A2 October 2021 Q5
5. The parabola \(C\) has equation
\[y^2 = 32x\]and the hyperbola \(H\) has equation
\[\frac{x^2}{36} - \frac{y^2}{9} = 1\]The line \(l_1\) is normal to \(C\) and parallel to the asymptote of \(H\) with positive gradient.
The line \(l_2\) is normal to \(C\) and parallel to the asymptote of \(H\) with negative gradient.
The lines \(l_1\) and \(l_2\) meet \(H\) at the points \(P\) and \(Q\) respectively.
| Scheme | Marks | AO |
|---|---|---|
| Equations of asymptotes of \(H\) are \(y = \pm\dfrac{1}{2}x\) oe e.g \(y = \pm\dfrac{3}{6}x\) or \(\dfrac{x}{6} = \pm\dfrac{y}{3}\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct equations for the asymptotes of \(H\) seen or implied, any form and need not be simplified.
| Scheme | Marks | AO |
|---|---|---|
| For parabola \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 32 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) or \(y = \sqrt{32x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \dfrac{2a}{2at} = \ldots\) | M1 | 2.1 |
| Finds the gradient of the normal using \(m_N = \dfrac{-1}{\text{their } \dfrac{\mathrm{d}y}{\mathrm{d}x}}\) \(m_N = -\dfrac{y}{16}\) or \(-t\) or \(-\dfrac{\sqrt{x}}{2\sqrt{2}}\) so \(m_N = (\pm)\dfrac{1}{2} \Rightarrow y = (\pm)8, x = 2\) | M1 | 3.1a |
| Finds the equation of either \(l_1\) or \(l_2\) \(y - \text{“}8\text{”} = \text{“their } m_N\text{”}(x - \text{“}2\text{”})\) or \(y - \text{“}{-8}\text{”} = \text{“their } m_N\text{”}(x - \text{“}2\text{”})\) | M1 | 1.1b |
| \(l_1\) is \(y + 8 = \dfrac{1}{2}(x - 2)\) and \(l_2\) is \(y - 8 = -\dfrac{1}{2}(x - 2)\) oe \(y = \dfrac{1}{2}x - 9\) and \(y = -\dfrac{1}{2}x + 9\) | A1 | 1.1b |
| (4) |
Notes
Note M1 M1 A1 A1 on ePen
M1: A correct method to find the gradient of the parabola.
M1: Finds the gradient of the normal and sets their normal gradient equal to their asymptote gradient to obtain at least one point on \(C\) where normal is parallel to an asymptote
M1: Finds the equation of either \(l_1\) or \(l_2\)
A1: Correct equation for each normal, \(y + 8 = \dfrac{1}{2}(x - 2)\) and \(y - 8 = -\dfrac{1}{2}(x - 2)\). Ignore labelling.
| Scheme | Marks | AO |
|---|---|---|
| Meet \(H \Rightarrow \dfrac{x^2}{36} - \dfrac{\left(\pm\left(\frac{1}{2}x - 9\right)\right)^2}{9} = 1 \Rightarrow \dfrac{x^2}{36} - \dfrac{\frac{1}{4}x^2 - 9x + 81}{9} = 1 \Rightarrow x = \ldots\) or \(\dfrac{(18 \pm 2y)^2}{36} - \dfrac{y^2}{9} = 1 \Rightarrow \dfrac{81 \pm 18y + y^2}{9} - \dfrac{y^2}{9} = 1 \Rightarrow y = \ldots\) | M1 | 2.1 |
| One correct point of intersection \((10, \pm 4)\) | A1 | 2.2a |
| Area \(OPQ\) is \(\dfrac{1}{2} \times 10 \times (4 - (-4)) = \ldots\) \(-\dfrac{1}{2}\begin{vmatrix}10 & 4 & 0\\ 10 & -4 & 0\end{vmatrix} = -\dfrac{1}{2}\left[-40 - 40\right]\) \(-\dfrac{1}{2}\begin{vmatrix}0 & 0 & 1\\ 10 & 4 & 1\\ 10 & -4 & 1\end{vmatrix} = -\dfrac{1}{2}\left[0 - 0 + \left[(10 \times -4) - (10 \times 4)\right]\right]\) | dM1 | 1.1b |
| \(= 40\) | A1 | 1.1b |
| (4) | ||
| (9 marks) |
Notes
M1: Substitutes for \(x\) or \(y\) into the equation of the hyperbola and solves for their variable.
A1: Achieves one correct coordinate \(x = 10\) and \(y = \pm 4\)
dM1: Dependent on previous method. Correct method for the area of their triangle e.g, \(\dfrac{1}{2} \times\) their \(10 \times\) twice their 4 or equivalent determinant methods.
A1: Area is 40. Correct answer only.