C4 June 2006 Q1
1. A curve \(C\) is described by the equation
\[3x^2 - 2y^2 + 2x - 3y + 5 = 0.\]
Find an equation of the normal to \(C\) at the point \((0, 1)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (7)
| Scheme | Marks |
|---|---|
| \(6x - 4y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2 - 3\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 A1 |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6x + 2}{4y + 3}\right\}\) not necessarily required. | |
| At \((0, 1)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{0 + 2}{4 + 3} = \dfrac{2}{7}\) | dM1; A1 cso |
| Hence \(m(\mathbf{N}) = -\dfrac{7}{2}\) or \(\dfrac{-1}{\frac{2}{7}}\) | A1ft oe. |
| Either \(\mathbf{N}\): \(y - 1 = -\frac{7}{2}(x - 0)\) or \(\mathbf{N}\): \(y = -\frac{7}{2}x + 1\) | M1; |
| \(\mathbf{N}\): \(7x + 2y - 2 = 0\) | A1 oe cso |
| (7 marks) |
Notes
M1 Differentiates implicitly to include either \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \right)\).)
A1 Correct equation.
dM1 Substituting \(x = 0\) & \(y = 1\) into an equation involving \(\frac{\mathrm{d}y}{\mathrm{d}x}\);
A1 to give \(\frac{2}{7}\) or \(\frac{-2}{-7}\)
A1ft Uses \(m(\mathbf{T})\) to ‘correctly’ find \(m(\mathbf{N})\). Can be ft from “their tangent gradient”.
M1 \(y - 1 = m(x - 0)\) with ‘their tangent or normal gradient’; or uses \(y = mx + 1\) with ‘their tangent or normal gradient’;
A1 Correct equation in the form ‘\(ax + by + c = 0\)’, where \(a\), \(b\) and \(c\) are integers.
Beware: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{7}\) does not necessarily imply the award of all the first four marks in this question. So please ensure that you check candidates’ initial differentiation before awarding the first A1 mark.
Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0.
Beware: A candidate finding an \(m(\mathbf{T}) = 0\) can obtain A1ft for \(m(\mathbf{N}) = \infty\), but obtains M0 if they write \(y - 1 = \infty(x - 0)\). If they write, however, \(\mathbf{N}\): \(x = 0\), then can score M1.
Beware: A candidate finding an \(m(\mathbf{T}) = \infty\) can obtain A1ft for \(m(\mathbf{N}) = 0\), and also obtains M1 if they write \(y - 1 = 0(x - 0)\) or \(y = 1\).
Beware: The final cso refers to the whole question.
Aliter Way 2
| \(6x\dfrac{\mathrm{d}x}{\mathrm{d}y} - 4y + 2\dfrac{\mathrm{d}x}{\mathrm{d}y} - 3 = 0\) | M1 A1 |
| \(\left\{\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{4y + 3}{6x + 2}\right\}\) not necessarily required. | |
| At \((0, 1)\), \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{4 + 3}{0 + 2} = \dfrac{7}{2}\) | dM1; A1 cso |
| Hence \(m(\mathbf{N}) = -\dfrac{7}{2}\) or \(\dfrac{-1}{\frac{2}{7}}\) | A1ft oe. |
| Either \(\mathbf{N}\): \(y - 1 = -\frac{7}{2}(x - 0)\) or \(\mathbf{N}\): \(y = -\frac{7}{2}x + 1\) | M1; |
| \(\mathbf{N}\): \(7x + 2y - 2 = 0\) | A1 oe cso |
M1 Differentiates implicitly to include either \(\pm kx\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\pm 2\dfrac{\mathrm{d}x}{\mathrm{d}y}\). (Ignore \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \right)\).)
A1 Correct equation.
dM1 Substituting \(x = 0\) & \(y = 1\) into an equation involving \(\frac{\mathrm{d}x}{\mathrm{d}y}\);
A1 to give \(\frac{7}{2}\)
A1ft Uses \(m(\mathbf{T})\) or \(\frac{\mathrm{d}x}{\mathrm{d}y}\) to ‘correctly’ find \(m(\mathbf{N})\). Can be ft using “\(-1 \cdot \frac{\mathrm{d}x}{\mathrm{d}y}\)”.
M1 \(y - 1 = m(x - 0)\) with ‘their tangent, \(\frac{\mathrm{d}x}{\mathrm{d}y}\) or normal gradient’; or uses \(y = mx + 1\) with ‘their tangent, \(\frac{\mathrm{d}x}{\mathrm{d}y}\) or normal gradient’;
A1 Correct equation in the form ‘\(ax + by + c = 0\)’, where \(a\), \(b\) and \(c\) are integers.
Aliter Way 3
| \(2y^2 + 3y - 3x^2 - 2x - 5 = 0\) \(\left(y + \frac{3}{4}\right)^2 - \frac{9}{16} = \frac{3x^2}{2} + x + \frac{5}{2}\) \(y = \sqrt{\left(\frac{3x^2}{2} + x + \frac{49}{16}\right)} - \frac{3}{4}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(\frac{3x^2}{2} + x + \frac{49}{16}\right)^{-\frac{1}{2}}(3x + 1)\) | M1; A1 oe |
| At \((0, 1)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(\dfrac{49}{16}\right)^{-\frac{1}{2}} = \dfrac{1}{2}\left(\dfrac{4}{7}\right) = \dfrac{2}{7}\) | dM1 A1 cso |
| Hence \(m(\mathbf{N}) = -\dfrac{7}{2}\) | A1ft |
| Either \(\mathbf{N}\): \(y - 1 = -\frac{7}{2}(x - 0)\) or \(\mathbf{N}\): \(y = -\frac{7}{2}x + 1\) | M1 |
| \(\mathbf{N}\): \(7x + 2y - 2 = 0\) | A1 oe |
| [7] |
M1; A1 Differentiates using the chain rule; Correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
dM1 A1 Substituting \(x = 0\) into an equation involving \(\frac{\mathrm{d}y}{\mathrm{d}x}\); to give \(\frac{2}{7}\) or \(\frac{-2}{-7}\)
A1ft Uses \(m(\mathbf{T})\) to ‘correctly’ find \(m(\mathbf{N})\). Can be ft from “their tangent gradient”.
M1 \(y - 1 = m(x - 0)\) with ‘their tangent or normal gradient’; or uses \(y = mx + 1\) with ‘their tangent or normal gradient’
A1 Correct equation in the form ‘\(ax + by + c = 0\)’, where \(a\), \(b\) and \(c\) are integers.
CHECK (corrected from the printed mark scheme: in Way 3 the second form of the normal is printed as \(y = -\frac{2}{7}x + 1\); the gradient of the normal is \(-\frac{7}{2}\).)