A2 June 2024 Q8
8. The parabola \(P\) has equation \(y^2 = 4ax\), where \(a\) is a positive constant.
The point \(A\left(at^2, 2at\right)\), where \(t \neq 0\), lies on \(P\).
The point \(B\left(2k^2, 4k\right)\) and the point \(C\left(2k^2, -4k\right)\), where \(k\) is a constant, lie on \(P\).
The tangent to \(P\) at \(B\) and the tangent to \(P\) at \(C\) intersect at the point \(D\).
Given that the area of the triangle \(BCD\) is 432
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{2at} = \dfrac{1}{t}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4a}{2(2at)} = \dfrac{1}{t}\) or \(y = 2\sqrt{a}\sqrt{x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sqrt{a} \times \dfrac{1}{2\sqrt{x}} = \dfrac{\sqrt{a}}{\sqrt{at^2}} = \dfrac{1}{t}\) | B1 | 1.1b |
| \(y - 2at = \dfrac{1}{t}\left(x - at^2\right)\) | M1 | 1.1b |
| \(yt = x + at^2\ *\) | A1* | 2.1 |
| (3) |
Notes
B1: Correct derivative in terms of \(t\) from a correct calculus method. Just stating \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) with no supporting working is B0.
M1: Uses \(y - 2at = \text{their ‘}\dfrac{1}{t}\text{’}\left(x - at^2\right)\), or uses \(y = \text{‘}\dfrac{1}{t}\text{’}x + c\) with \(\left(at^2, 2at\right)\) to find \(c\). They may have just stated the \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) for this mark.
A1*: Achieves the printed equation with no errors, but allow if the \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) is stated without working (so B0M1A1 is possible).
| Scheme | Marks | AO |
|---|---|---|
| Equations of the tangents \(yk = x + 2k^2\) and/or \(-yk = x + 2k^2\) | B1 | 2.2a |
| Solves simultaneously \(\left.\begin{aligned}yk &= x + 2k^2\\ -yk &= x + 2k^2\end{aligned}\right\} \Rightarrow x = \ldots\left\{-2k^2\right\},\ y = \ldots\{0\}\) | M1 | 3.1a |
| Finds the area of the triangle, sets equal to 432 and solves to find a value for \(k\) \(\dfrac{8k \times \left(2k^2 - \text{their ‘}-2k^2\text{’}\right)}{2} = 432\) leading to \(k = \ldots\) | M1 | 3.1a |
| \(k = 3\) | A1 | 1.1b |
| \((18, 12)\) and \((18, -12)\) | A1 | 2.2a |
| (5) | ||
| (8 marks) |
Notes
B1: Deduces a correct equation for one of the tangents. May find both but accept for one correct. If working in terms of \(a\) or \(a\) and \(t\) they must identify \(a = 2\) or \(at^2 = -2k^2\) (oe as relevant) at some stage to gain this mark following a correct tangent in terms of \(a\) or \(a\) and \(t\).
M1: Finds the intersection of the tangent with the \(x\)-axis, or solves both tangent equations simultaneously to find at least the \(x\) coordinate of the point of intersection. May be in terms of \(a\) and \(t\) for this mark.
Note the first 2 marks can be implied from stating a correct point of intersection
M1: A complete method to find the value of \(k\). E.g. uses \(\dfrac{8k \times \left(2k^2 - \text{their ‘}-2k^2\text{’}\right)}{2} = 432\) and solves to find a value for \(k\). Alternative approaches are possible, such as
\(432 = \dfrac{1}{2}\left|\overrightarrow{DC} \times \overrightarrow{DB}\right| = \dfrac{1}{2}\left\|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4k^2 & 4k & 0\\ 4k^2 & -4k & 0\end{matrix}\right\| = \dfrac{1}{2}\left|-32k^3\mathbf{k}\right| = 16k^3 \Rightarrow k = \ldots\) or
\(432 = \dfrac{1}{2}\begin{vmatrix}-2k & 2k & 2k & -2k\\ 0 & 4k^2 & -4k^2 & 0\end{vmatrix} = \dfrac{1}{2}\left|\left(-8k^3 - 8k^3 + 0 - 0 - 8k^3 - 8k^3\right)\right| = 16k^3 \Rightarrow k = \ldots\)
A1: \(k = 3\)
A1 Deduces the correct coordinates \((18, 12)\) and \((18, -12)\) Do not be concerned with the labelling.