FP1 June 2011 Q8

EdexcelOld spec10 marksConic Sections

8. The parabola \(C\) has equation \(y^2 = 48x\).

The point \(P(12t^2,\ 24t)\) is a general point on \(C\).

(a) Find the equation of the directrix of \(C\). (2)

(b) Show that the equation of the tangent to \(C\) at \(P(12t^2,\ 24t)\) is \[x - ty + 12t^2 = 0\] (4)

The tangent to \(C\) at the point \((3,\ 12)\) meets the directrix of \(C\) at the point \(X\).

Mark scheme (a)

Scheme	Marks
\(C: y^2 = 48x\) with general point \(P(12t^2,\ 24t)\).
\(y^2 = 4ax \Rightarrow a = \tfrac{48}{4} = 12\) Using \(y^2 = 4ax\) to find \(a\).	M1
So, directrix has the equation \(x + 12 = 0\) \(x + 12 = 0\)	A1 oe
	(2)

Correct answer with no working allow full marks

Mark scheme (b)

Scheme	Marks
\(y = \sqrt{48}\,x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\sqrt{48}\,x^{-\frac{1}{2}}\ \left(= 2\sqrt{3}\,x^{-\frac{1}{2}}\right)\) or (implicitly) \(y^2 = 48x \Rightarrow 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 48\) or (chain rule) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x} = 24 \times \dfrac{1}{24t}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm k\,x^{-\frac{1}{2}}\) \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \times \left(\dfrac{1}{\text{their }\frac{\mathrm{d}x}{\mathrm{d}t}}\right)\)	M1
When \(x = 12t^2\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{48}}{2\sqrt{12t^2}} = \dfrac{\sqrt{4}}{2t} = \dfrac{1}{t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{48}{2y} = \dfrac{48}{48t} = \dfrac{1}{t}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\)	A1
\(\mathbf{T}\): \(y - 24t = \dfrac{1}{t}(x - 12t^2)\) Applies \(y - 24t = \text{their } m_T(x - 12t^2)\) or \(y = (\text{their } m_T)x + c\) using \(x = 12t^2\) and \(y = 24t\) in an attempt to find \(c\). Their \(m_T\) must be a function of \(t\).	M1
\(\mathbf{T}\): \(ty - 24t^2 = x - 12t^2\)
\(\mathbf{T}\): \(x - ty + 12t^2 = 0\) Correct solution.	A1 cso *
	(4)

Special case: If the gradient is quoted as \(1/t\), this can score M0A0M1A1

Mark scheme (c)

Scheme	Marks
Compare \(P(12t^2,\ 24t)\) with \((3,\ 12)\) gives \(t = \tfrac{1}{2}\). \(t = \tfrac{1}{2}\)	B1
NB \(x - ty + 12t^2 = 0\) with \(x = 3\) and \(y = 12\) gives \(4t^2 - 4t + 1 = 0 = (2t - 1)^2 \Rightarrow t = \dfrac{1}{2}\)
\(t = \tfrac{1}{2}\) into \(\mathbf{T}\) gives \(x - \tfrac{1}{2}y + 3 = 0\) Substitutes their \(t\) into \(\mathbf{T}\).	M1
At \(X\), \(x = -12 \Rightarrow -12 - \tfrac{1}{2}y + 3 = 0\) Substitutes their \(x\) from (a) into \(\mathbf{T}\).	M1
So, \(-9 = \tfrac{1}{2}y \Rightarrow y = -18\)
So the coordinates of \(X\) are \((-12,\ -18)\). \((-12,\ -18)\)	A1
	(4)
	(10 marks)

The coordinates must be together at the end for the final A1 e.g. as above or \(x = -12\), \(y = -18\)

See Appendix for an alternative approach to find the tangent

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sqrt{3}\,x^{-\frac{1}{2}} = \dfrac{2\sqrt{3}}{\sqrt{3}} = 2\)	B1
Gives \(y - 12 = 2(x - 3)\) Uses \((3,\ 12)\) and their “2” to find the equation of the tangent.	M1
\(x = -12 \Rightarrow y - 12 = 2(-12 - 3)\) Substitutes their \(x\) from (a) into their tangent	M1
\(y = -18\)
So the coordinates of \(X\) are \((-12,\ -18)\).	A1
	(4)