| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Standard +0.3 This is a straightforward 'show that' question requiring standard differentiation to find the tangent equation, followed by algebraic manipulation to find intersection points. Part (a) is routine calculus with no problem-solving, and part (b) involves setting two tangent equations equal and simplifying—mechanical algebra with a predictable structure. Slightly above average only due to the symbolic parameters and two-part nature. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 2ax + b\) | M1 | Allow \(2at + b\) |
| \(y - (at^2 + bt + c) = (2at + b)(x - t)\) or \(at^2 + bt + c = (2at + b)t + k\) | M1 | Use of equation of straight line with their gradient and \(x = t\) |
| \(y = (2at + b)x - 2at^2 - bt + at^2 + bt + c\), so \(y = (2at + b)x - at^2 + c\) | A1 | Convincing completion to correct result |
| Alternative: Line and curve cross where \(ax^2 + bx + c = (2at+b)x - at^2 + c\) | M1 | — |
| \(ax^2 - 2atx + at^2 = 0\) | M1 | Getting everything to one side |
| \(a(x-t)^2 = 0\), line touches curve when \(x = t\) | A1 | Convincing completion to correct result |
| Answer | Marks | Guidance |
|---|---|---|
| \((2ax_P + b)x - ax_P^2 + c = (2ax_Q + b)x - ax_Q^2 + c\) | M1 | Use of tangent formula with distinct values of \(t\) |
| \((2ax_P - 2ax_Q)x = ax_P^2 - ax_Q^2\) | M1 | Getting terms in \(x\) on one side |
| \(x = \frac{x_P^2 - x_Q^2}{2(x_P - x_Q)} = \frac{(x_P + x_Q)(x_P - x_Q)}{2(x_P - x_Q)} = \frac{x_P + x_Q}{2}\) | A1 | Convincing completion to given result |
## Question 15(a):
$\frac{dy}{dx} = 2ax + b$ | M1 | Allow $2at + b$
$y - (at^2 + bt + c) = (2at + b)(x - t)$ **or** $at^2 + bt + c = (2at + b)t + k$ | M1 | Use of equation of straight line with their gradient and $x = t$
$y = (2at + b)x - 2at^2 - bt + at^2 + bt + c$, so $y = (2at + b)x - at^2 + c$ | A1 | Convincing completion to correct result
**Alternative:** Line and curve cross where $ax^2 + bx + c = (2at+b)x - at^2 + c$ | M1 | —
$ax^2 - 2atx + at^2 = 0$ | M1 | Getting everything to one side
$a(x-t)^2 = 0$, line touches curve when $x = t$ | A1 | Convincing completion to correct result
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## Question 15(b):
$(2ax_P + b)x - ax_P^2 + c = (2ax_Q + b)x - ax_Q^2 + c$ | M1 | Use of tangent formula with distinct values of $t$
$(2ax_P - 2ax_Q)x = ax_P^2 - ax_Q^2$ | M1 | Getting terms in $x$ on one side
$x = \frac{x_P^2 - x_Q^2}{2(x_P - x_Q)} = \frac{(x_P + x_Q)(x_P - x_Q)}{2(x_P - x_Q)} = \frac{x_P + x_Q}{2}$ | A1 | Convincing completion to given result
---15
\begin{enumerate}[label=(\alph*)]
\item Show that, for the curve $y = a x ^ { 2 } + b x + c$, the equation of the tangent at the point with $x$-coordinate $t$ is $\mathrm { y } = ( 2 \mathrm { at } + \mathrm { b } ) \mathrm { x } - \mathrm { at } ^ { 2 } + \mathrm { c }$.
\item Hence show that for the curve with equation $y = a x ^ { 2 } + b x + c$, the tangents at two points, $P$ and Q , on the curve cross at a point which has $x$-coordinate equal to the mean of the $x$-coordinates of points P and Q , as given in lines 11 to 14 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q15 [6]}}