AS June 2023 Q3

EdexcelAS paperCurrent spec5 marksConic Sections

3. The rectangular hyperbola \(H\) has equation \(xy = c^2\) where \(c\) is a positive constant.

The line \(l\) has equation \(x - 2y = c\)

The points \(P\) and \(Q\) are the points of intersection of \(H\) and \(l\)

(a) Determine, in terms of \(c\), the coordinates of \(P\) and the coordinates of \(Q\) (3)

The point \(R\) is the midpoint of \(PQ\)

(b) Show that, as \(c\) varies, the coordinates of \(R\) satisfy the equation\[xy = -\frac{c^2}{a}\]where \(a\) is a constant to be determined. (2)

Mark scheme (a)

Scheme	Marks	AO
\(xy = c^2\) and \(x - 2y = c \Rightarrow\) \(2y^2 + cy = c^2\) or \(\dfrac{1}{2}(x^2 - cx) = c^2\)	M1	1.1b
\(2y^2 + cy - c^2 = 0 \Rightarrow y = -c, \dfrac{c}{2} \Rightarrow x = \ldots\) or \(x^2 - cx - 2c^2 = 0 \Rightarrow x = -c, 2c \Rightarrow y = \ldots\)	dM1	1.1b
\((-c, -c)\) and \(\left(2c, \dfrac{c}{2}\right)\)	A1	2.2a
	(3)

Notes

M1: Solve simultaneously \(xy = c^2\) and \(x - 2y = c\) leading to a quadratic in either \(x\) or \(y\) (and \(c\)). It is a method mark so allow if e.g. there is a miscopy if the intent is clear.
Alt: Uses parametric equations and substitutes into \(x - 2y = c\) leading to a quadratic in \(t\) (and \(c\)).

dM1: Dependent on the first method mark. Solves their 3TQ (usual rules) and proceeds to find at least one set of coordinates for either \(P\) or \(Q\).

A1: Deduces the correct coordinates for \(P\) and \(Q\) (need not be named). Accept as \(x = \ldots, y = \ldots\) as long as the coordinates are clearly paired.

Alt

Scheme	Marks	AO
General point is \(\left(ct, \dfrac{c}{t}\right) \Rightarrow ct - \dfrac{2c}{t} = c \Rightarrow ct^2 - 2c = ct\)	M1	1.1b
\(\Rightarrow t^2 - t - 2 = 0 \Rightarrow t = 2, -1 \Rightarrow \left(2c, \dfrac{c}{2}\right)\) or \((-c, -c)\)	dM1	1.1b
\((-c, -c)\) and \(\left(2c, \dfrac{c}{2}\right)\)	A1	2.2a
	(3)

(Corrected from the printed mark scheme: the second line of the Alt is printed as \(t^2 - t - 2t = 0\).)

(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)

Mark scheme (b)

Scheme	Marks	AO
Midpoint \(= \left(\dfrac{-c + 2c}{2}, \dfrac{-c + \frac{c}{2}}{2}\right) = \ldots\)	M1	1.1b
\(x = \dfrac{c}{2}\) and \(y = -\dfrac{c}{4} \Rightarrow xy = \dfrac{c}{2} \times -\dfrac{c}{4}\) leading to \(xy = -\dfrac{c^2}{8}\)	A1cso	2.1
	(2)
	(5 marks)

Notes

M1: Find the midpoint of their \(P\) and \(Q\), provided that \(P\) and \(Q\) are not symmetric in the \(y\)-axis (ie. midpoint is not \((0,0)\)). May be implied by one correct coordinate if no method shown. Simplification is not required.

A1cso: Uses the correct coordinates of the midpoint from correct work to show that \(xy = -\dfrac{c^2}{8}\). Must be an equation, not just the value of \(a\). There must be no contrary statements.