FP1 June 2012 Q8
8. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where \(c\) is a positive constant.
The point \(P\left(ct,\ \dfrac{c}{t}\right)\), \(t \neq 0\), is a general point on \(H\).
(a) Show that an equation for the tangent to \(H\) at \(P\) is \[x + t^2y = 2ct\] (4)
The tangent to \(H\) at the point \(P\) meets the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\).
Given that the area of the triangle \(OAB\), where \(O\) is the origin, is 36,
(b) find the exact value of \(c\), expressing your answer in the form \(k\sqrt{2}\), where \(k\) is an integer. (4)
| Scheme | Marks |
|---|---|
| \(xy = c^2\) at \(\left(ct,\ \tfrac{c}{t}\right)\). | |
| \(y = \dfrac{c^2}{x} = c^2x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2} = -\dfrac{c^2}{x^2}\) \(xy = c^2 \Rightarrow x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}.\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{c}{t^2}.\dfrac{1}{c}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\,x^{-2}\) Correct use of product rule. The sum of two terms, one of which is correct and rhs = 0 their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \times \left(\dfrac{1}{\text{their }\frac{\mathrm{d}x}{\mathrm{d}t}}\right)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2}\) or \(x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-c}{t^2}.\dfrac{1}{c}\) or equivalent expressions Correct differentiation | A1 |
| So, \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) \(-\dfrac{1}{t^2}\) | |
| \(y - \dfrac{c}{t} = -\dfrac{1}{t^2}(x - ct)\quad (\times t^2)\) \(y - \dfrac{c}{t} = \text{their } m_T(x - ct)\) or \(y = mx + c\) with their \(m_T\) and \(\left(ct, \dfrac{c}{t}\right)\) in an attempt to find ‘\(c\)’. Their \(m_T\) must have come from calculus and should be a function of \(t\) or \(c\) or both \(c\) and \(t\). | M1 |
| \(x + t^2y = 2ct\) (Allow \(t^2y + x = 2ct\)) Correct solution. | A1 * |
| [4] |
Notes
(a) Candidates who derive \(x + t^2y = 2ct\), by stating that \(m_T = -\dfrac{1}{t^2}\), with no justification score no marks in (a).
| Scheme | Marks |
|---|---|
| \(y = 0 \Rightarrow x = 2ct \Rightarrow A(2ct,\ 0)\). \(x = 2ct\), seen or implied. | B1 |
| \(x = 0 \Rightarrow y = \dfrac{2ct}{t^2} \Rightarrow B\left(0,\ \dfrac{2c}{t}\right)\). \(y = \dfrac{2ct}{t^2}\) or \(\dfrac{2c}{t}\), seen or implied. | B1 |
| Area \(OAB = 36 \Rightarrow \dfrac{1}{2}(2ct)\left(\dfrac{2c}{t}\right) = 36\) Applies \(\dfrac{1}{2}(\text{their } x)(\text{their } y) = 36\) where \(x\) and \(y\) are functions of \(c\) or \(t\) or both (not \(x\) or \(y\)) and some attempt was made to substitute both \(x = 0\) and \(y = 0\) in the tangent to find \(A\) and \(B\). | M1 |
| \(\Rightarrow 2c^2 = 36 \Rightarrow c^2 = 18 \Rightarrow c = 3\sqrt{2}\) \(c = 3\sqrt{2}\) Do not allow \(c = \pm 3\sqrt{2}\) | A1 |
| [4] | |
| 8 marks |
Notes
Do not allow the \(x\) and \(y\) coordinates of P to be used for the dimensions of the triangle.