FP1 January 2010 Q4
4.

Figure 1 shows a sketch of part of the parabola with equation \(y^2 = 12x\).
The point \(P\) on the parabola has \(x\)-coordinate \(\dfrac{1}{3}\).
The point \(S\) is the focus of the parabola.
(a) Write down the coordinates of \(S\). (1)
The points \(A\) and \(B\) lie on the directrix of the parabola.
The point \(A\) is on the \(x\)-axis and the \(y\)-coordinate of \(B\) is positive.
Given that \(ABPS\) is a trapezium,
(b) calculate the perimeter of \(ABPS\). (5)
| Scheme | Marks |
|---|---|
| \((3,\ 0)\) cao | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(P\!: \quad x = \dfrac{1}{3} \quad \Rightarrow \quad y = 2\) | B1 |
| \(A\) and \(B\) lie on \(x = -3\) | B1 |
| \(PB = PS\) or a correct method to find both \(PB\) and \(PS\) | M1 |
| Perimeter \(= 6 + 2 + 3\tfrac{1}{3} + 3\tfrac{1}{3} = 14\tfrac{2}{3}\) | M1 A1 |
| (5) | |
| [6] |
Notes
(b) Both B marks can be implied by correct diagram with lengths labelled or coordinates of vertices stated.
Second M1 for their four values added together.
\(14\tfrac{2}{3}\) or awrt 14.7 for final A1