AS October 2020 Q4

EdexcelAS paperCurrent spec7 marksConic Sections

Figure 2: parabola C opening to the right with vertex at the origin O, focus S on the positive x-axis, and point P on the upper branch — Figure 2

Figure 2 shows a sketch of the parabola \(C\) with equation \(y^2 = 4ax\), where \(a\) is a positive constant. The point \(S\) is the focus of \(C\) and the point \(P(ap^2, 2ap)\) lies on \(C\) where \(p \gt 0\)

(a) Write down the coordinates of \(S\). (1)

(b) Write down the length of \(SP\) in terms of \(a\) and \(p\). (1)

The point \(Q(aq^2, 2aq)\), where \(p \neq q\), also lies on \(C\).
The point \(M\) is the midpoint of \(PQ\).

Given that \(pq = -1\)

Scheme	Marks	AO
\((a, 0)\)	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(SP = ap^2 + a\) Note that if focus-directrix property not used may use Pythagoras: E.g. \(SP = \sqrt{4a^2p^2 + (ap^2 - a)^2} = \ldots = ap^2 + a\)	B1	1.1b
	(1)

Notes

B1: Correct expression

Mark scheme (c)

Scheme	Marks	AO
\(M\) has coordinates \(\left(\dfrac{ap^2 + aq^2}{2}, \dfrac{2ap + 2aq}{2}\right)\)	B1	1.1b
\(y^2 = a^2\left(p^2 + 2pq + q^2\right)\)	M1	1.1b
\(y^2 = a^2\left(p^2 - 2 + q^2\right)\)	A1	2.1
\(2a(x - a) = 2a\left(\dfrac{1}{2}ap^2 + \dfrac{1}{2}aq^2 - a\right) = a^2\left(p^2 + q^2 - 2\right)\)	M1	1.1b
\(\Rightarrow y^2 = 2a(x - a)\ *\)	A1*	2.1
	(5)
	(7 marks)

Notes

B1: Correct coordinates for the midpoint

M1: Squares their y coordinate of the midpoint

A1: Uses \(pq = -1\) to obtain a correct expression for \(y^2\)

M1: Attempts \(2a(x - a)\) using the \(x\) coordinate of their midpoint and attempts to simplify

A1*: Fully correct completion to show \(y^2 = 2a(x - a)\)

Alternative for (c)

Scheme	Marks	AO
\(M\) has coordinates \(\left(\dfrac{ap^2 + aq^2}{2}, \dfrac{2ap + 2aq}{2}\right)\)	B1	1.1b
\(\dfrac{y}{a} = p + q\)	M1	1.1b
\(\dfrac{y^2}{a^2} = p^2 + q^2 + 2pq = p^2 + q^2 - 2\)	A1	2.1
\(\dfrac{2x}{a} = p^2 + q^2\)	M1	1.1b
\(\dfrac{y^2}{a^2} = \dfrac{2x}{a} - 2 \Rightarrow y^2 = 2a(x - a)\ *\)	A1*	2.1
	(5)

B1: Correct coordinates for the midpoint

M1: Uses their \(y\) coordinate of the midpoint to find \(p + q\)

A1: Square and uses \(pq = -1\) to obtain a correct expression for \(y^2/a^2\)

M1: Uses the \(x\) coordinate of their midpoint to find \(p^2 + q^2\)

A1*: Fully correct completion to show \(y^2 = 2a(x - a)\)