A2 October 2020 Q7
7. The points \(P(9p^2, 18p)\) and \(Q(9q^2, 18q)\), \(p \neq q\), lie on the parabola \(C\) with equation
\[y^2 = 36x\]The line \(l\) passes through the points \(P\) and \(Q\)
The normal to \(C\) at \(P\) and the normal to \(C\) at \(Q\) meet at the point \(A\).
Given that the points \(P\) and \(Q\) vary such that \(l\) always passes through the point \((12, 0)\)
| Scheme | Marks | AO |
|---|---|---|
| Gradient of \(PQ = \dfrac{18q - 18p}{9q^2 - 9p^2} = \dfrac{2}{p + q}\) or \(\left.\begin{aligned}18p &= 9p^2m + c\\ 18q &= 9q^2m + c\end{aligned}\right\} \Rightarrow 18p - 18q = 9p^2m - 9q^2m \Rightarrow m = \dfrac{2}{p + q}\) | B1 | 2.2a |
| Equation of \(l\) is \(y - 18p = \text{“}\dfrac{2}{p + q}\text{”}(x - 9p^2)\) or \(18p = 9p^2\left(\dfrac{2}{p + q}\right) + c \Rightarrow c = \ldots\) | M1 | 1.1b |
| Leading to \((p + q)y = 2(x + 9pq)\ *\) | A1* | 2.1 |
| (3) |
Notes
B1: Deduces gradient is \(\dfrac{2}{p + q}\). May be implied by correct simplification of equation if the unsimplified form is used to start with.
M1: Correct method for the equation of the line (gradient need not be simplified/correct for this method, as long as it is clearly an attempt at the gradient).
A1*: Completes to the correct equation with no errors seen.
| Scheme | Marks | AO |
|---|---|---|
| Complete method to find equation of both normals and attempts to solve simultaneously | M1 | 3.1a |
| E.g. \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 36 \Rightarrow m_T = \dfrac{36}{36p} \Rightarrow m_N = -p\) | B1 | 1.1b |
| Normal at \(P\) is \(y - 18p = -p(x - 9p^2)\) or normal at \(Q\) is \(y - 18q = -q(x - 9q^2)\) (oe) | M1 | 2.1 |
| Both normals correct \(y - 18p = -p(x - 9p^2)\) or \(y = -px + 9p^3 + 18p\) (o.e.) \(y - 18q = -q(x - 9q^2)\) or \(y = -qx + 9q^3 + 18q\) (o.e.) | A1 | 2.2a |
| E.g. \(18p - px + 9p^3 - 18q = -qx + 9q^3 \Rightarrow x = \ldots\) | M1 | 1.1b |
| Need to show that \(\left(9p^3 - 9q^3 + 18p - 18q\right) = 9\left(p^2 + q^2 + pq + 2\right)(p - q)\) or \(p^3 - q^3 = \left(p^2 + pq + q^2\right)(p - q)\) Leading to \(x_A = 9(p^2 + q^2 + pq + 2)\ *\) | A1* | 2.2a |
| \(y = -9p(p^2 + q^2 + pq + 2) + 9p^3 + 18p = -9p^2q - 9pq^2\) Leading to \(y_A = -9pq(p + q)\ *\) | A1* | 2.2a |
| (7) |
Notes
M1: A correct overall method – must find both normals and attempt to solve simultaneously. They do not need to reach \(x =\) or \(y =\) as long as they have eliminated one variable.
B1: Correct gradient of normal found from any correct method or just stated.
M1: A full correct method to find the equation of at least one of the normals with justification of the gradient shown.
A1: Deduces equation of the second normal – so both correct.
M1: Solves the two normal equations simultaneously leading to either \(x = \ldots\) or \(y = \ldots\)
A1*: Need to show that \(\left(9p^3 - 9q^3 + 18p - 18q\right) = 9\left(p^2 + q^2 + pq + 2\right)(p - q)\) leading to correct \(x\) coordinate with no errors seen. This could by long division or factorising.
A1*: Correct \(y\) coordinate with no errors seen.
(Corrected from the printed mark scheme: the factorisation is printed as \(\left(9p^2 + q^2 + pq + 2\right)(p - q)\), in the scheme and in the note; the factor 9 belongs outside the bracket.)
| Scheme | Marks | AO |
|---|---|---|
| \((12, 0)\) on \(l \Rightarrow pq = -\dfrac{4}{3}\) (oe) | B1 | 3.1a |
| Hence \(x_A = 9\left(p^2 + q^2 + \dfrac{2}{3}\right)\) and \(y_A = 12(p + q)\) | M1 | 1.1b |
| \(y^2 = 144(p^2 + q^2 + 2pq) = 144\left(\dfrac{x}{9} - \dfrac{2}{3} + 2\left(-\dfrac{4}{3}\right)\right)\) | M1 | 3.1a |
| \(y^2 = 16(x - 30)\) or \(y^2 = 16x - 480\) | A1 | 1.1b |
| (4) | ||
| (14 marks) |
Notes
B1: Uses the condition on \(l\) to establish the relationship between \(p\) and \(q\)
M1: Uses their relationship between \(p\) and \(q\) to simplify the expressions
M1: Any complete method for relating \(x\) and \(y\) independently of \(p\) and \(q\)
A1: \(y^2 = 16(x - 30)\) or \(y^2 = 16x - 480\)