AS & A Level · Practice questions · Intersections of lines and curves

Substitute, then read the discriminant.

Six original Cambridge-style questions on intersections of lines with curves and circles, including sets of values of a constant and a projectile on a hill.

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

The method

One equation, one quadratic, one discriminant.

Substitute the line into the curve or circle.
Collect everything on one side: \(ax^{2} + bx + c = 0\), where \(a\), \(b\), \(c\) may contain the constant.
Decide: two points \(> 0\); tangent \(= 0\); no meeting \(< 0\).
Solve for the constant, then find any point of contact from the repeated root.
State the conclusion in words, e.g. 'so the line is a tangent'.

Recall

[3 marks]

Find the coordinates of the points of intersection of \(y = x^{2} - 4x + 1\) and \(y = 2x - 4\).

\(x^{2} - 4x + 1 = 2x - 4 \Rightarrow x^{2} - 6x + 5 = 0\). ✓
\((x - 1)(x - 5) = 0 \Rightarrow x = 1\) or \(x = 5\). ✓
Points \((1, -2)\) and \((5, 6)\). ✓

Calculation

[3 marks]

The line \(y = 5x + k\) is a tangent to the curve \(y = x^{2} + x + 2\). Find \(k\) and the coordinates of the point of contact.

\(x^{2} + x + 2 = 5x + k \Rightarrow x^{2} - 4x + (2 - k) = 0\). ✓
Tangent: \(16 - 4(2 - k) = 0 \Rightarrow 8 + 4k = 0 \Rightarrow k = -2\). ✓
\(x^{2} - 4x + 4 = 0 \Rightarrow x = 2\), \(y = 5(2) - 2 = 8\). Point \((2, 8)\). ✓

Calculation

[3 marks]

Find the set of values of \(m\) for which the line \(y = mx + 3\) does not meet the curve \(y = 2x^{2} + 5\).

\(2x^{2} + 5 = mx + 3 \Rightarrow 2x^{2} - mx + 2 = 0\). ✓
No meeting: \(m^{2} - 4(2)(2) < 0 \Rightarrow m^{2} < 16\). ✓
\(-4 < m < 4\). ✓

Analysis

[5 marks]

The circle \(x^{2} + y^{2} - 6x + 4y + 8 = 0\) is given.

(a) Find its centre and radius.

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

(a) \((x - 3)^{2} + (y + 2)^{2} = 9 + 4 - 8 = 5\): centre \((3, -2)\), radius \(\sqrt{5}\). ✓
(b) \(x^{2} + (2x + c)^{2} - 6x + 4(2x + c) + 8 = 0\). ✓
\(5x^{2} + (4c + 2)x + (c^{2} + 4c + 8) = 0\). ✓
Tangent: \((4c + 2)^{2} - 20(c^{2} + 4c + 8) = 0 \Rightarrow -4c^{2} - 64c - 156 = 0 \Rightarrow c^{2} + 16c + 39 = 0\). ✓
\((c + 3)(c + 13) = 0\): \(c = -3\) or \(c = -13\). ✓

Proof

[3 marks]

Show that the line \(y = 4 - x\) does not meet the curve \(y = x^{2} + x + 6\).

\(x^{2} + x + 6 = 4 - x \Rightarrow x^{2} + 2x + 2 = 0\). ✓
Discriminant \(= 2^{2} - 4(1)(2) = 4 - 8 = -4 < 0\). ✓
No real roots, so the line and the curve do not meet. ✓

Analysis

[5 marks]

A ball is kicked from the foot of a hill. Taking the kick as the origin, with \(x\) and \(y\) in metres, the ball's path is \(y = x - \dfrac{x^{2}}{40}\) and the hillside is the line \(y = \dfrac{x}{4}\).

(a) Find where the ball lands on the hillside.

(b) Find the value of \(h\) for which the line \(y = \dfrac{x}{4} + h\) is a tangent to the path, and explain what \(h\) represents.

(a) \(x - \dfrac{x^{2}}{40} = \dfrac{x}{4} \Rightarrow \dfrac{3x}{4} - \dfrac{x^{2}}{40} = 0 \Rightarrow x\left(30 - x\right) = 0\) (after multiplying by 40). ✓
\(x = 30\) (\(x = 0\) is the kick), \(y = 7.5\): the ball lands at \((30, 7.5)\). ✓
(b) \(x - \dfrac{x^{2}}{40} = \dfrac{x}{4} + h \Rightarrow \dfrac{x^{2}}{40} - \dfrac{3x}{4} + h = 0 \Rightarrow x^{2} - 30x + 40h = 0\). ✓
Tangent: \(900 - 160h = 0 \Rightarrow h = \dfrac{45}{8} = 5.625\), touching at \(x = 15\). ✓
\(h\) is the greatest vertical height of the ball above the hillside: 5.625 m, when \(x = 15\) m. ✓

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.