FP1 June 2013 Q4
4. The rectangular hyperbola \(H\) has Cartesian equation \(xy = 4\)
The point \(P\left(2t,\ \dfrac{2}{t}\right)\) lies on \(H\), where \(t \neq 0\)
(a) Show that an equation of the normal to \(H\) at the point \(P\) is \[ty - t^3x = 2 - 2t^4\] (5)
The normal to \(H\) at the point where \(t = -\dfrac{1}{2}\) meets \(H\) again at the point \(Q\).
(b) Find the coordinates of the point \(Q\). (4)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{4}{x} = 4x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -4x^{-2} = -\dfrac{4}{x^2}\) \(xy = 4 \Rightarrow x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}.\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{2}{t^2}.\dfrac{1}{2}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\,x^{-2}\) Use of the product rule. The sum of two terms including \(\mathrm{d}y/\mathrm{d}x\), one of which is correct. their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \times \left(\dfrac{1}{\text{their }\frac{\mathrm{d}x}{\mathrm{d}t}}\right)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4x^{-2}\) or \(x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = 0\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2}{t^2}.\dfrac{1}{2}\) or equivalent expressions Correct derivative \(-4x^{-2}\), \(-\dfrac{y}{x}\) or \(\dfrac{-1}{t^2}\) | A1 |
| So, \(m_N = t^2\) Perpendicular gradient rule \(m_N m_T = -1\) | M1 |
| \(y - \dfrac{2}{t} = t^2(x - 2t)\) \(y - \dfrac{2}{t} = \text{their } m_N(x - 2t)\) or \(y = mx + c\) with their \(m_N\) and \(\left(2t, \dfrac{2}{t}\right)\) in an attempt to find ‘\(c\)’. Their gradient of the normal must be different from their gradient of the tangent and have come from calculus and should be a function of \(t\). | M1 |
| \(ty - t^3x = 2 - 2t^4\) * | A1* cso |
| (5) |
| Scheme | Marks |
|---|---|
| \(t = -\dfrac{1}{2} \Rightarrow -\dfrac{1}{2}y - \left(-\dfrac{1}{2}\right)^3x = 2 - 2\left(-\dfrac{1}{2}\right)^4\) Substitutes the given value of \(t\) into the normal | M1 |
| \(4y - x + 15 = 0\) | |
| \(y = \dfrac{4}{x} \Rightarrow x^2 - 15x - 16 = 0\) or \(\left(2t, \dfrac{2}{t}\right) \to \dfrac{8}{t} - 2t + 15 = 0 \Rightarrow 2t^2 - 15t - 8 = 0\) or \(x = \dfrac{4}{y} \Rightarrow 4y^2 + 15y - 4 = 0\). Substitutes to give a quadratic | M1 |
| \((x + 1)(x - 16) = 0 \Rightarrow x = \) or \((2t + 1)(t - 8) = 0 \Rightarrow t = \) or \((4y - 1)(y + 4) = 0 \Rightarrow y = \) Solves their 3TQ | M1 |
| \((P: x = -1, y = -4)\ (Q:)\ x = 16,\ y = \dfrac{1}{4}\) Correct values for \(x\) and \(y\) | A1 |
| (4) | |
| Total 9 |