Home / A-Level Maths 9709 / Coordinate geometry / Intersections of lines and curves
Cambridge International AS & A Level · Topic 1.3
A-Level Maths 9709 / Topic 1.3 / AS & A Level

Cross, touch or miss.

Where two graphs meet, their equations are true together, so points of intersection are solutions of an equation. Substitute, form a quadratic, and its discriminant tells you whether a line cuts a curve, touches it or misses it, and for which values of a constant.

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The key idea

Substitute the line into the curve to get a quadratic. Its discriminant decides the picture: \(b^{2} - 4ac > 0\), the line cuts the curve at two points; \(= 0\), the line is a tangent; \(< 0\), the line does not meet the curve.

x y (2, 3) y = x² − 2x + 3 y = 2x + 2: two points k > −1 y = 2x − 1: tangent k = −1 y = 2x − 3: no meeting k < −1 x² − 4x + 3 − k = 0, Δ = 4 + 4k
Fig. 1: lines y = 2x + k against the curve y = x² - 2x + 3. Substituting gives x² - 4x + 3 - k = 0 with discriminant 4 + 4k, so k = -1 gives the tangent at (2, 3)
Section 01

Intersections are solutions.

In the lab, draw a curve and a line and read off where they cross: those \(x\)-values are exactly the solutions of the equation formed by setting the two expressions equal. The reverse also works: to solve an equation with a graph you already have, rearrange it so one side is that graph.

Worked example Line meets parabola
Find the points where the line \(y = x - 1\) meets the curve \(y = x^{2} - 3x + 2\).
  1. At an intersection both equations hold: \(x^{2} - 3x + 2 = x - 1\).
  2. \(x^{2} - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0\).
  3. \(x = 1\): \(y = 0\). \(x = 3\): \(y = 2\).
Answer: \((1,\ 0)\) and \((3,\ 2)\)
Worked example Which line to draw
The graph of \(y = x^{2} - 5x + 4\) has been drawn. Which straight line should be added to solve \(x^{2} - 5x + 1 = 0\)?
  1. Make the left-hand side match the graph: \(x^{2} - 5x + 1 = 0 \Leftrightarrow x^{2} - 5x + 4 = 3\).
  2. So the solutions are where \(y = x^{2} - 5x + 4\) meets \(y = 3\).
Answer: The line \(y = 3\); the \(x\)-coordinates of the crossing points are the roots
Examiner trap

Points of intersection need both coordinates. After finding \(x\), substitute into the LINEAR equation for \(y\) (it is simpler and avoids pairing errors).

Section 02

Tangent, two points or none.

For a line containing an unknown constant, substitute and look at the discriminant of the resulting quadratic. Fig. 1 shows the family \(y = 2x + k\): one value of \(k\) gives a tangent, larger values cut the curve twice, smaller values miss it.

Worked example A family of parallel lines
Find the set of values of \(k\) for which the line \(y = 2x + k\) meets the curve \(y = x^{2} - 2x + 3\) at two distinct points, and find the point of contact when the line is a tangent.
  1. \(x^{2} - 2x + 3 = 2x + k \Rightarrow x^{2} - 4x + (3 - k) = 0\).
  2. Two distinct points: \(16 - 4(3 - k) > 0 \Rightarrow 4 + 4k > 0 \Rightarrow k > -1\).
  3. Tangent when \(k = -1\): \(x^{2} - 4x + 4 = (x - 2)^{2} = 0\), so \(x = 2\), \(y = 2(2) - 1 = 3\).
Answer: \(k > -1\); tangent at \((2,\ 3)\) when \(k = -1\)
Worked example Tangents with an unknown gradient
The line \(y = mx + 1\) is a tangent to \(y = x^{2} + 2x + 5\). Find the possible values of \(m\) and the points of contact.
  1. \(x^{2} + 2x + 5 = mx + 1 \Rightarrow x^{2} + (2 - m)x + 4 = 0\).
  2. Tangent: \((2 - m)^{2} - 16 = 0 \Rightarrow 2 - m = \pm 4 \Rightarrow m = -2\) or \(m = 6\).
  3. \(m = 6\): \(x^{2} - 4x + 4 = 0\), \(x = 2\), \(y = 13\).
  4. \(m = -2\): \(x^{2} + 4x + 4 = 0\), \(x = -2\), \(y = 5\).
Answer: \(m = 6\), touching at \((2, 13)\); \(m = -2\), touching at \((-2, 5)\)
Your turn
Find the value of \(k\) for which \(y = 5x + k\) is a tangent to \(y = x^{2} + x + 2\).
Examiner trap

Collect EVERYTHING on one side before reading \(a\), \(b\), \(c\): in \(x^{2} + (2 - m)x + 4 = 0\) the coefficient \(b\) is the whole bracket \((2 - m)\), not 2 and not \(-m\).

Stage 1 · Learn

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The \(x\)-coordinates of the points where \(y = x^{2}\) meets \(y = 2x + 3\) satisfy

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After substituting a line into a curve, the quadratic has a negative discriminant. This means

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The graph of \(y = x^{2} - 5x + 4\) is drawn. Which line should be added to solve \(x^{2} - 5x + 1 = 0\)?

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For which value of \(k\) is the line \(y = k\) a tangent to \(y = x^{2} - 6x + 10\)?

Section 03

Lines and circles.

The same method works for circles: substitute the line into the circle's equation. A tangent to a circle gives a repeated root, and the point of contact can be checked with 'tangent perpendicular to radius'.

Worked example Tangents to a circle
Find the values of \(c\) for which the line \(y = x + c\) is a tangent to the circle \(x^{2} + y^{2} = 8\).
  1. \(x^{2} + (x + c)^{2} = 8 \Rightarrow 2x^{2} + 2cx + c^{2} - 8 = 0\).
  2. Tangent: \((2c)^{2} - 4(2)(c^{2} - 8) = 0 \Rightarrow 4c^{2} - 8c^{2} + 64 = 0\).
  3. \(4c^{2} = 64 \Rightarrow c^{2} = 16\).
Answer: \(c = 4\) or \(c = -4\)
Worked example Point of contact
Show that the line \(x + 2y = 15\) is a tangent to the circle \((x - 1)^{2} + (y - 2)^{2} = 20\), and find the point of contact.
  1. \(x = 15 - 2y\): \((14 - 2y)^{2} + (y - 2)^{2} = 20\).
  2. \(196 - 56y + 4y^{2} + y^{2} - 4y + 4 = 20 \Rightarrow 5y^{2} - 60y + 180 = 0 \Rightarrow y^{2} - 12y + 36 = 0\).
  3. \((y - 6)^{2} = 0\): a repeated root, so the line is a tangent. \(y = 6\), \(x = 15 - 12 = 3\).
  4. Check: radius from \((1, 2)\) to \((3, 6)\) has gradient 2; the line has gradient \(-\frac{1}{2}\); product \(-1\).
Answer: Tangent at \((3,\ 6)\)
Examiner trap

Showing a line is a tangent needs a stated conclusion: '\(b^{2} - 4ac = 0\), so there is a repeated root, so the line is a tangent'. Finding a single root without saying why it means tangency loses the explanation mark.

Stage 2 · Exam

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Find the set of values of \(k\) for which the line \(y = x + k\) meets the curve \(y = x^{2} - 3x + 5\) at two distinct points.

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The line \(y = mx - 2\) is a tangent to the curve \(y = x^{2} + 2\). Find the possible values of \(m\).

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Find the \(x\)-coordinates of the points where the line \(y = 3 - x\) meets the circle \((x - 2)^{2} + (y - 3)^{2} = 8\).

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Find the values of \(c\) for which the line \(y = x + c\) is a tangent to the circle \(x^{2} + y^{2} = 8\).

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Six original Cambridge-style questions
Intersections of a line and a parabola, a tangent with an unknown intercept, a line that misses a curve, tangents to a circle, a proof of no intersection, and the greatest height of a ball above a slope.