IGCSE · Practice questions · Simultaneous equations: a line and a curve

Every root has a partner.

Six original Cambridge-style questions on solving a linear and a non-linear equation simultaneously: substitution, careful expansion, paired answers, the chord between two intersection points and the tangent case.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange, substitute, solve, pair.

Rearrange the linear equation for the letter with coefficient 1, so no fractions appear.
Substitute with brackets: \(y^{2}\) becomes \((x + 1)^{2}\), \(xy\) becomes \(x(x + 1)\).
Expand and collect into a quadratic \(= 0\); divide through by any common factor.
Pair: put each root into the LINEAR equation and write the answers as \((x, y)\).
Interpret: two roots, two points; a repeated root, a tangent; then use the midpoint and distance formulae for a chord.

Recall

[3 marks]

Solve the simultaneous equations \(y = x + 1\) and \(y = x^{2} - 2x - 3\).

\(x^{2} - 2x - 3 = x + 1\), so \(x^{2} - 3x - 4 = 0\). ✓
\((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\). ✓
\(x = 4\): \(y = 5\). \(x = -1\): \(y = 0\). The solutions are \((4, 5)\) and \((-1, 0)\). ✓

Calculation

[4 marks]

Solve the simultaneous equations \(y - x + 2 = 0\) and \(x^{2} - xy + y^{2} = 7\).

From the linear equation, \(y = x - 2\). ✓
Substitute: \(x^{2} - x(x - 2) + (x - 2)^{2} = 7\), so \(x^{2} - x^{2} + 2x + x^{2} - 4x + 4 = 7\). ✓
\(x^{2} - 2x - 3 = 0\), so \((x - 3)(x + 1) = 0\) and \(x = 3\) or \(x = -1\). ✓
\(x = 3\): \(y = 1\). \(x = -1\): \(y = -3\). ✓
Check: \(9 - 3 + 1 = 7\) and \(1 - 3 + 9 = 7\). The solutions are \(x = 3,\ y = 1\) and \(x = -1,\ y = -3\). ✓

Calculation

[4 marks]

Solve the simultaneous equations \(x + 2y = 4\) and \(x^{2} + 2y^{2} = 6\), giving your answers as exact fractions where necessary.

\(x\) has coefficient 1, so \(x = 4 - 2y\). ✓
\((4 - 2y)^{2} + 2y^{2} = 6\), so \(16 - 16y + 4y^{2} + 2y^{2} = 6\) and \(6y^{2} - 16y + 10 = 0\). ✓
Divide by 2: \(3y^{2} - 8y + 5 = 0\), so \((3y - 5)(y - 1) = 0\) and \(y = 1\) or \(y = \dfrac{5}{3}\). ✓
\(y = 1\): \(x = 2\). \(y = \dfrac{5}{3}\): \(x = 4 - \dfrac{10}{3} = \dfrac{2}{3}\). ✓
The solutions are \(x = 2,\ y = 1\) and \(x = \dfrac{2}{3},\ y = \dfrac{5}{3}\). ✓

Analysis

[5 marks]

The line \(y = x + 3\) meets the circle \(x^{2} + y^{2} - 2x - 4y - 15 = 0\) at the points \(A\) and \(B\).

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

(b) Find the coordinates of the midpoint of \(AB\) and the exact length of \(AB\).

(a) Substitute: \(x^{2} + (x + 3)^{2} - 2x - 4(x + 3) - 15 = 0\). ✓
\(x^{2} + x^{2} + 6x + 9 - 2x - 4x - 12 - 15 = 0\), so \(2x^{2} - 18 = 0\) and \(x^{2} = 9\). ✓
\(x = -3\): \(y = 0\). \(x = 3\): \(y = 6\). So \(A(-3, 0)\) and \(B(3, 6)\). ✓
(b) Midpoint \(\left(\dfrac{-3 + 3}{2}, \dfrac{0 + 6}{2}\right) = (0, 3)\). ✓
\(AB = \sqrt{6^{2} + 6^{2}} = \sqrt{72} = 6\sqrt{2}\). ✓

Proof

[4 marks]

Show that the line \(x + y = 4\) is a tangent to the curve \(y = x^{2} - 5x + 8\), and find the coordinates of the point of contact.

From the line, \(y = 4 - x\). Substitute: \(x^{2} - 5x + 8 = 4 - x\). ✓
\(x^{2} - 4x + 4 = 0\), so \((x - 2)^{2} = 0\). ✓
There is one repeated root, \(x = 2\), so the line meets the curve at exactly one point: it is a tangent. ✓
\(y = 4 - 2 = 2\). The point of contact is \((2, 2)\). ✓

Analysis

[7 marks]

The curve \(y = x^{2} - 4x + 5\) and the line \(y = x + 1\) meet at the points \(A\) and \(B\).

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

(b) Find the exact length of \(AB\).

(c) The line \(y = x + k\) is a tangent to the curve. Find the value of \(k\) and the coordinates of the point of contact.

(a) \(x^{2} - 4x + 5 = x + 1\), so \(x^{2} - 5x + 4 = 0\), \((x - 1)(x - 4) = 0\). ✓
\(x = 1\): \(y = 2\). \(x = 4\): \(y = 5\). So \(A(1, 2)\) and \(B(4, 5)\). ✓
(b) \(AB = \sqrt{3^{2} + 3^{2}} = \sqrt{18} = 3\sqrt{2}\). ✓
(c) \(x^{2} - 4x + 5 = x + k\) gives \(x^{2} - 5x + (5 - k) = 0\). A tangent needs a repeated root, so (topic 2.3) \(b^{2} - 4ac = 0\): \(25 - 4(5 - k) = 0\). ✓
\(25 - 20 + 4k = 0\), so \(k = -\dfrac{5}{4}\). ✓
Then \(x^{2} - 5x + \dfrac{25}{4} = \left(x - \dfrac{5}{2}\right)^{2} = 0\), so \(x = \dfrac{5}{2}\) and \(y = \dfrac{5}{2} - \dfrac{5}{4} = \dfrac{5}{4}\). ✓
The point of contact is \(\left(\dfrac{5}{2}, \dfrac{5}{4}\right)\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.