A2 June 2021 Paper 2 Q6

AQACurrent spec8 marksGraphs & Inequalities

6 The ellipse \(E_1\) has equation

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

\(E_1\) is translated by the vector \(\begin{bmatrix} 3 \\ 0 \end{bmatrix}\) to give the ellipse \(E_2\)

(a) Write down the equation of \(E_2\) [1 mark]

(b) The ellipse \(E_3\) has equation\[\frac{x^2}{4} + (y - 3)^2 = 1\]

Describe the transformation that maps \(E_2\) to \(E_3\) [1 mark]

\(L_A\) is closer to the origin than \(L_B\)

\(E_2\) and \(E_3\) both lie between \(L_A\) and \(L_B\)

Sketch and label \(E_2\), \(E_3\), \(L_A\) and \(L_B\) on the axes below.

You do not need to show the values of the axis intercepts for \(L_A\) and \(L_B\) [4 marks]

(d) Explain, without doing any calculations, why \(L_A\) has an equation of the form\[x + y = c\]

where \(c\) is a constant. [2 marks]

Scheme	Marks	AO
Obtains the correct equation of \(E_2\)	B1	1.1b
	(1)

Scheme	Marks	AO
Obtains a correct answer Condone a correct sequence of transformations	B1	1.1b
	(1)

Sketch: ellipse E3 centred on the y-axis between y = 2 and y = 4, ellipse E2 centred on the x-axis between x = 2 and x = 4, and two parallel lines of gradient −1, LA nearer the origin and LB further out, each touching both ellipses — Mark scheme (c)

Scheme	Marks	AO
Two ellipses, one crossing the positive \(x\)-axis and the other crossing the positive \(y\)-axis	B1	1.1b
Correct axis intercepts shown for both ellipses	B1	1.1b
At least one correct tangent drawn Condone \(x = 2\) or \(y = 2\)	B1	1.1b
Both lines drawn and labelled correctly	B1	2.2b
	(4)

Mark scheme (d)

Scheme	Marks	AO
Uses the fact that \(E_3\) is a reflection of \(E_2\) in the line \(y = x\)	E1	2.4
Explains that the tangent is perpendicular to the line \(y = x\) and concludes that its equation is \(x + y = c\)	E1	2.4
	(2)
	(8 marks)

Typical solution

Points on \(E_2\) and \(E_3\) joined by \(L_A\) are symmetrical about \(y = x\), therefore the line is perpendicular to \(y = x\) and has a gradient of \(-1\) and is of the form \(y = -x + c\) or \(x + y = c\)