FP1 January 2012 Q3
3. A parabola \(C\) has cartesian equation \(y^2 = 16x\). The point \(P(4t^2,\ 8t)\) is a general point on \(C\).
(a) Write down the coordinates of the focus \(F\) and the equation of the directrix of \(C\). (3)
(b) Show that the equation of the normal to \(C\) at \(P\) is \(y + tx = 8t + 4t^3\). (5)
| Scheme | Marks |
|---|---|
| Focus \((4,\ 0)\) | B1 |
| Directrix \(x + 4 = 0\) \(x + \text{“4”} = 0\) or \(x = -\text{“4”}\) \(x + 4 = 0\) or \(x = -4\) | M1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(y = 4x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-\frac{1}{2}}\) \(y^2 = 16x \Rightarrow 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 16\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t}.\dfrac{\mathrm{d}t}{\mathrm{d}x} = 8.\dfrac{1}{8t}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\,x^{-\frac{1}{2}}\) \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \times \left(\dfrac{1}{\text{their }\frac{\mathrm{d}x}{\mathrm{d}t}}\right)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-\frac{1}{2}}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 16\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8.\dfrac{1}{8t}\) Correct differentiation | A1 |
| At \(P\), gradient of normal \(= -t\) Correct normal gradient with no errors seen. | A1 |
| \(y - 8t = -t(x - 4t^2)\) Applies \(y - 8t = \text{their } m_N(x - 4t^2)\) or \(y = (\text{their } m_N)x + c\) using \(x = 4t^2\) and \(y = 8t\) in an attempt to find \(c\). Their \(m_N\) must be different from their \(m_T\) and must be a function of \(t\). | M1 |
| \(y + tx = 8t + 4t^3\) * cso **given answer** | A1 |
| (5) | |
| (8 marks) |
Notes
Special case – if the correct gradient is quoted could score M0A0A0M1A1