FP1 January 2013 Q9

EdexcelOld spec9 marksConic Sections

Figure 1: part of the parabola y squared = 36x from O, point P on the curve joined to the focus S and to N on the x-axis — Figure 1

Figure 1 shows a sketch of part of the parabola with equation \(y^2 = 36x\).

The point \(P\ (4,\ 12)\) lies on the parabola.

(a) Find an equation for the normal to the parabola at \(P\). (5)

This normal meets the \(x\)-axis at the point \(N\) and \(S\) is the focus of the parabola, as shown in Figure 1.

(b) Find the area of triangle \(PSN\). (4)

Mark scheme (a)

Scheme	Marks
\(y = 6x^{\frac{1}{2}}\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{-\frac{1}{2}}\)	M1
Gradient when \(x = 4\) is \(\tfrac{3}{2}\) and gradient of normal is \(-\tfrac{2}{3}\)	M1 A1
So equation of normal is \((y - 12) = -\tfrac{2}{3}(x - 4)\) (or \(3y + 2x = 44\))	M1 A1
	(5)

Notes

Alternatives:
First M1 for \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = 36\) or for
\(x = 9t^2,\ y = 18t \to \dfrac{\mathrm{d}x}{\mathrm{d}t} = 18t,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = 18 \to \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\)

(a) First M for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ax^{-\frac{1}{2}}\),

Second M for substituting \(x = 4\) (or \(y = 12\) or \(t = 2/3\) if alternative used) into their gradient and applying negative reciprocal.

First A for \(-\dfrac{2}{3}\)

Third M for \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) and attempt to substitute a changed gradient AND (4,12)

Second A for \(3y + 2x = 44\) or any equivalent equation

Mark scheme (b)

Scheme	Marks
\(S\) is at point (9,0)	B1
\(N\) is at (22,0), found by substituting \(y = 0\) into their part (a) Both B marks can be implied or on diagram.	B1ft
So area is \(\tfrac{1}{2} \times 12 \times (22 - 9) = 78\)	M1 A1 cao
	(4)
	[9]

Notes

(b) M for Area\(= \dfrac{1}{2}\) base x height and attempt to substitute including their numerical ‘(22-9)’ or equivalent complete method to find area of triangle \(PSN\).