FP1 June 2010 Q8
8. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where c is a positive constant.
The point \(A\) on \(H\) has \(x\)-coordinate \(3c\).
The normal to \(H\) at \(A\) meets \(H\) again at the point \(B\).
| Scheme | Marks |
|---|---|
| \(\dfrac{c}{3}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(y = \dfrac{c^2}{x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2}\), or \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) or \(\dot{x} = c,\ \dot{y} = -\dfrac{c}{t^2}\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) | B1 |
| and at \(A\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{c^2}{(3c)^2} = -\dfrac{1}{9}\) so gradient of normal is 9 | M1 A1 |
| Either \(y - \dfrac{c}{3} = 9(x - 3c)\) or \(y = 9x + k\) and use \(x = 3c,\ y = \dfrac{c}{3}\) | M1 |
| \(\Rightarrow \quad 3y = 27x - 80c\) (*) | A1 |
| (5) |
Notes
(b) B1: Any valid method of differentiation but must get to correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Substitutes values and uses negative reciprocal (needs to follow calculus)
A1: 9 cao (needs to follow calculus)
M1: Finds equation of line through \(A\) with any gradient (other than 0 and \(\infty\))
A1: Correct work throughout – obtaining printed answer.
| Scheme | Marks |
|---|---|
| \(\dfrac{c^2}{x} = \dfrac{27x - 80c}{3}\) | M1 |
| \(3c^2 = 27x^2 - 80cx\) | A1 |
| \((x - 3c)(27x + c) = 0\) so \(x =\) | M1 |
| \(x = -\dfrac{c}{27}, \quad y = -27c\) | A1, A1 |
| (5) | |
| 11 marks |
Alternative (in \(y\))
| Scheme | Marks |
|---|---|
| \(\dfrac{c^2}{y} = \dfrac{3y + 80c}{27}\) | M1 |
| \(27c^2 = 3y^2 + 80cy\) | A1 |
| \((y + 27c)(3y - c) = 0\) so \(y =\) | M1 |
| \(x = -\dfrac{c}{27}, \quad y = -27c\) | A1, A1 |
Alternative (in \(t\))
| Scheme | Marks |
|---|---|
| \(3\dfrac{c}{t} = 27ct - 80c\) | M1 |
| \(3c = 27ct^2 - 80ct\) | A1 |
| \((t - 3)(27t + 1) = 0\) so \(t =\) | M1 |
| \(\left(t = -\dfrac{1}{27}\text{ and so}\right)\) \(x = -\dfrac{c}{27}, \quad y = -27c\) | A1, A1 |
Notes
(c) M1: Obtains equation in one variable (\(x\), \(y\) or \(t\))
A1: Writes as correct three term quadratic (any equivalent form)
M1: Attempts to solve three term quadratic to obtain \(x =\) or \(y =\) or \(t =\)
A1: \(x\) coordinate, A1: \(y\) coordinate. (cao but allow recovery following slips)