FP1 June 2015 Q5
5. The rectangular hyperbola \(H\) has equation \(xy = 9\)
The point \(A\) on \(H\) has coordinates \(\left(6, \dfrac{3}{2}\right)\).
The normal at \(A\) meets \(H\) again at the point \(B\).
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{9}{x^2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) | M1 |
| so gradient at \(x = 6\) or \(t = 2\) is \(-\dfrac{9}{36}\) or \(-\dfrac{\frac{3}{2}}{6}\) or \(-\dfrac{1}{4}\) o.e. | A1 |
| Gradient of normal is \(-\dfrac{1}{m}\ \ (= 4)\) | M1 |
| Equation of normal is \(y - \tfrac{3}{2} = 4(x - 6)\) | dM1 |
| So \(2y - 8x + 45 = 0\) **given answer** | A1 * |
| (5) |
Notes
M1: Differentiates to obtain \(\dfrac{k}{x^2}\) and substitutes \(x = 6\)
or uses implicit differentiation \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) and substitutes \(x\) and \(y\)
or uses parametric differentiation \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) and substitutes \(t = 2\)
A1: For grad of tangent – accept any equivalent i.e. \(-0.25\) etc
M1: Uses negative reciprocal of their gradient.
dM1: \(y - y_1 = m(x - x_1)\) with \(\left(6, \dfrac{3}{2}\right)\) or \(y = mx + c\) and sub \(\left(6, \dfrac{3}{2}\right)\) to find \(c=\).
A1: cso: Correct answer with no errors seen in the solution.
| Scheme | Marks |
|---|---|
| \(\dfrac{18}{x} - 8x + 45 = 0\) or \(2y - \dfrac{72}{y} + 45 = 0\) or \(x(4x - 22.5) = 9\) or \(y\left(\dfrac{y}{4} + \dfrac{45}{8}\right) = 9\) o.e. | M1 |
| \(8x^2 - 45x - 18 = 0\) or \(2y^2 + 45y - 72 = 0\) | |
| So \(x = -\dfrac{3}{8}\) or \(y = -24\) | A1 |
| Finds other ordinate: \(\left(-\dfrac{3}{8}, -24\right)\) | M1 A1 |
| ALT | |
| Sub \(\left(3t, \dfrac{3}{t}\right)\) in \(2y - 8x + 45 = 0 \Rightarrow t = -\dfrac{1}{8}\) | M1A1 |
| Sub \(t = -\dfrac{1}{8}\) in \(\left(3t, \dfrac{3}{t}\right) \Rightarrow \left(-\dfrac{3}{8}, -24\right)\) | M1A1 |
| (4) | |
| (9 marks) |
Notes
M1: Obtains equation in one variable, \(x\) or \(y\)
A1: Correct value of \(x\) or correct value of \(y\)
M1: Finds second coordinate using \(xy = 9\) or solving second quadratic or equation of the normal
A1: Correct coordinates that can be written as \(x = \ldots, y = \ldots\)