FP1 January 2011 Q6
6.
Figure 1 shows a sketch of the parabola \(C\) with equation \(y^2 = 36x\).
The point \(S\) is the focus of \(C\).
(a) Find the coordinates of \(S\). (1)
(b) Write down the equation of the directrix of \(C\). (1)
Figure 1 shows the point \(P\) which lies on \(C\), where \(y > 0\), and the point \(Q\) which lies on the directrix of \(C\). The line segment \(QP\) is parallel to the \(x\)-axis.
Given that the distance \(PS\) is 25,
(c) write down the distance \(QP\), (1)
(d) find the coordinates of \(P\), (3)
(e) find the area of the trapezium \(OSPQ\). (2)
| Scheme | Marks |
|---|---|
| \(C\!: y^2 = 36x \quad \Rightarrow \quad a = \tfrac{36}{4} = 9\) | |
| \(S(9,\ 0)\) (9, 0) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(x + 9 = 0\) or \(x = -9\) \(x + 9 = 0\) or \(x = -9\) or ft using their \(a\) from part (a). | B1ft |
| (1) |
| Scheme | Marks |
|---|---|
| \(PS = 25 \Rightarrow \underline{QP = 25}\) Either 25 by itself or \(PQ = 25\). Do not award if just \(PS = 25\) is seen. | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(x\)-coordinate of \(P \Rightarrow x = 25 - 9 = 16\) \(x = 16\) | B1ft |
| \(y^2 = 36(16)\) Substitutes their \(x\)-coordinate into equation of \(C\). | M1 |
| \(\underline{y} = \sqrt{576} = \underline{24}\) Therefore \(P(16,\ 24)\) \(\underline{y = 24}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Area \(OSPQ = \tfrac{1}{2}(9 + 25)24\) \(\tfrac{1}{2}(\text{their } a + 25)(\text{their } y)\) or rectangle and 2 distinct triangles, correct for their values. | M1 |
| \(= \underline{408}\ (\text{units})^2\) 408 | A1 |
| (2) | |
| [8] |