C3 June 2010 Q2
2. A curve \(C\) has equation
\[y = \frac{3}{(5 - 3x)^2}, \quad x \ne \frac{5}{3}\]The point \(P\) on \(C\) has \(x\)-coordinate 2. Find an equation of the normal to \(C\) at \(P\) in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (7)
| Scheme | Marks |
|---|---|
| At \(P\), \(y = \underline{3}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{3(-2)(5 - 3x)^{-3}(-3)}\ \left\{\text{or } \dfrac{18}{(5 - 3x)^3}\right\}\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{18}{(5 - 3(2))^3}\ \{= -18\}\) | M1 |
| \(\mathrm{m}(\mathbf{N}) = \dfrac{-1}{-18}\) or \(\dfrac{1}{18}\) | M1 |
| \(\mathbf{N}\): \(y - 3 = \tfrac{1}{18}(x - 2)\) | M1 |
| \(\mathbf{N}\): \(\underline{x - 18y + 52 = 0}\) | A1 |
| (7 marks) |
Notes
1st M1: \(\pm k(5 - 3x)^{-3}\) can be implied. See appendix for application of the quotient rule.
2nd M1: Substituting \(x = 2\) into an equation involving their \(\frac{\mathrm{d}y}{\mathrm{d}x}\);
3rd M1: Uses \(\mathrm{m}(\mathbf{N}) = -\dfrac{1}{\text{their } \mathrm{m}(\mathbf{T})}\).
4th M1: \(y - y_1 = m(x - 2)\) with ‘their NORMAL gradient’ or a “changed” tangent gradient and their \(y_1\). Or uses a complete method to express the equation of the tangent in the form \(y = mx + c\) with ‘their NORMAL (“changed” numerical) gradient’, their \(y_1\) and \(x = 2\).
Note: To gain the final A1 mark all the previous 6 marks in this question need to be earned. Also there must be a completely correct solution given.