October 2021 Paper 2 Q8
8. The curve \(C\) has equation
\[px^3 + qxy + 3y^2 = 26\]where \(p\) and \(q\) are constants.
Given that
- the point \(P(-1,\ -4)\) lies on \(C\)
- the normal to \(C\) at \(P\) has equation \(19x + 26y + 123 = 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(3y^2\right) = 6y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}}{\mathrm{d}x}(qxy) = qx\dfrac{\mathrm{d}y}{\mathrm{d}x} + qy\) | M1 | 2.1 |
| \(3px^2 + qx\dfrac{\mathrm{d}y}{\mathrm{d}x} + qy + 6y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | A1 | 1.1b |
| \((qx + 6y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3px^2 - qy \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) | dM1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-3px^2 - qy}{qx + 6y}\) | A1 | 1.1b |
| (4) |
Notes
M1: For selecting the appropriate method of differentiating:
Allow this mark for either \(3y^2 \rightarrow \alpha y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(qxy \rightarrow \alpha x\dfrac{\mathrm{d}y}{\mathrm{d}x} + \beta y\)
A1: Fully correct differentiation. Ignore any spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
dM1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with 2 terms only in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) coming from \(qxy\) and \(3y^2\)
Depends on the first method mark.
A1: Fully correct expression
| Scheme | Marks | AO |
|---|---|---|
| \(p(-1)^3 + q(-1)(-4) + 3(-4)^2 = 26\) | M1 | 1.1b |
| \(19x + 26y + 123 = 0 \Rightarrow m = -\dfrac{19}{26}\) | B1 | 2.2a |
| \(\dfrac{-3p(-1)^2 - q(-4)}{q(-1) + 6(-4)} = \dfrac{26}{19}\) or \(\dfrac{q(-1) + 6(-4)}{3p(-1)^2 + q(-4)} = -\dfrac{19}{26}\) | M1 | 3.1a |
| \(p - 4q = 22,\ 57p - 102q = 624 \Rightarrow p = \ldots,\ q = \ldots\) | dM1 | 1.1b |
| \(p = 2,\ q = -5\) | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Notes
M1: Uses \(x = -1\) and \(y = -4\) in the equation of \(C\) to obtain an equation in \(p\) and \(q\)
B1: Deduces the correct gradient of the given normal.
This may be implied by e.g.
\(19x + 26y + 123 = 0 \Rightarrow y = -\dfrac{19}{26}x + \ldots \Rightarrow\) Tangent equation is \(y = \dfrac{26}{19}x + \ldots\)
M1: Fully correct strategy to establish an equation connecting \(p\) and \(q\) using \(x = -1\) and \(y = -4\) in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and the gradient of the normal. E.g. \((a) = -1 \div\) their \(-\dfrac{19}{26}\) or \(-1 \div (a) =\) their \(-\dfrac{19}{26}\)
dM1: Solves simultaneously to obtain values for \(p\) and \(q\).
Depends on both previous method marks.
A1: Correct values
Alternative for (b)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-3p + 4q}{-q - 24} \Rightarrow y + 4 = \dfrac{q + 24}{4q - 3p}(x + 1)\) M1A1
\(\Rightarrow y(4q - 3p) + 4(4q - 3p) = (q + 24)x + q + 24\) M1
\(19x + 26y + 123 = 0 \Rightarrow q + 24 = 19 \Rightarrow q = -5\)
\(3p - 4q = 26 \Rightarrow 3p + 20 = 26 \Rightarrow p = 2\) M1A1
M1: Uses \((-1,\ -4)\) in the tangent gradient and attempts to form normal equation
A1: Correct equation for normal
M1: Multiplies up so that coefficients can be compared
dM1: Full method comparing coefficients to find values for \(p\) and \(q\)
A1: Correct values