FP1 June 2014 Q6
6. The rectangular hyperbola \(H\) has cartesian equation \(xy = c^2\).
The point \(P\left(ct, \dfrac{c}{t}\right)\), \(t > 0\), is a general point on \(H\).
(a) Show that an equation of the tangent to \(H\) at the point \(P\) is \[t^2y + x = 2ct\] (4)
An equation of the normal to \(H\) at the point \(P\) is \(t^3x - ty = ct^4 - c\)
Given that the normal to \(H\) at \(P\) meets the \(x\)-axis at the point \(A\) and the tangent to \(H\) at \(P\) meets the \(x\)-axis at the point \(B\),
(b) find, in terms of \(c\) and \(t\), the coordinates of \(A\) and the coordinates of \(B\). (2)
Given that \(c = 4\),
(c) find, in terms of \(t\), the area of the triangle \(APB\). Give your answer in its simplest form. (3)
| Scheme | Marks |
|---|---|
| \(xy = c^2\) at \(\left(ct, \tfrac{c}{t}\right)\). | |
| \(y = \dfrac{c^2}{x} = c^2x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2} = -\dfrac{c^2}{x^2}\) or \(xy = c^2 \Rightarrow x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \cdot \dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{c}{t^2} \cdot \dfrac{1}{c}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = kx^{-2}\) or correct use of product rule. The sum of two terms, one of which is correct and rhs = 0 or 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 \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-c}{t^2} \cdot \dfrac{1}{c}\) or equivalent expressions Correct differentiation | A1 |
| \(y - \dfrac{c}{t} = -\dfrac{1}{t^2}(x - ct) \qquad (\times t^2)\) \(y - \dfrac{c}{t} = \text{their } m_T(x - ct)\) 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 \(t\). | dM1 |
| \(t^2y + x = 2ct\) (Allow \(x + t^2y = 2ct\)) Correct solution only. | 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 |
|---|---|
| \(y = 0 \Rightarrow x = \dfrac{ct^4 - c}{t^3} \Rightarrow A\left(\dfrac{ct^4 - c}{t^3}, 0\right)\) \(\dfrac{ct^4 - c}{t^3}\) or equivalent form | B1 |
| \(y = 0 \Rightarrow x = 2ct \Rightarrow B(2ct, 0)\). \(2ct\) | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(AB = \text{"}2ct\text{"} - \text{"}\dfrac{ct^4 - c}{t^3}\text{"}\) or \(PA = ct^{-3}\sqrt{t^4 + 1}\) and \(PB = ct^{-1}\sqrt{t^4 + 1}\) Attempt to subtract their \(x\)-coordinates either way around. | M1 |
| Area \(APB = \dfrac{1}{2} \times \text{their } AB \times \dfrac{c}{t}\) Valid complete method for the area of the triangle in terms of \(t\) or \(c\) and \(t\). | M1 |
| \(= \dfrac{1}{2}\left(2ct - \dfrac{ct^4 - c}{t^3}\right)\dfrac{c}{t} = \dfrac{c^2(t^4 + 1)}{2t^4}\) | |
| \(= 8\left(1 + \dfrac{1}{t^4}\right)\) or \(\dfrac{8(t^4 + 1)}{t^4}\) or \(\dfrac{8t^4 + 8}{t^4}\) or \(8 + \dfrac{8}{t^4}\) Use of \(c = 4\) and completes to one of the given forms oe simplest form. Final answer should be positive for A mark. | A1 |
| (3) | |
| (9 marks) |