FP1 January 2013 Q7
7. The rectangular hyperbola, \(H\), has cartesian equation \(xy = 25\)
The point \(P\left(5p,\ \dfrac{5}{p}\right)\), and the point \(Q\left(5q,\ \dfrac{5}{q}\right)\), where \(p, q \neq 0\), \(p \neq q\), are points on the rectangular hyperbola \(H\).
The tangents at \(P\) and \(Q\) meet at the point \(N\).
Given \(p + q \neq 0\),
The line joining \(N\) to the origin is perpendicular to the line \(PQ\).
| Scheme | Marks |
|---|---|
| \(y = \dfrac{25}{x}\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -25x^{-2}\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{25}{(5p)^2} = -\dfrac{1}{p^2}\) | A1 |
| \(y - \dfrac{5}{p} = -\dfrac{1}{p^2}(x - 5p) \Rightarrow p^2y + x = 10p\quad (*)\) | M1 A1 |
| (4) |
Notes
Alternatives for first M1 A1 in part (a)
| Scheme | Marks |
|---|---|
| \(x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) | M1 |
| So at \(P\) gradient \(= \dfrac{-\frac{5}{p}}{5p} = -\dfrac{1}{p^2}\) | A1 |
| Or \(x = 5t,\ y = \dfrac{5}{t} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = 5,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = -\dfrac{5}{t^2}\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \) | M1 |
| \(\dfrac{-\frac{5}{t^2}}{5} = -\dfrac{1}{t^2}\) so at \(P\) gradient \(= -\dfrac{1}{p^2}\) | A1 |
(a) First M for attempt at explicit, implicit or parametric differentiation not using \(p\) or \(q\) as an initial parameter, first A for \(\dfrac{-1}{p^2}\) or equivalent. Quoting gradient award first M0A0. Second M for using \(y - y_1 = m(x - x_1)\) and attempt to substitute or \(y = mx + c\) and attempt to find c; gradient in terms of \(p\) only and using \(\left(5p, \dfrac{5}{p}\right)\), second A for correct solution only.
| Scheme | Marks |
|---|---|
| \(q^2y + x = 10q\) only | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \((p^2 - q^2)y = 10(p - q)\) so \(y = \dfrac{10(p - q)}{(p^2 - q^2)} = \dfrac{10}{p + q}\) | M1 A1cso |
| \(x = 10p - p^2\dfrac{10}{p + q} = \dfrac{10pq}{p + q}\) | M1 A1 cso |
| (4) |
Notes
(c) First M for eliminating \(x\) and reaching \(y = \mathrm{f}(p, q)\), second M for eliminating \(y\) and reaching \(x = \mathrm{f}(p, q)\), both As for given answers. Minimum amount of working given in the main scheme above for 4/4, but do not award accuracy if any errors are made.
| Scheme | Marks |
|---|---|
| Line \(PQ\) has gradient \(\dfrac{\frac{5}{p} - \frac{5}{q}}{5p - 5q}\ \left(= -\dfrac{1}{pq}\right)\) | M1 A1 |
| \(ON\) has gradient \(\dfrac{\frac{10}{p + q}}{\frac{10pq}{p + q}}\ \left(= \dfrac{1}{pq}\right)\) or \(\dfrac{-1}{\frac{-1}{pq}}\ (= pq)\) could be as unsimplified equivalents seen anywhere | B1 |
| As these lines are perpendicular \(\dfrac{1}{pq} \times -\dfrac{1}{pq} = -1\) so \(p^2q^2 = 1\) OR for \(ON\) \(y - y_1 = m(x - x_1)\) with gradient (equivalent to) \(pq\) and sub in points \(O\) AND \(N\) to give \(p^2q^2 = 1\) OR for \(PQ\) \(y - y_1 = m(x - x_1)\) with gradient (equivalent to) \(-pq\) and sub in points \(P\) AND \(Q\) to give \(p^2q^2 = 1\). NB \(-pq\) used as gradient of \(PQ\) implies first M1A1 | M1 A1 |
| (5) | |
| [14] |
Notes
(d) First M for use of \(\dfrac{y_2 - y_1}{x_2 - x_1}\) and substituting, first A for \(\dfrac{-1}{pq}\) or unsimplified equivalent .
Second M for their product of gradients\(= -1\) (or equating equivalent gradients of \(ON\) or equating equivalent gradients of \(PQ\)), second A for correct answer only.