FP1 June 2015 Q8
8. The point \(P(3p^2, 6p)\) lies on the parabola with equation \(y^2 = 12x\) and the point \(S\) is the focus of this parabola.
The point \(Q(3q^2, 6q)\), \(p \neq q\), also lies on this parabola.
The tangent to the parabola at the point \(P\) and the tangent to the parabola at the point \(Q\) meet at the point \(R\).
| Scheme | Marks |
|---|---|
| \(SP = \sqrt{(3p^2 - a)^2 + 36p^2}\), with \(a = 3\) | M1, B1 |
| \(SP = \sqrt{9p^4 + 18p^2 + 9} \quad = 3(1 + p^2)\) **given answer** | A1 * |
| ALT | |
| For parabola, perpendicular distance from \(P\) to directrix \(= SP\) | M1 |
| Directrix \(x = -3\) | B1 |
| So \(SP = 3 + 3p^2 = 3(1 + p^2)\) | A1 |
| (3) |
Notes
M1: Uses distance between two points or states perpendicular distance from \(P\) to directrix required.
B1: States or uses focus at \((3,0)\) or focus at \(a = 3\) or directrix as \(x = -3\)
A1: cso
| Scheme | Marks |
|---|---|
| \(y^2 = 12x \Rightarrow 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12\) or \(y = \sqrt{12x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{3}x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}y}{\mathrm{d}p}}{\frac{\mathrm{d}x}{\mathrm{d}p}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}y}{\mathrm{d}q}}{\frac{\mathrm{d}x}{\mathrm{d}q}}\) | M1 |
| The tangent at \(P\) has gradient \(= \dfrac{1}{p}\) or the tangent at \(Q\) has gradient \(\dfrac{1}{q}\) | A1 |
| and equation is \(y - 6p = \dfrac{1}{p}(x - 3p^2)\) or \(py = x + 3p^2\) o.e. | A1 |
| Tangent at \(Q\) is \(y - 6q = \dfrac{1}{q}(x - 3q^2)\) or \(qy = x + 3q^2\) o.e. | B1 |
| Eliminate \(x\) or \(y\): So \(x = 3pq\) or \(y = 3(p + q) = 3p + 3q\) | M1 A1 |
| Substitute for second variable so \(x = 3pq\) and \(y = 3(p + q) = 3p + 3q\) | M1 A1 |
| (8) |
Notes
M1: Calculus method for finding gradient and substitutes \(x\) value at either point
A1: Either correct. Accept unsimplified.
A1: One equation of tangent correct. B1: Both correct
M1: Eliminate \(x\) or \(y\). A1: Obtain first variable
M1: Substitute or eliminate again. A1: Both variables correct in simplest form as above.
| Scheme | Marks |
|---|---|
| \(SR^2 = (3 - 3pq)^2 + (3p + 3q)^2\ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\) | M1 |
| \(SP.SQ = 3(1 + p^2)\ 3(1 + q^2)\ \ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\) | M1 |
| So \(SR^2 = SP.SQ\) as required | A1 |
| (3) | |
| (14 marks) |
Notes
M1: Find their \(SR^2 = (3 - 3pq)^2 + (3p + 3q)^2\ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\)
M1: Find their \(SP.SQ = 3(1 + p^2)\ 3(1 + q^2)\ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\)
A1: Deduce equal after no errors seen. Concluding statement required cso.