FP1 January 2010 Q7
7. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where \(c\) is a constant.
The point \(P\left(ct,\ \dfrac{c}{t}\right)\) is a general point on \(H\).
(a) Show that the tangent to \(H\) at \(P\) has equation \[t^2y + x = 2ct\] (4)
The tangents to \(H\) at the points \(A\) and \(B\) meet at the point \((15c,\ -c)\).
(b) Find, in terms of \(c\), the coordinates of \(A\) and \(B\). (5)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{c^2}{x} \qquad \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{c^2}{(ct)^2} = -\dfrac{1}{t^2}\) without \(x\) or \(y\) | M1 |
| \(y - \dfrac{c}{t} = -\dfrac{1}{t^2}(x - ct) \quad \Rightarrow \quad t^2y + x = 2ct\) (*) | M1 A1cso |
| (4) |
Notes
(a) Use of \(y - y_1 = m(x - x_1)\) where \(m\) is their gradient expression in terms of \(c\) and / or \(t\) only for second M1. Accept \(y = mx + k\) and attempt to find \(k\) for second M1.
Alternatives
| Scheme | Marks |
|---|---|
| (a) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = c\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -ct^{-2}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \div \dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{1}{t^2}\), then as in main scheme. | M1 |
| (a) \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x} = -\dfrac{1}{t^2}\), then as in main scheme. | M1 |
| Scheme | Marks |
|---|---|
| Substitute \((15c,\ -c)\): \(\quad -ct^2 + 15c = 2ct\) | M1 |
| \(t^2 + 2t - 15 = 0\) | A1 |
| \((t + 5)(t - 3) = 0 \quad \Rightarrow \quad t = -5 \quad t = 3\) | M1 A1 |
| Points are \(\left(-5c,\ -\dfrac{c}{5}\right)\) and \(\left(3c,\ \dfrac{c}{3}\right)\) both | A1 |
| (5) | |
| [9] |
Notes
(b) Correct absolute factors for their constant for second M1.
Accept correct use of quadratic formula for second M1.