Intersections of lines and curves

AS · P1 · 17 min

The last idea in the coordinate geometry section of the syllabus ties the whole subject together: a graph is a picture of an equation, so the points where two graphs meet are exactly the solutions of their equations taken together. That one idea lets you find intersections, count them without solving, and find the values of a constant for which a line "intersects, touches or does not meet" a curve, which is the syllabus's own example. This note shows the principle across every kind of curve on Paper 1 and then puts it to work in longer coordinate geometry problems.

Graphs and equations

A point lies on the graph of $y = f(x)$ exactly when its coordinates satisfy $y = f(x)$ . So a point lies on both $y = f(x)$ and $y = g(x)$ exactly when

f(x) = g(x)

Key result

The $x$ -coordinates of the points of intersection of $y = f(x)$ and $y = g(x)$ are the roots of $f(x) = g(x)$ .
The number of points of intersection equals the number of distinct real roots of the equation formed by eliminating one variable.
When that equation is a quadratic $ax^2 + bx + c = 0$ :

$b^2 - 4ac$	Line and curve
$> 0$	intersect at two distinct points
$= 0$	touch at one point: the line is a tangent
$< 0$	do not meet

The same reasoning works backwards. To count the solutions of an awkward equation, rewrite it as $f(x) = g(x)$ with two graphs you can sketch, and count the crossings.

The algebra of substitution is in Simultaneous linear and quadratic equations, and the discriminant conditions in The discriminant. Circles have their own note, Lines and circles. Here the focus is on what the conditions mean on a graph and how to use them.

Families of lines

Many questions give a line with an unknown constant. It helps to see the whole family of lines at once.

Parallel lines: $y = mx + k$ with $m$ fixed

Changing $k$ slides the line up and down without turning it. Far below a $\cup$ -shaped parabola the line cuts it twice; sliding down, the two points merge into one (the tangent); below that, the line misses. So there is exactly one tangent in the family, and the answer to "two points" is a single inequality such as $k > 3$ .

y = x^2 - 3x + 7 y = x + 3 y = x + 6 y = x (2, 5)

For $y = x^2 - 3x + 7$ and $y = x + k$ : the line $y = x + 3$ touches at $(2, 5)$ , lines above it cut the curve twice, and lines below it miss.

Lines through a fixed point: $y = mx + c$ with $c$ fixed

Changing $m$ rotates the line about the fixed point $(0, c)$ . If that point is below a $\cup$ -shaped parabola, the line misses the curve when it is nearly horizontal, and as it steepens it touches the curve on one side and then cuts it twice. There are two tangents, one on each side, and the "two points" condition is usually a pair of inequalities such as $m < -2$ or $m > 6$ .

y = x^2 + 2x y = 6x - 4 y = -2x - 4 (0, -4) (2, 8) (-2, 0)

The lines $y = mx - 4$ all pass through $(0, -4)$ . The tangents to $y = x^2 + 2x$ have $m = 6$ and $m = -2$ .

Intersects, touches or does not meet

Eliminate $y$ (or $x$ ): set the two expressions equal, or substitute the line into the curve.
Clear any fractions and collect into $ax^2 + bx + c = 0$ , with the unknown constant inside $a$ , $b$ or $c$ .
Write the condition: $b^2 - 4ac > 0$ (two points), $= 0$ (tangent), $< 0$ (no meeting), $\ge 0$ ("meets").
Solve the resulting equation or quadratic inequality in the constant, using a sketch for inequalities.
Check special cases: a value of the constant that makes $a = 0$ , or that a value of $x$ you divided by (such as $x = 0$ ) is excluded.

Curves other than parabolas

The method needs only that eliminating one variable gives a quadratic. That happens for:

parabolas $y = ax^2 + bx + c$ and sideways parabolas such as $y^2 = x + 3$ (eliminate $x$ instead, and get a quadratic in $y$ );
hyperbolas $y = \dfrac{k}{x}$ or $xy = k$ (multiply through by $x$ , noting $x \neq 0$ );
two parabolas $y = f(x)$ and $y = g(x)$ , since $f(x) - g(x)$ is quadratic (or linear, if the $x^2$ terms cancel);
circles with lines.

Worked examples

Two curves

Find the coordinates of the points where the curves $y = 2x^2 - 5x$ and $y = x^2 - x + 5$ meet, and the equation of the line through them.

Solution

Set the expressions equal:

2x^2 - 5x = x^2 - x + 5 \quad\Rightarrow\quad x^2 - 4x - 5 = 0 \quad\Rightarrow\quad (x - 5)(x + 1) = 0

$x = 5$ : $y = 50 - 25 = 25$ . $x = -1$ : $y = 2 + 5 = 7$ . The points are $(5, 25)$ and $(-1, 7)$ .

The line through them has gradient $\dfrac{25 - 7}{5 - (-1)} = 3$ , so $y - 7 = 3(x + 1)$ , i.e. $y = 3x + 10$ .

A family of parallel lines

Find the set of values of $k$ for which the line $y = x + k$ meets the curve $y = x^2 - 3x + 7$ at two distinct points. State the value of $k$ for which the line is a tangent and the point of contact.

Solution

x^2 - 3x + 7 = x + k \quad\Rightarrow\quad x^2 - 4x + (7 - k) = 0

$b^2 - 4ac = 16 - 4(7 - k) = 4k - 12$ .

Two distinct points: $4k - 12 > 0$ , so $k > 3$ .

Tangent: $k = 3$ . Then $x^2 - 4x + 4 = 0$ , $x = 2$ , $y = 2 + 3 = 5$ : the point of contact is $(2, 5)$ .

Lines through a fixed point

The line $y = mx - 4$ meets the curve $y = x^2 + 2x$ .

(a) Find the set of values of $m$ for which the line meets the curve at two distinct points.

(b) For each value of $m$ for which the line is a tangent, find the coordinates of the point of contact.

Solution

(a)

x^2 + 2x = mx - 4 \quad\Rightarrow\quad x^2 + (2 - m)x + 4 = 0

Two distinct points: $(2 - m)^2 - 16 > 0$ . The boundary values are $2 - m = \pm 4$ , i.e. $m = -2$ or $m = 6$ . The expression $(2 - m)^2 - 16$ is a $\cup$ -shaped quadratic in $m$ , positive outside its roots:

m < -2 \quad\text{or}\quad m > 6

(b) $m = 6$ : $x^2 - 4x + 4 = 0$ , so $x = 2$ , $y = 8$ . Contact at $(2, 8)$ .

$m = -2$ : $x^2 + 4x + 4 = 0$ , so $x = -2$ , $y = 0$ . Contact at $(-2, 0)$ .

A tangent to a sideways parabola

The line $4y = x + c$ , where $c$ is a constant, is a tangent to the curve $y^2 = x + 3$ at the point $P$ . Find the value of $c$ and the coordinates of $P$ .

Solution

The curve is a parabola lying on its side, so eliminate $x$ : from the line, $x = 4y - c$ .

y^2 = 4y - c + 3 \quad\Rightarrow\quad y^2 - 4y + (c - 3) = 0

Tangent: $16 - 4(c - 3) = 0$ , so $c = 7$ .

Then $y^2 - 4y + 4 = 0$ , so $y = 2$ and $x = 4(2) - 7 = 1$ . $P = (1, 2)$ .

Check: $2^2 = 4 = 1 + 3$ . Correct.

y^2 = x + 3 4y = x + 7 (1, 2)

A hyperbola and a special case

Find the set of values of $m$ for which the line $y = mx + 4$ meets the curve $y = -\dfrac{1}{x}$ at two distinct points.

Solution

mx + 4 = -\frac{1}{x} \quad\Rightarrow\quad mx^2 + 4x + 1 = 0 \quad (x \neq 0)

$x = 0$ is never a root (it would give $1 = 0$ ), so multiplying by $x$ lost nothing.

If $m \neq 0$ this is a quadratic, with two distinct roots when $16 - 4m > 0$ , i.e. $m < 4$ .

If $m = 0$ the equation is $4x + 1 = 0$ , which has only one root: the horizontal line $y = 4$ meets the curve once, at $\left(-\tfrac{1}{4}, 4\right)$ .

So the set of values is $m < 4$ , $m \neq 0$ .

A chord, its perpendicular bisector and an area

The line $x + y = 5$ meets the curve $y = \dfrac{4}{x}$ at the points $A$ and $B$ .

(a) Find the coordinates of $A$ and $B$ .

(b) Find the equation of the perpendicular bisector of $AB$ and show that it passes through the origin $O$ .

Solution

(a) $y = 5 - x$ , so $x(5 - x) = 4$ , giving $x^2 - 5x + 4 = 0$ and $(x - 1)(x - 4) = 0$ . The points are $A(1, 4)$ and $B(4, 1)$ .

(b) Midpoint $M = \left(\tfrac{5}{2}, \tfrac{5}{2}\right)$ . $m_{AB} = \dfrac{1 - 4}{4 - 1} = -1$ , so the bisector has gradient $1$ :

y - \tfrac{5}{2} = 1\left(x - \tfrac{5}{2}\right) \quad\Rightarrow\quad y = x

$(0, 0)$ satisfies $y = x$ , so the bisector passes through $O$ . (This is the line of symmetry of the hyperbola.)

AB = \sqrt{3^2 + 3^2} = 3\sqrt{2}, \qquad OM = \sqrt{\left(\tfrac{5}{2}\right)^2 + \left(\tfrac{5}{2}\right)^2} = \tfrac{5}{2}\sqrt{2}

\text{Area} = \tfrac{1}{2} \times 3\sqrt{2} \times \tfrac{5}{2}\sqrt{2} = \tfrac{15}{2}

Watch out

Solving the boundary equation and stopping. "The set of values of $m$ for which the line meets the curve at two points" needs an inequality, not just $m = -2$ and $m = 6$ . Sketch the quadratic in $m$ to choose "outside" or "between".

Using $b^2 - 4ac$ before collecting terms. With $x^2 + 2x = mx - 4$ , the coefficient of $x$ is $2 - m$ , not $2$ . Collect into $ax^2 + bx + c = 0$ first.

Forgetting the case $a = 0$ . If the constant appears in the $x^2$ coefficient, the equation stops being quadratic for one value. Check that value separately.

Eliminating the wrong variable. For $y^2 = x + 3$ , substitute for $x$ ; substituting for $y$ gives a square root.

Writing $\ge 0$ for "two distinct points". Distinct means $> 0$ ; "meets" or "intersects" without "distinct" usually means $\ge 0$ .

Exam tip

Read the wording. "Intersects at two distinct points" ( $> 0$ ), "touches" or "is a tangent" ( $= 0$ ), "does not meet" ( $< 0$ ), "meets" ( $\ge 0$ ). The syllabus phrase "intersects, touches or does not meet" signals exactly this.
Show the quadratic. Mark schemes award a method mark for the correct three-term quadratic with all terms on one side, and another for applying the discriminant correctly to it.
Quadratic inequalities. Find the critical values, then decide the region with a sketch. Write the final answer as an inequality, not just critical values.
Follow-on parts. After finding a constant, questions often ask for the point of contact, the midpoint of a chord, the equation of a perpendicular bisector or an area. Keep coordinates exact.

Summary

Points of intersection of two graphs $\leftrightarrow$ roots of the equation formed by eliminating a variable.
Line and quadratic curve: discriminant $> 0$ two points, $= 0$ tangent, $< 0$ none.
Parallel family $y = mx + k$ : one tangent value of $k$ ; two points for $k$ on one side of it.
Fixed-point family $y = mx + c$ : usually two tangent gradients; two points for $m$ outside them (if the point is below a $\cup$ -shaped curve).
Sideways parabolas: eliminate $x$ . Hyperbolas: multiply by $x$ and check $x \neq 0$ . Check any value that makes the leading coefficient zero.
Finish with the follow-on geometry: midpoints, perpendicular bisectors, lengths and areas.

Practice questions

Question

Find the coordinates of the points where the line $y = 3x + 1$ meets the curve $y = 2x^2 - x - 5$ .
Find the set of values of $k$ for which the line $y = kx + 2$ meets the curve $y = x^2 - 6x + 11$ at two distinct points.
Find the points of intersection of the curves $y = x^2 - 4$ and $y = 2x - x^2$ .
Lines through the origin are drawn as tangents to the curve $y = x^2 + 4$ . Find their equations, the points of contact, and the area of the triangle formed by the origin and the two points of contact.
The line $y = x + c$ is a tangent to the curve $y^2 = 8x$ . Find $c$ and the point of contact.
Find the values of $k$ for which the line $x + y = k$ is a tangent to the curve $xy = 9$ , and the points of contact.
The line $y = 2x + k$ and the curve $y = x^2 - 6x + 5$ are given. (a) Find the value of $k$ for which the line is a tangent to the curve, and the point of contact. (b) State the set of values of $k$ for which the line does not meet the curve.
The line $y = 2x + 1$ meets the curve $y = x^2 - 2x + 4$ at $A$ and $B$ . The perpendicular bisector of $AB$ meets the $y$ -axis at $C$ . Find the coordinates of $A$ , $B$ and $C$ , and the area of triangle $ABC$ .
The line $y = mx + 3$ and the curve $y = 2x^2 + x + 5$ are given. (a) Find the set of values of $m$ for which the line does not meet the curve. (b) Find the equations of the two tangents in this family and their points of contact. (c) Find the equation of the line through the two points of contact.
The line $y = 2x - 4$ meets the curve $y^2 = 4x$ at $A$ and $B$ . (a) Find the coordinates of $A$ and $B$ and the exact length of $AB$ . (b) Find the value of $c$ for which the line $y = 2x + c$ is a tangent to $y^2 = 4x$ , and the point of contact.

Answers

$2x^2 - x - 5 = 3x + 1$ gives $2x^2 - 4x - 6 = 0$ , $x^2 - 2x - 3 = 0$ , $(x - 3)(x + 1) = 0$ . Points $(3, 10)$ and $(-1, -2)$ .
$x^2 - (6 + k)x + 9 = 0$ . Two distinct points: $(6 + k)^2 - 36 > 0$ . Critical values $6 + k = \pm 6$ , i.e. $k = 0$ or $k = -12$ . The answer is $k < -12$ or $k > 0$ .
$x^2 - 4 = 2x - x^2$ gives $2x^2 - 2x - 4 = 0$ , $(x - 2)(x + 1) = 0$ . Points $(2, 0)$ and $(-1, -3)$ .
$y = mx$ : $x^2 - mx + 4 = 0$ . Tangent: $m^2 - 16 = 0$ , $m = \pm 4$ . $y = 4x$ touches at $x = 2$ , i.e. $(2, 8)$ ; $y = -4x$ touches at $(-2, 8)$ . The triangle has base $4$ (from $(-2, 8)$ to $(2, 8)$ ) and height $8$ : area $16$ .
$(x + c)^2 = 8x$ gives $x^2 + (2c - 8)x + c^2 = 0$ . Tangent: $(2c - 8)^2 - 4c^2 = 0$ , so $64 - 32c = 0$ and $c = 2$ . Then $x^2 - 4x + 4 = 0$ , $x = 2$ , $y = 4$ . Contact $(2, 4)$ .
$x(k - x) = 9$ gives $x^2 - kx + 9 = 0$ . Tangent: $k^2 = 36$ , $k = \pm 6$ . $k = 6$ : $(x - 3)^2 = 0$ , contact $(3, 3)$ . $k = -6$ : contact $(-3, -3)$ .
(a) $x^2 - 8x + (5 - k) = 0$ . Tangent: $64 - 4(5 - k) = 0$ , so $k = -11$ . Then $(x - 4)^2 = 0$ : contact $(4, -3)$ . (b) No meeting: $64 - 4(5 - k) < 0$ , i.e. $44 + 4k < 0$ , so $k < -11$ .
$x^2 - 2x + 4 = 2x + 1$ gives $x^2 - 4x + 3 = 0$ , so $A(1, 3)$ and $B(3, 7)$ . Midpoint $(2, 5)$ ; $m_{AB} = 2$ , so the bisector is $y - 5 = -\tfrac{1}{2}(x - 2)$ , i.e. $x + 2y = 12$ . At $x = 0$ , $C = (0, 6)$ . $AB = \sqrt{20}$ and $CM = \sqrt{4 + 1} = \sqrt{5}$ , with $CM \perp AB$ : area $= \tfrac{1}{2}\sqrt{20}\sqrt{5} = 5$ .
(a) $2x^2 + (1 - m)x + 2 = 0$ . No meeting: $(1 - m)^2 - 16 < 0$ , so $-4 < 1 - m < 4$ , i.e. $-3 < m < 5$ . (b) Tangents at $m = 5$ and $m = -3$ . $m = 5$ : $2x^2 - 4x + 2 = 0$ , $x = 1$ , contact $(1, 8)$ ; tangent $y = 5x + 3$ . $m = -3$ : $2x^2 + 4x + 2 = 0$ , $x = -1$ , contact $(-1, 6)$ ; tangent $y = -3x + 3$ . (c) Gradient $\dfrac{8 - 6}{1 - (-1)} = 1$ : $y = x + 7$ .
(a) $(2x - 4)^2 = 4x$ gives $4x^2 - 20x + 16 = 0$ , $x^2 - 5x + 4 = 0$ , so $x = 1$ or $4$ . $A(1, -2)$ , $B(4, 4)$ . $AB = \sqrt{9 + 36} = 3\sqrt{5}$ . (b) $(2x + c)^2 = 4x$ gives $4x^2 + (4c - 4)x + c^2 = 0$ . Tangent: $(4c - 4)^2 - 16c^2 = 0$ , so $16 - 32c = 0$ and $c = \tfrac{1}{2}$ . Then $4x^2 - 2x + \tfrac{1}{4} = 0$ , i.e. $\left(2x - \tfrac{1}{2}\right)^2 = 0$ , $x = \tfrac{1}{4}$ and $y = 1$ . Contact $\left(\tfrac{1}{4}, 1\right)$ .

Graphs and equations

Families of lines

Parallel lines: y=mx+ky = mx + ky=mx+k with mmm fixed

Lines through a fixed point: y=mx+cy = mx + cy=mx+c with ccc fixed

Curves other than parabolas

Worked examples

Practice questions

Console

Parallel lines: $y = mx + k$ with $m$ fixed

Lines through a fixed point: $y = mx + c$ with $c$ fixed