FP1 June 2016 Q5
5. Points \(P(ap^2, 2ap)\) and \(Q(aq^2, 2aq)\), where \(p^2 \neq q^2\), lie on the parabola \(y^2 = 4ax\).
Given that this chord passes through the focus of the parabola,
| Scheme | Marks |
|---|---|
| Gradient \(= \dfrac{2ap - 2aq}{ap^2 - aq^2}\) seen | B1 |
| \(\left(\dfrac{2ap - 2aq}{ap^2 - aq^2} = \dfrac{2a(p - q)}{a(p - q)(p + q)}\right) = \dfrac{2}{p + q}\) or seen in an equation | B1 |
| uses \(y - y_1 = m(x - x_1)\) to give \((y - 2aq) = \text{"}m\text{"}(x - aq^2)\) or equivalent with \(p\) or uses \(y = mx + c\) to give \(y = \text{"}m\text{"}x + c\) and substitute a point to find \(c\) or uses \(\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}\) to give \(\dfrac{(y - 2aq)}{2ap - 2aq} = \dfrac{(x - aq^2)}{ap^2 - aq^2}\) or equivalent with \(p\) | M1 |
| so \((y - 2aq) = \dfrac{2}{p + q}(x - aq^2)\) or \((y - 2ap) = \dfrac{2}{p + q}(x - ap^2)\) or \(y = \dfrac{2}{p + q}x + \dfrac{2apq}{p + q}\) or \(\dfrac{(y - 2aq)}{2a} = \dfrac{(x - aq^2)}{a(p + q)}\) | A1 |
| See \(2aq^2\) or \(2ap^2\) term appear and disappear to give \(y(p + q) = 2x + 2apq\) * | A1 cso |
| (5) |
Notes
B1: Correct statement for gradient (isw) B1: \(\dfrac{2}{p + q}\) - can be seen later in the solution.
M1: use of a correct formula for a line equation through \(P\) or \(Q\) with their gradient. Must be finding a chord, not a tangent or a normal.
A1: for a correct line equation with simplified gradient in any equivalent form
A1: cso (as given answer)
| Scheme | Marks |
|---|---|
| Substitute \((a, 0)\) into line equation, to give \(0 = 2a + 2apq\) so \(pq = -1\) | B1 |
| (1) |
Notes
B1: For using \((a, 0)\) to show that \(pq = -1\)
| 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 \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}p} \times \dfrac{\mathrm{d}p}{\mathrm{d}x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2a \times \dfrac{1}{2ap}\) | M1 |
| So at \(P\) tangent gradient \(= \dfrac{1}{p}\) | A1 |
| (2) |
Notes
M1: Use calculus to find an expression for \(\mathrm{d}y/\mathrm{d}x\) and substitute coordinates of \(P\). They may use chord gradient and let \(p\) tend to \(q\).
| Scheme | Marks |
|---|---|
| At \(Q\) tangent gradient \(= \dfrac{1}{q}\) | B1 |
| \(\dfrac{1}{p} \times \dfrac{1}{q} = \dfrac{1}{pq} = \dfrac{1}{-1} = -1\) with at least one intermediate step, the tangents are perpendicular or at right angles | B1cso |
| (2) | |
| (10 marks) |
Notes
B1: \(1/q\) seen B1: \(\dfrac{1}{p} \times \dfrac{1}{q} = -1\) or \(\dfrac{1}{p} = -\dfrac{1}{\frac{1}{q}}\) or \(\dfrac{1}{q} = -\dfrac{1}{\frac{1}{p}}\) and at least words in bold with no errors seen.