FP1 June 2013 (R) Q7
7. The parabola \(C\) has equation \(y^2 = 4ax\), where \(a\) is a positive constant.
The point \(P(at^2, 2at)\) is a general point on \(C\).
(a) Show that the equation of the tangent to \(C\) at \(P(at^2, 2at)\) is \[ty = x + at^2\] (4)
The tangent to \(C\) at \(P\) meets the \(y\)-axis at a point \(Q\).
(b) Find the coordinates of \(Q\). (1)
Given that the point \(S\) is the focus of \(C\),
(c) show that \(PQ\) is perpendicular to \(SQ\). (3)
| Scheme | Marks |
|---|---|
| \(y^2 = 4ax\), at \(P(at^2, 2at)\). | |
| \(y = 2\sqrt{a}\,x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{a}\,x^{-\frac{1}{2}}\) or (implicitly) \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a\) or (chain rule) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2a \times \dfrac{1}{2at}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm k\,x^{-\frac{1}{2}}\) or \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) or \(\dfrac{\text{their } \frac{\mathrm{d}y}{\mathrm{d}t}}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}t}}\) | M1 |
| When \(x = at^2\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{a}}{\sqrt{at^2}} = \dfrac{\sqrt{a}}{\sqrt{a}\,t} = \dfrac{1}{t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4a}{2(2at)} = \dfrac{1}{t}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) | A1 |
| \(\mathbf{T}\text{: } y - 2at = \dfrac{1}{t}(x - at^2)\) Applies \(y - 2at = \text{their } m_T(x - at^2)\) Their \(m_T\) must be a function of \(t\) from calculus. | M1 |
| \(\mathbf{T}\text{: } ty - 2at^2 = x - at^2\) | |
| \(\mathbf{T}\text{: } ty = x + at^2\) Correct solution. | A1 cso * |
| (4) |
| Scheme | Marks |
|---|---|
| At \(Q\), \(x = 0 \Rightarrow y = \dfrac{at^2}{t} = at \Rightarrow Q(0, at)\) \(y = at\) or \(Q(0, at)\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(S(a, 0)\) | |
| \(\mathrm{m}(PQ) = \dfrac{at - 2at}{0 - at^2} = \dfrac{-at}{-at^2} = \dfrac{1}{t}\) A correct method for finding either \(\mathrm{m}(PQ)\) or \(\mathrm{m}(SQ)\) for their \(Q\) or \(S\). | M1 |
| \(\mathrm{m}(SQ) = \dfrac{at - 0}{0 - a} = \dfrac{at}{-a} = -t\) \(\mathrm{m}(PQ) = \dfrac{1}{t}\) and \(\mathrm{m}(SQ) = -t\) | A1 |
| \(\mathrm{m}(PQ) \times \mathrm{m}(SQ) = \dfrac{1}{t} \times -t = -1 \Rightarrow PQ \perp SQ\) Shows \(\mathrm{m}(PQ) \times \mathrm{m}(SQ) = -1\) and conclusion. | A1 cso |
| (3) | |
| (8 marks) |