A2 June 2019 Q8
8. The hyperbola \(H\) has equation
\[\frac{x^2}{16} - \frac{y^2}{9} = 1\]The line \(l_1\) is the tangent to \(H\) at the point \(P(4\cosh\theta, 3\sinh\theta)\).
The line \(l_1\) meets the \(x\)-axis at the point \(A\).
The line \(l_2\) is the tangent to \(H\) at the point \((4, 0)\).
The lines \(l_1\) and \(l_2\) meet at the point \(B\) and the midpoint of \(AB\) is the point \(M\).
Let \(S\) be the focus of \(H\) that lies on the positive \(x\)-axis.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 \Rightarrow \dfrac{x}{8} - \dfrac{2yy^{\prime}}{9} = 0 \Rightarrow y^{\prime} = \dfrac{9x}{16y} = \dfrac{36\cosh\theta}{48\sinh\theta}\) or \(x = 4\cosh\theta,\ y = 3\sinh\theta \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3\cosh\theta}{4\sinh\theta}\) | M1 | 3.1a |
| \(y - 3\sinh\theta = \dfrac{3\cosh\theta}{4\sinh\theta}\left(x - 4\cosh\theta\right)\) | M1 | 3.1a |
| \(y = 0 \Rightarrow x = \dfrac{4}{\cosh\theta}\) | A1 | 2.2a |
| line \(l_2\) has equation \(x = 4\) | B1 | 2.2a |
| \(x = 4 \Rightarrow y - 3\sinh\theta = \dfrac{3\cosh\theta}{4\sinh\theta}\left(4 - 4\cosh\theta\right)\) | M1 | 2.1 |
| \(y = \dfrac{3\cosh\theta - 3}{\sinh\theta}\) | A1 | 2.2a |
| \(M\) is \(\left(\dfrac{1}{2}\left(4 + \dfrac{4}{\cosh\theta}\right), \dfrac{1}{2}\left(\dfrac{3\cosh\theta - 3}{\sinh\theta}\right)\right)\) | M1 | 1.1b |
| \(\begin{aligned}&x = 2 + \dfrac{2}{\cosh\theta} \Rightarrow \cosh\theta = \dfrac{2}{x - 2}\\[6pt] &\Rightarrow y^2 = \dfrac{9\left(\cosh\theta - 1\right)^2}{4\sinh^2\theta} = \dfrac{9\left(\dfrac{2}{x - 2} - 1\right)^2}{4\left(\left(\dfrac{2}{x - 2}\right)^2 - 1\right)}\end{aligned}\) | M1 | 3.1a |
| \(= \dfrac{9\left(\dfrac{2}{x - 2} - 1\right)^2}{4\left(\dfrac{2}{x - 2} - 1\right)\left(\dfrac{2}{x - 2} + 1\right)} = \dfrac{9\left(\dfrac{2}{x - 2} - 1\right)}{4\left(\dfrac{2}{x - 2} + 1\right)} = \dfrac{9(4 - x)}{4x}\ *\) | A1* | 1.1b |
| Alternative for M1A1: \(y^2 = \dfrac{9\left(\cosh\theta - 1\right)^2}{4\sinh^2\theta} = \dfrac{9\left(\cosh\theta - 1\right)^2}{4\left(\cosh\theta - 1\right)\left(\cosh\theta + 1\right)} = \dfrac{9\left(\cosh\theta - 1\right)}{4\left(\cosh\theta + 1\right)}\) \(\dfrac{9(4 - x)}{4x} = \dfrac{9\left(4 - 2 - \dfrac{2}{\cosh\theta}\right)}{8 + \dfrac{8}{\cosh\theta}} = \dfrac{9\left(\cosh\theta - 1\right)}{4\left(\cosh\theta + 1\right)} \Rightarrow y^2 = \dfrac{9(4 - x)}{4x}\) | ||
| \(p = 2\) or \(q = 4\) | M1 | 3.1a |
| \(p = 2\) and \(q = 4\) | A1 | 1.1b |
| (11) |
Notes
M1: Attempts to solve the problem by using differentiation to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). Allow this mark for \(\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 \Rightarrow \alpha x - \beta yy^{\prime} = 0 \Rightarrow y^{\prime} = \ldots\) or an attempt to differentiate \(x\) and \(y\) wrt \(\theta\) and then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}\theta} \div \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \ldots\)
M1: Correct straight line method using the coordinates of \(P\) and their gradient in terms of \(\theta\)
Allow the results for the first 2 M marks to be “quoted”, but any statements must be correct to score the marks.
A1: Uses \(y = 0\) to deduce the correct coordinates (or value of \(x\)) for the point \(A\). Allow in any form, simplified or unsimplified (e.g. unsimplified: \(x = 4\cosh\theta - 4\tanh\theta\sinh\theta\))
B1: Deduces that the equation of \(l_2\) is \(x = 4\) (may be implied by \(x = 4\) used to find \(y\) coordinate of \(B\))
M1: Realises that \(x = 4\) is all that is needed for the second line and substitutes this into the first line in order to find the point \(B\)
A1: Deduces the correct coordinates or \(y\) value for \(B\)
(e.g. unsimplified \(y = \dfrac{3}{\tanh\theta} - \dfrac{3\cosh\theta}{\tanh\theta} + 3\sinh\theta\))
M1: Uses a correct method for the midpoint of \(AB\) (coordinates must be the right way round). This may be seen as the coordinates written separately e.g. \(x = \ldots\), \(y = \ldots\)
M1: Having found the midpoint, identifies a correct strategy that will enable a Cartesian equation to be found. E.g. find \(\cosh\theta\) in terms of \(x\) and substitutes into \(y\) or \(y^2\) to obtain an equation in terms of \(y\) and \(x\) only. Mark positively here, so allow the mark if the candidate makes progress in eliminating \(\theta\) even if there are slips in the working.
A1*: Obtains the printed answer with no errors
Alternative for the previous 2 marks: Substitutes the coordinates of their midpoint into both sides of the given equation in an attempt to show they are equal. Again mark positively but having made the substitution, some progress needs to be made in showing that both sides are equal. For this method there must be a minimal conclusion for the A1 e.g. tick, hence true etc.
Note that these 2 marks can also be attempted by expressing the midpoint in terms of exponentials – if you are in doubt whether to award marks seek advice from your Team Leader.
M1: For \(p = 2\) or \(q = 4\)
A1: For \(p = 2\) and \(q = 4\)
| Scheme | Marks | AO |
|---|---|---|
| \(b^2 = a^2\left(e^2 - 1\right) \Rightarrow 9 = 16\left(e^2 - 1\right) \Rightarrow e = \dfrac{5}{4}\) Focus is at \(x = ae = 4 \times \dfrac{5}{4} = 5\) | M1 | 1.1b |
| \(d \gt \text{“}5\text{”} - 4 = \ldots\) | M1 | 3.1a |
| \(d \gt 1\ *\) | A1* | 1.1b |
| (3) | ||
| (14 marks) |
Notes
M1: A complete method for finding the \(x\) coordinate of the focus using a correct eccentricity formula to find a value for e and then calculating \(4e\)
M1: Completes the problem by subtracting 4 from the \(x\) coordinate of the focus
A1*: Correct answer
If you come across correct attempts using Pythagoras to prove the result send to review.