A2 June 2025 Paper 1 Q8

AQACurrent spec3 marksGraphs & Inequalities

8 The curve \(C_1\) has equation

\[\frac{x^2}{4} + \frac{y^2}{25} = 1\]

The curve \(C_1\) is translated by the vector \(\begin{bmatrix} 1 \\ 3 \end{bmatrix}\) to give the curve \(C_2\)

Find the coordinates of the points where the curve \(C_2\) intersects the \(x\)-axis. [3 marks]

Mark scheme

Scheme	Marks	AO
Obtains \(\dfrac{(x - 1)^2}{4} + \dfrac{(y - 3)^2}{25} = 1\)	B1	1.2
Substitutes \(y = 0\) into their equation for \(C_2\) and solves to obtain at least one value of \(x\)	M1	1.1a
Obtains \(\left(\dfrac{13}{5}, 0\right)\) and \(\left(-\dfrac{3}{5}, 0\right)\) Must be coordinates.	A1	1.1b
	(3 marks)

Typical solution

\[C_2 : \frac{(x - 1)^2}{4} + \frac{(y - 3)^2}{25} = 1\]\[y = 0 \Rightarrow \frac{(x - 1)^2}{4} + \frac{9}{25} = 1\]\[25(x - 1)^2 + 36 = 100\]\[25(x - 1)^2 = 64\]\[(x - 1)^2 = \frac{64}{25}\]\[x = 1 \pm \frac{8}{5}\]\[\left(\frac{13}{5}, 0\right), \left(-\frac{3}{5}, 0\right)\]