FP1 June 2014 (R) Q7
7. The parabola \(C\) has cartesian equation \(y^2 = 4ax\), \(a > 0\)
The points \(P(ap^2, 2ap)\) and \(P^{\prime}(ap^2, -2ap)\) lie on \(C\).
(a) Show that an equation of the normal to \(C\) at the point \(P\) is \[y + px = 2ap + ap^3\] (5)
(b) Write down an equation of the normal to \(C\) at the point \(P^{\prime}\). (1)
The normal to \(C\) at \(P\) meets the normal to \(C\) at \(P^{\prime}\) at the point \(Q\).
(c) Find, in terms of \(a\) and \(p\), the coordinates of \(Q\). (2)
Given that \(S\) is the focus of the parabola,
(d) find the area of the quadrilateral \(SPQP^{\prime}\). (3)
| Scheme | Marks |
|---|---|
| \(y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(y^2 = 4ax \Rightarrow 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}p} \cdot \dfrac{\mathrm{d}p}{\mathrm{d}x} = 2a \cdot \dfrac{1}{2ap}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = kx^{-\frac{1}{2}}\) or \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) their \(\dfrac{\mathrm{d}y}{\mathrm{d}p} \times \left(\dfrac{1}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}p}}\right)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2a \cdot \dfrac{1}{2ap}\) Correct differentiation | A1 |
| At \(P\), gradient of normal \(= -p\) Correct normal gradient with no errors seen. | A1 |
| \(y - 2ap = -p(x - ap^2)\) Applies \(y - 2ap = \text{their } m_N(x - ap^2)\) or \(y = (\text{their } m_N)x + c\) using \(x = ap^2\) and \(y = 2ap\) in an attempt to find c. Their \(m_N\) must be different from their \(m_T\) and must be a function of \(p\). | M1 |
| \(y + px = 2ap + ap^3\) * cso **given answer** | A1* |
| (5) |
| Scheme | Marks |
|---|---|
| \(y - px = -2ap - ap^3\) oe | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(y = 0 \Rightarrow x = 2a + ap^2\) M1: \(y = 0\) in either normal or solves simultaneously to find \(x\) A1: \(y = 0\) and correct \(x\) coordinate. | M1A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(S\) is \((a, 0)\) Can be implied below | B1 |
| Area \(SPQP^{\prime} = \dfrac{1}{2} \times (\text{"}2a + ap^2\text{"} - a) \times 2ap \times 2\) Correct method for the area of the quadrilateral. | M1 |
| \(= 2a^2p(1 + p^2)\) Any equivalent form | A1 |
| (3) | |
| (11 marks) |