FP1 January 2011 Q10
10. The point \(P\left(6t,\ \dfrac{6}{t}\right),\ t \neq 0\), lies on the rectangular hyperbola \(H\) with equation \(xy = 36\).
(a) Show that an equation for the tangent to \(H\) at \(P\) is \[y = -\frac{1}{t^2}x + \frac{12}{t}\] (5)
The tangent to \(H\) at the point \(A\) and the tangent to \(H\) at the point \(B\) meet at the point \((-9,\ 12)\).
(b) Find the coordinates of \(A\) and \(B\). (7)
| Scheme | Marks |
|---|---|
| \(xy = 36\) at \(\left(6t,\ \tfrac{6}{t}\right)\). | |
| \(y = \dfrac{36}{x} = 36x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -36x^{-2} = -\dfrac{36}{x^2}\) An attempt at \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) | M1 |
| At \(\left(6t,\ \tfrac{6}{t}\right)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{36}{(6t)^2}\) An attempt at \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). in terms of \(t\) | M1 |
| So, \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) * Must see working to award here | A1 |
| \(\mathbf{T}\!: \ y - \dfrac{6}{t} = -\dfrac{1}{t^2}(x - 6t)\) Applies \(y - \dfrac{6}{t} = \) their \(m_T(x - 6t)\) | M1 |
| \(\mathbf{T}\!: \ y - \dfrac{6}{t} = -\dfrac{1}{t^2}x + \dfrac{6}{t}\) \(\mathbf{T}\!: \ y = -\dfrac{1}{t^2}x + \dfrac{6}{t} + \dfrac{6}{t}\) | |
| \(\mathbf{T}\!: \ y = -\dfrac{1}{t^2}x + \dfrac{12}{t}\) * Correct solution. | A1 cso |
| (5) |
| Scheme | Marks |
|---|---|
| Both \(\mathbf{T}\) meet at \((-9,\ 12)\) gives | |
| \(12 = -\dfrac{1}{t^2}(-9) + \dfrac{12}{t}\) Substituting (−9,12) into \(\mathbf{T}\). | M1 |
| \(12 = \dfrac{9}{t^2} + \dfrac{12}{t} \quad (\times t^2)\) \(12t^2 = 9 + 12t\) | |
| \(12t^2 - 12t - 9 = 0\) \(4t^2 - 4t - 3 = 0\) An attempt to form a “3 term quadratic” | M1 |
| \((2t - 3)(2t + 1) = 0\) An attempt to factorise. | M1 |
| \(t = \tfrac{3}{2},\ -\tfrac{1}{2}\) \(t = \tfrac{3}{2},\ -\tfrac{1}{2}\) | A1 |
| \(t = \tfrac{3}{2} \Rightarrow x = 6\left(\tfrac{3}{2}\right) = 9,\ \ y = \dfrac{6}{\left(\frac{3}{2}\right)} = 4 \Rightarrow (9,\ 4)\) An attempt to substitute either their \(t = \tfrac{3}{2}\) or their \(t = -\tfrac{1}{2}\) into \(x\) and \(y\). | M1 |
| \(t = -\tfrac{1}{2} \Rightarrow x = 6\left(-\tfrac{1}{2}\right) = -3,\) At least one of \((9,\ 4)\) or \((-3,\ -12)\). | A1 |
| \(y = \dfrac{6}{\left(-\frac{1}{2}\right)} = -12 \Rightarrow (-3,\ -12)\) Both \((9,\ 4)\) and \((-3,\ -12)\). | A1 |
| (7) | |
| [12] |