Find the range of possible values of $k$ .
2
$(k x + 1) (x + 1) = 2$ — Multiply both sides by $(x + 1)$ .
3
$k x^{2} + (k + 1) x - 1 = 0$ — Expand and collect: $k x^{2} + k x + x + 1 - 2$ .
4
$(k + 1)^{2} - 4 (k) (- 1) \geq 0$ — Meets at least once: discriminant greater than or equal to 0.
5
$k^{2} + 6 k + 1 \geq 0$ — $k^{2} + 2 k + 1 + 4 k$ .
6
$k = \frac{- 6 \pm 32}{2} = - 3 \pm 22$ — Critical values. $32 = 42$ .
7
$k \leq - 3 - 22 or k \geq - 3 + 22$ — Greater than or equal to 0: the two outside parts.

Worked example

1
The line $y = c - x$ does not meet the curve $y = \frac{3}{x - 1}$ . Find the range of possible values of $c$ .
2
$(c - x) (x - 1) = 3$ — Multiply both sides by $(x - 1)$ .
3
$x^{2} - (c + 1) x + c + 3 = 0$ — Expand and collect: $c x - c - x^{2} + x - 3 = 0$ , then times $- 1$ .
4
$(c + 1)^{2} - 4 (c + 3) < 0$ — Does not meet: discriminant less than 0.
5
$c^{2} - 2 c - 11 < 0$ — Expand: $c^{2} + 2 c + 1 - 4 c - 12$ .
6
$c = 1 \pm 23$ — Formula: $\frac{2 \pm 48}{2}$ , and $48 = 43$ .
7
$1 - 23 < c < 1 + 23$ — Less than 0: the part between.

Rule

The mark scheme gives a method mark for choosing the right region (inside or outside). The final answer must be exact: a decimal like $c < 4.46$ loses the last mark.

Common mistake

Solving $b^{2} - 4 a c = 0$ and stopping there. If the question asks for a range, the answer must be an inequality. Writing $= 0$ in a “show that … > 0” part also loses the mark unless you put the sign back.

Key idea

“At least once” means $\geq 0$ . “Does not meet” means $< 0$ . The surd critical values are the two tangent cases.

One point known

Some questions give you one meeting point and ask for the other. You do not need the discriminant here. The known point gives you $k$ , and the known root gives you a factor.

Worked example

1
The curve $y = \frac{4}{x + 1}$ and the line $y = k x + 6$ meet at P, where $x = 1$ , and at a second point Q. Find $k$ and the coordinates of Q.
2
$\frac{4}{2} = k (1) + 6$ — P is on both graphs, so put $x = 1$ into both.
3
$k = - 4$ — $2 = k + 6$ .
4
$4 = (- 4 x + 6) (x + 1)$ — Now equate in general and multiply by $(x + 1)$ .
5
$4 x^{2} - 2 x - 2 = 0 \Rightarrow 2 x^{2} - x - 1 = 0$ — Collect and divide by 2.
6
$(x - 1) (2 x + 1) = 0$ — You know $x = 1$ is a root, so $(x - 1)$ is a factor.
7
$x = - \frac{1}{2}, y = - 4 (- \frac{1}{2}) + 6 = 8$ — Q is $(- \frac{1}{2}, 8)$ . Use the line for y: it is easier.

Key idea

Check: $\frac{4}{- \frac{1}{2} + 1} = \frac{4}{\frac{1}{2}} = 8$ . Same y from the curve, so Q is right.

Rule

When a part says “using algebra”, a calculator equation solver alone gets no marks. Show the quadratic and how you solved it.

Key idea

One root known: one factor known. The other factor gives the second point.

Four meeting points

Some equations are quadratics in $x^{2}$ . Put $u = x^{2}$ . Each positive $u$ gives two points, at $x = u$ and $x = - u$ . So four points need two different positive values of $u$ .

Memory trick

A mirror on the y-axis: each positive $u$ gives a point and its reflection. Two positive $u$ values, four points.

Worked example

1
The curves $y = 8 - 2 x^{2}$ and $y = \frac{A}{x ^{2}}$ meet at 4 distinct points. Find the range of possible values of $A$ .
2
$8 x^{2} - 2 x^{4} = A$ — Equate and multiply by $x^{2}$ .
3
$2 u^{2} - 8 u + A = 0$ — Collect, with $u = x^{2}$ .
4
$64 - 8 A > 0 \Rightarrow A < 8$ — Two different values of $u$ : discriminant greater than 0.
5
$u = \frac{8 - 64 - 8 A}{4} > 0$ — The smaller $u$ must be positive too.
6
$64 - 8 A < 8 \Rightarrow A > 0$ — Square both sides: $64 - 8 A < 64$ .
7
$0 < A < 8$ — Both conditions together.

Common mistake

Giving only $A < 8$ . The mark scheme gives a separate mark for $A > 0$ . With $A = - 1$ , one $u$ is negative and there are only 2 points.

Rule

A quadratic in $x^{3}$ is different. Every real $u$ has exactly one cube root, so no sign condition is needed: $x^{6} - 7 x^{3} - 8 = 0$ gives $x^{3} = 8$ or $- 1$ , so $x = 2$ or $- 1$ .

Key idea

Four points from a quadratic in $x^{2}$ : discriminant $> 0$ and both roots positive.

Which method? Read the question

The first move is always the same: set the curves equal and clear the fractions. The question words then tell you the test.

You see	You do
“tangent”, “touches”, “meet at a single point”	Quadratic, then $b^{2} - 4 a c = 0$ ; solve for $k$
“two distinct points”	$b^{2} - 4 a c > 0$ ; outside region
“does not meet”, “no real roots”	$b^{2} - 4 a c < 0$ ; region between
“meets at least once”	$b^{2} - 4 a c \geq 0$
“line parallel to … touches”	Fix the gradient, write $y = m x + c$ , find $c$ with $= 0$
“meet at P, where x = …”, find Q	Put P in to find $k$ ; factorise using the known root
“4 distinct points”, terms in $x^{4}$ and $x^{2}$	$u = x^{2}$ : discriminant $> 0$ and both $u$ positive

Common mistake

Writing a two-part answer as “ $A > 0$ or $A < 8$ ”, or with just a comma. Both must hold, so write $0 < A < 8$ or “ $A > 0$ and $A < 8$ ”.

Key idea

In 1637, before calculus, René Descartes found tangents with algebra. He drew a circle that meets the curve and asked when the meeting equation has a repeated root. Then the circle only touches the curve. A repeated root still means “touches” today.

Key idea

Clear the fraction, collect to 0, read the question words, then pick the sign.

Practise lines meeting curves: the discriminant now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on lines meeting curves: the discriminant

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.