FP3 June 2015 Q6
6. The hyperbola \(H\) is given by the equation \(x^2 - y^2 = 1\)
The tangent at \(P\) meets the asymptotes of \(H\) at the points \(Q\) and \(R\).
| Scheme | Marks |
|---|---|
| \(y = x,\ y = -x\) | B1 |
| (1) |
Notes
B1: Both required. Accept \(y = \pm x\) and \(x = \pm y\)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\cosh t}{\sinh t}\) | B1 Note M1 on ePEN |
| \(y - \sinh t = \dfrac{\cosh t}{\sinh t}(x - \cosh t)\) | M1 |
| \(y\sinh t = x\cosh t - (\cosh^2 t - \sinh^2 t)\) | |
| \(y\sinh t = x\cosh t - 1\) * | A1* cso |
| (3) |
Notes
B1: Correct gradient
M1: Correct straight line method. For \(y = mx + c\) method, \(c\) must be found
A1* cso: Obtains the printed answer with at least one intermediate step.
| Scheme | Marks |
|---|---|
| \(y = x \Rightarrow x = \dfrac{1}{\cosh t - \sinh t},\ y = \dfrac{1}{\cosh t - \sinh t}\) \(y = -x \Rightarrow x = \dfrac{1}{\cosh t + \sinh t},\ y = \dfrac{-1}{\cosh t + \sinh t}\) | B1 |
| \(X = \dfrac{1}{2}\left(\dfrac{1}{\cosh t - \sinh t} + \dfrac{1}{\cosh t + \sinh t}\right)\) or \(Y = \dfrac{1}{2}\left(\dfrac{1}{\cosh t - \sinh t} + \dfrac{-1}{\cosh t + \sinh t}\right)\) | M1 |
| \(X = \dfrac{1}{2}\left(\dfrac{\cosh t + \sinh t + \cosh t - \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \cosh t\) \(Y = \dfrac{1}{2}\left(\dfrac{\cosh t + \sinh t - \cosh t + \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \sinh t\) | A1cso |
| (3) |
Notes
B1: All four values correct. May be in exponential form e.g. \(\left(e^t,\ e^t\right)\) and \(\left(e^{-t},\ -e^{-t}\right)\)
M1: Correct attempt at \(X\) or \(Y\). May be in exponential form e.g. \(\left(\dfrac{e^t + e^{-t}}{2},\ \dfrac{e^t - e^{-t}}{2}\right)\)
A1cso: Obtains \(X = \cosh t\) and \(Y = \sinh t\). May be shown using exponentials as above.
| Scheme | Marks |
|---|---|
| \(A = \dfrac{1}{2}\sqrt{\dfrac{2}{(\cosh t - \sinh t)^2}}\cdot\sqrt{\dfrac{2}{(\cosh t + \sinh t)^2}}\) Or e.g. \(\dfrac{1}{2}\sqrt{2e^{2t}}\sqrt{2e^{-2t}}\) | M1 |
| \(= \dfrac{1}{\cosh^2 t - \sinh^2 t} = 1\) | A1 |
| So area is independent of \(t\) | A1ft |
| (3) | |
| (10 marks) |
Notes
M1: Correct triangle area method
A1: Obtains an area of 1
A1ft: Concludes independence of \(t\) having obtained a constant area. Conclusion must include the word independent (or not dependent) (but not e.g. just QED)
Alternative area method
If \(A\left(\frac{1}{\cosh t}, 0\right)\) is the intersection of \(QR\) with the \(x\)-axis
Area OAR + Area OAQ \(= \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \dfrac{1}{\cosh t - \sinh t} + \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \dfrac{1}{\cosh t + \sinh t}\)
\(= \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \left(\dfrac{1}{\cosh t - \sinh t} + \dfrac{1}{\cosh t + \sinh t}\right) = \dfrac{1}{2\cosh t} \times 2\cosh t = 1\)