AS June 2018 Q5
5. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where \(c\) is a non-zero constant.
The point \(P\left(cp, \dfrac{c}{p}\right)\), where \(p \neq 0\), lies on \(H\).
The normal to \(H\) at the point \(P\) meets \(H\) again at the point \(Q\).
| Scheme | Marks | AO |
|---|---|---|
| \(H : xy = c^2,\ c \neq 0;\ P\left(cp, \dfrac{c}{p}\right),\ p \neq 0\), lies on \(H\) | ||
| Either \(y = \dfrac{c^2}{x} = c^2x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2}\) or \(-\dfrac{c^2}{x^2}\) or \(xy = c^2 \Rightarrow x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) or \(x = ct,\ y = \dfrac{c}{t} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}.\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\left(\dfrac{c}{t^2}\right)\left(\dfrac{1}{c}\right)\) and so, at \(P\left(cp, \dfrac{c}{p}\right)\), \(m_T = -\dfrac{1}{p^2}\) | M1 | 2.1 |
| So, \(m_N = p^2\) | A1 | 2.2a |
| \(y - \dfrac{c}{p} = \text{“}p^2\text{”}(x - cp)\) or \(\dfrac{c}{p} = \text{“}p^2\text{”}(cp) + b \Rightarrow y = \text{“}p^2\text{”}x + \text{their } b\) | M1 | 1.1b |
| correct algebra leading to \(p^3x - py + c(1 - p^4) = 0\ *\) | A1* | 2.1 |
| (4) |
Notes
M1: Starts the process of establishing the gradient of the normal by differentiating \(xy = c^2\)
- to give \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm kx^{-2};\ k \neq 0\), or
- by the product rule to give \(\pm x\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm y\), or
- by parametric differentiation to give \(\left(\text{their } \dfrac{\mathrm{d}y}{\mathrm{d}t}\right) \times \dfrac{1}{\left(\text{their } \dfrac{\mathrm{d}x}{\mathrm{d}t}\right)}\), condoning \(t \equiv p\)
and attempt to use \(P\left(cp, \dfrac{c}{p}\right)\) to write down the gradient of the tangent to the curve in terms of \(p\)
A1: Deduces the correct normal gradient \(p^2\) from their tangent gradient which is found using calculus
M1: Correct straight line method for an equation of a normal where \(m_N\ (\neq m_T)\) is found by using calculus. Note: \(m_N\) must be a function of \(p\) for this mark
A1*: Obtains \(p^3x - py + c(1 - p^4) = 0\), by correct solution only
| Scheme | Marks | AO |
|---|---|---|
| \(y = \dfrac{c^2}{x} \Rightarrow p^3x - p\dfrac{c^2}{x} + c(1 - p^4) = 0\) or \(x = \dfrac{c^2}{y} \Rightarrow p^3\dfrac{c^2}{y} - py + c(1 - p^4) = 0\) | M1 | 3.1a |
| \(p^3x^2 + c(1 - p^4)x - c^2p = 0\) or \(py^2 - c(1 - p^4)y - c^2p^3 = 0\) | A1 | 1.1b |
| \((x - cp)(p^3x + c) = 0 \Rightarrow x = \ldots\) or \(\left(y - \dfrac{c}{p}\right)\left(yp + cp^4\right) = 0 \Rightarrow y = \ldots\) | M1 | 3.1a |
| \(x = -\dfrac{c}{p^3}\) and \(y = -cp^3\) or \(\{Q\}\left(-\dfrac{c}{p^3}, -cp^3\right)\) | A1 | 1.1b |
| Midpoint is \(\left(\dfrac{1}{2}\left(cp - \dfrac{c}{p^3}\right), \dfrac{1}{2}\left(\dfrac{c}{p} - cp^3\right)\right)\) | M1 A1 | 1.1b 1.1b |
| (6) | ||
| (10 marks) |
Notes
M1: Substitutes \(y = \dfrac{c^2}{x}\) or \(x = \dfrac{c^2}{y}\) into the printed equation to obtain an equation in \(x\), \(c\) and \(p\) only or in \(y\), \(c\) and \(p\) only
A1: Obtains a 3TQ equation in \(x\) or a 3TQ equation in \(y\)
Note: E.g. \(p^3x^2 + cx - cp^4x = c^2p\) or \(py^2 = cy - cp^4y + c^2p^3\) are acceptable for the 1st A mark
M1: Recognises that one solution of the quadratic equation is already known and uses a correct factorisation method of solving a 3TQ to give either \(x = \ldots\) or \(y = \ldots\)
Alternatively applies a correct quadratic formula method for solving a 3TQ
A1: Correct coordinates for \(Q\), which can be simplified or un-simplified
Allow \(x = -\dfrac{c}{p^3}\) and \(y = -cp^3\)
M1: Uses \(\left(cp, \dfrac{c}{p}\right)\) and their \((x_Q, y_Q)\) and applies \(\left(\dfrac{cp + \text{their } x_Q}{2}, \dfrac{\frac{c}{p} + \text{their } y_Q}{2}\right)\) to give \((x_M, y_M)\), where \(x_M\) and \(y_M\) are both in terms of \(c\) and \(p\) only
A1: Correct coordinates \(\left(\dfrac{1}{2}\left(cp - \dfrac{c}{p^3}\right), \dfrac{1}{2}\left(\dfrac{c}{p} - cp^3\right)\right)\). Condone \(\left(\dfrac{cp - \frac{c}{p^3}}{2}, \dfrac{\frac{c}{p} - cp^3}{2}\right)\)
Note: Condone \(x = \dfrac{1}{2}\left(cp - \dfrac{c}{p^3}\right)\) and \(y = \dfrac{1}{2}\left(\dfrac{c}{p} - cp^3\right)\) for the final A mark
Note: You can apply isw after correctly stated coordinates for the midpoint of \(P\) and \(Q\)
(b) Alt 1
| Scheme | Marks | AO |
|---|---|---|
| Let \(Q\) be \(\left(cq, \dfrac{c}{q}\right)\), so \(p^3cq - p\dfrac{c}{q} + c(1 - p^4) = 0\) | M1 | 3.1a |
| \(p^3cq^2 - pc + c(1 - p^4)q = 0 \Rightarrow p^3q^2 + (1 - p^4)q - p = 0\) | A1 | 1.1b |
| \((q - p)(p^3q + 1) = 0 \Rightarrow q = \ldots\) | M1 | 3.1a |
| \(\{Q\}\left(-\dfrac{c}{p^3}, -cp^3\right)\) or \(x = -\dfrac{c}{p^3}\) and \(y = -cp^3\) | A1 | 1.1b |
Alt 1 (for the first 4 marks)
M1: Substitutes \(x = cq\) and \(y = \dfrac{c}{q}\) into the printed equation to obtain an equation in only \(p\), \(c\) and \(q\)
A1: Eliminates \(c\) and obtains a correct quadratic equation in \(q\)
Note: E.g. \(p^3q^2 + q - p^4q = p\) is acceptable for the 1st A mark
M1: Recognises that one solution of the quadratic equation is already known and uses a correct factorisation method of solving a 3TQ to give \(q = \ldots\)
Alternatively applies a correct quadratic formula method for solving a 3TQ in \(q\)
A1: Correct coordinates for \(Q\), which can be simplified or un-simplified
Allow \(x = -\dfrac{c}{p^3}\) and \(y = -cp^3\)