FP1 June 2009 Q6

EdexcelOld spec11 marksConic Sections

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

(a) Verify that the point \(P(4t^2,\ 8t)\) is a general point on \(C\). (1)

(b) Write down the coordinates of the focus \(S\) of \(C\). (1)

The normal to \(C\) at \(P\) meets the \(x\)-axis at the point \(N\).

(d) Find the area of triangle \(PSN\) in terms of \(t\), giving your answer in its simplest form. (4)

Scheme	Marks
\(y^2 = (8t)^2 = 64t^2\) and \(16x = 16 \times 4t^2 = 64t^2\) Or identifies that \(a = 4\) and uses general coordinates \((at^2,\ 2at)\)	B1
	(1)

Scheme	Marks
\((4,\ 0)\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(y = 4x^{\frac{1}{2}} \qquad \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-\frac{1}{2}}\)	B1
Replaces \(x\) by \(4t^2\) to give gradient \(\left[2(4t^2)^{-\frac{1}{2}} = \dfrac{2}{2t} = \dfrac{1}{t}\right]\)	M1,
Uses Gradient of normal is \(-\dfrac{1}{\text{gradient of curve}}\) \([-t]\)	M1
\(y - 8t = -t(x - 4t^2) \quad \Rightarrow \quad y + tx = 8t + 4t^3\) (*)	M1 A1cso
	(5)

Alternatives

Scheme	Marks
(c) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 8t\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 8\)	B1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \div \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{t}\), then as in main scheme.	M1
(c) \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 16\) (or uses \(x = \dfrac{y^2}{8}\) to give \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{2y}{8}\))	B1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{y} = \dfrac{8}{8t} = \dfrac{1}{t}\), then as in main scheme.	M1

Notes

Third M1 requires linear equation (not fraction) and should include the parameter \(t\) but could be given for equation of tangent (So tangent equation loses 2 marks only and could gain B1M1M0M1A0)

Mark scheme (d)

Scheme	Marks
At \(N\), \(y = 0\), so \(x = 8 + 4t^2\) or \(\dfrac{8t + 4t^3}{t}\)	B1
Base \(SN = (8 + 4t^2) - 4 \quad (= 4 + 4t^2)\)	B1ft
Area of \(\Delta\,PSN = \dfrac{1}{2}(4 + 4t^2)(8t) = 16t(1 + t^2)\) or \(16t + 16t^3\) for \(t > 0\)	M1 A1
{Also Area of \(\Delta\,PSN = \dfrac{1}{2}(4 + 4t^2)(-8t) = -16t(1 + t^2)\) for \(t \lt 0\)} this is not required
	(4)
	[11]

Notes

(d) Second B1 does not require simplification and may be a constant rather than an expression in \(t\).

M1 needs correct area of triangle formula using ½ ‘their \(SN\)’ \(\times 8t\)

Or may use two triangles in which case need \((4t^2 - 4)\) and \((4t^2 + 8 - 4t^2)\) for B1ft

Then Area of \(\Delta\,PSN = \dfrac{1}{2}(4t^2 - 4)(8t) + \dfrac{1}{2}(4t^2 + 8 - 4t^2)(8t) = 16t(1 + t^2)\) or \(16t + 16t^3\)