AS June 2025 Q5

EdexcelAS paperCurrent spec9 marksConic Sections

5. The parabola \(C\) has equation \(y^2 = 16x\)

The point \(P(4t^2, 8t)\) lies on \(C\).

(a) Show that an equation for the tangent to \(C\) at \(P\) is\[yt - x = 4t^2\] (3)

The line \(l\) passes through the origin and is perpendicular to the tangent to \(C\) at \(P\).
The line \(l\) and the tangent to \(C\) at \(P\) intersect at the point \(Q\).

(b) Determine, in simplest form in terms of \(t\), the coordinates of \(Q\). (3)

(c) Hence show that, as \(t\) varies, the point \(Q\) lies on the curve with equation\[y^2 = \frac{Ax^3}{x + B}\]where \(A\) and \(B\) are integers to be determined. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{8t} = \dfrac{1}{t}\) or \(y = 4\sqrt{x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{\sqrt{x}} = \dfrac{1}{t}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 16 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{y} = \dfrac{1}{t}\) Or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{y} = \dfrac{1}{t}\)	B1	1.1b
\(y - 8t = \dfrac{1}{t}(x - 4t^2)\) Or \(y = \dfrac{1}{t}x + c \Rightarrow 8t = \dfrac{1}{t} \times 4t^2 \Rightarrow c = \ldots\{4t\}\)	M1	1.1b
\(ty - 8t^2 = x - 4t^2 \Rightarrow yt - x = 4t^2\ \) Or \(y = \dfrac{1}{t}x + 4t \Rightarrow yt - x = 4t^2\ \)	A1*	2.1
	(3)

Notes

B1: Obtains the correct tangent gradient

M1: Correct strategy for the equation of the tangent

A1*: Correct proof with no errors

Mark scheme (b)

Scheme	Marks	AO
\(y = -tx\)	B1	2.2a
\(yt - x = 4t^2,\ y = -tx \Rightarrow -t^2x - x = 4t^2 \Rightarrow x = \ldots\) or \(yt - x = 4t^2,\ y = -tx \Rightarrow yt + \dfrac{y}{t} = 4t^2 \Rightarrow y = \ldots\)	M1	2.1
\(\left(\dfrac{-4t^2}{t^2 + 1},\ \dfrac{4t^3}{t^2 + 1}\right)\) or seen as \(x = \ldots\ y = \ldots\)	A1	1.1b
	(3)

Notes

B1: Deduces the correct equation of \(l\) (corrected from the printed mark scheme: this note is printed as M1, but the scheme gives this mark as B1)

M1: Solves simultaneously their equation of the normal (using a changed gradient) and the answer to (a) to find \(x\) or \(y\)

A1: Correct simplified coordinates

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{y}{x} = -t \Rightarrow x = \dfrac{-4\frac{y^2}{x^2}}{\frac{y^2}{x^2} + 1}\) or \(y = \dfrac{-4\frac{y^3}{x^3}}{\frac{y^2}{x^2} + 1}\)	M1	1.1b
\(\Rightarrow y^2 = \ldots\)	dM1	1.1b
\(y^2 = \dfrac{-x^3}{x + 4}\)	A1	1.1b
	(3)
	(9 marks)

Notes

M1: Eliminates \(t\) to obtain a Cartesian equation

dM1: Rearranges to obtain the form \(y^2 = \mathrm{f}(x)\)

A1: Correct equation

Alternative 1

Scheme	Marks	AO
\(y^2 = \left(\dfrac{4t^3}{1 + t^2}\right)^2 = \dfrac{16t^6}{(1 + t^2)^2} \qquad x^3 = \left(\dfrac{-4t^2}{1 + t^2}\right)^3 = \dfrac{-64t^6}{(1 + t^2)^3}\) Leading to \(y^2 = -\dfrac{1}{4}x^3(1 + t^2)\) or \(\dfrac{y^2}{x^3} = -\dfrac{1}{4}(1 + t^2)\)	M1	1.1b
\(x = -\dfrac{4t^2}{(1 + t^2)} \Rightarrow t^2 = -\dfrac{x}{x + 4} \Rightarrow t^2 + 1 = \dfrac{4}{x + 4}\) \(y^2 = -\dfrac{1}{4}x^3\left(\dfrac{4}{x + 4}\right)\) Or \(\dfrac{y^2}{x^3} = \dfrac{1 + t^2}{-4} = \dfrac{A}{x + B} = \dfrac{A}{\left(\dfrac{-4t^2}{1 + t^2}\right) + B} = \dfrac{A(1 + t^2)}{-4t^2 + B(1 + t^2)}\) Leading to values for \(A\) and \(B\)	dM1	1.1b
\(y^2 = \dfrac{-x^3}{x + 4}\)	A1	1.1b
	(3)

M1: Finds their \(y^2\) and their \(x^3\) substitutes into the given equation and simplifies

dM1: Uses their \(x\) to find an expression in terms of \(x\) for \(1 + t^2\) or compares the simplified \(\dfrac{y^2}{x^3}\) with \(\dfrac{A}{x + B}\) uses correct algebra to find as a single fraction and find values for \(A\) and \(B\)

A1: Achieves the correct equation

Alternative 2

Scheme	Marks	AO
\(y = -tx \Rightarrow y^2 = t^2x^2\) \(x = -\dfrac{4t^2}{t^2 + 1} \Rightarrow xt^2 + x = -4t^2 \Rightarrow t^2 = -\dfrac{x}{4 + x}\)	M1	1.1b
\(y^2 = \left(-\dfrac{x}{4 + x}\right)x^2\)	dM1	1.1b
\(y^2 = \dfrac{-x^3}{x + 4}\)	A1	1.1b
	(3)

M1: Finds \(y^2\) and uses the x coordinates to find \(t^2\) in terms of \(x\)

dM1: Substitutes their \(t^2\) into \(y^2\) to find in terms of \(x\) only

Alternative 3

Scheme	Marks	AO
\(y^2 = \left(\dfrac{4t^3}{1 + t^2}\right)^2 = \dfrac{16t^6}{(1 + t^2)^2} = \dfrac{A\left(\dfrac{-4t^2}{1 + t^2}\right)^3}{\left(\dfrac{-4t^2}{1 + t^2}\right) + B}\) \(\dfrac{A(-64t^6)}{(1 + t^2)^3} \div \dfrac{-4t^2 + B(1 + t^2)}{(1 + t^2)} = \dfrac{A(-64t^6)}{(1 + t^2)^2(-4t^2 + B + Bt^2)}\) Leading to \(B = 4\)	M1	1.1b
\(\dfrac{A(-64t^6)}{4(1 + t^2)^2} = \dfrac{16t^6}{(1 + t^2)^2} \Rightarrow A = \ldots\{-1\}\)	dM1	1.1b
Correct values for \(A\) and \(B\) and draws the conclusion that therefore \(y^2 = \dfrac{-x^3}{x + 4}\)	A1	1.1b
	(3)

M1: Substitutes their \(y^2\) and their \(x\) into given equation. Uses correct algebra to simply to a single fraction to deduce a value for \(B\).

dM1: Uses their value of \(B\) and equates to find a value for \(A\)

A1: Correct value for \(A\) and \(B\) and draws the conclusion therefore \(y^2 = \dfrac{-x^3}{x + 4}\)

Note: There may be other methods, please send to review if you are unsure

(Corrected from the printed mark scheme: in the first line of Alternative 3, the denominator is printed as \(\left(\dfrac{-4t}{1 + t^2}\right) + B\); it should be \(\left(\dfrac{-4t^2}{1 + t^2}\right) + B\), as used in the next line.)