June 2024 Paper 3 Q15
15 This question refers to the article on the Insert, “Tangents and normals to a quadratic curve”. The relevant extract (lines 11 to 15) is reproduced here.
The general quadratic curve has equation \(y = ax^2 + bx + c\). The tangents at any two points P and Q on this curve also cross at a point whose \(x\)-coordinate is equal to the mean of the \(x\)-coordinates of P and Q. So if P has \(x\)-coordinate \(x_\mathrm{P}\) and Q has \(x\)-coordinate \(x_\mathrm{Q}\) then the \(x\)-coordinate of the intersection point of the tangents is \(\dfrac{x_\mathrm{P} + x_\mathrm{Q}}{2}\). The \(y\)-coordinate of the intersection point can be shown to be \(ax_\mathrm{P}x_\mathrm{Q} + b\left(\frac{x_\mathrm{P}+x_\mathrm{Q}}{2}\right) + c\).
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b\) | M1 | 1.1 |
| \(y - (at^2 + bt + c) = (2at + b)(x - t)\) or \(at^2 + bt + c = (2at + b)t + \mathrm{k}\) | M1 | 1.1 |
| \(y = (2at + b)x - 2at^2 - bt + at^2 + bt + c\) So \(\boldsymbol{y = (2at + b)x - at^2 + c}\) | A1 | 2.1 |
| [3] |
Notes
M1: Allow \(2at + b\)
M1: Use of a form of equation of straight line with their gradient and \(x = t\)
A1: Convincing completion to correct result
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| The line and curve cross where \(\boldsymbol{ax^2 + bx + c = (2at + b)x - at^2 + c}\) | M1 | |
| \(\boldsymbol{ax^2 - 2atx + at^2 = 0}\) | M1 | |
| \(a(x - t)^2 = 0\) Line touches curve when \(x = t\) so it is the required tangent | A1 |
M1: Getting everything to one side
A1: Convincing completion to correct result
Additional guidance
The first M1 is for differentiating. We are allowing the gradient as 2at + b as well as 2ax + b. Some are just writing 2ax+b/2at+b and that isn’t convincing that they have differentiated – rather just copying from the given answer so give M0.
The second M1 is for substituting into the equation of a straight line using their gradient and x = t. They could use \((y - y_1) = m(x - x_1)\) or y= mx + c. Condone a lack of brackets around their m if recovered.
The final A1 is for getting to the correct answer. As the answer is given, we need to see some working here.
| Scheme | Marks | AO |
|---|---|---|
| Tangents cross where \(\boldsymbol{\left(2ax_\mathrm{P} + b\right)x - ax_\mathrm{P}^2 + c = \left(2ax_\mathrm{Q} + b\right)x - ax_\mathrm{Q}^2 + c}\) | M1 | 2.1 |
| \(\boldsymbol{\left(2ax_\mathrm{P} - 2ax_\mathrm{Q}\right)x = ax_\mathrm{P}^2 - ax_\mathrm{Q}^2}\) | M1 | 1.1 |
| \(x = \dfrac{x_\mathrm{P}^2 - x_\mathrm{Q}^2}{2\left(x_\mathrm{P} - x_\mathrm{Q}\right)}\) \(= \dfrac{\left(x_\mathrm{P} + x_\mathrm{Q}\right)\left(x_\mathrm{P} - x_\mathrm{Q}\right)}{2\left(x_\mathrm{P} - x_\mathrm{Q}\right)} = \dfrac{x_\mathrm{P} + x_\mathrm{Q}}{2}\) | A1 | 2.1 |
| [3] |
Notes
M1: Use of tangent formula with distinct values of \(t\)
Condone alternative notation for co-ordinates.
e.g. P and Q, m and n, and other subscript versions
M1: Getting terms in \(x\) on one side
A1: Convincing completion to given result.
Accept the final result in terms of their variables.
Additional guidance
The first M1 is for equating their two tangents with distinct values of t. These could be \(x_p\) and \(x_q\) as per the scheme, but they might choose other pairs of symbols – e.g. P and Q, m and n, \(t_1\) and \(t_2\).
The second M1 is for getting the x terms on one side and the non-x terms on the other.
To get the A1 they must show the working out including the difference of two squares and cancelling to get the required result. Allow the final result in terms of their variable.