| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Tangent/normal intersection problems |
| Difficulty | Standard +0.8 This is a substantial parametric calculus problem requiring dy/dx via chain rule (standard C4), verifying a tangent equation (routine), then finding where the tangent intersects the curve again (requires solving a cubic equation from substituting the line into parametric equations, which is algebraically demanding and goes beyond typical textbook exercises). The 7-mark final part indicates significant algebraic manipulation and problem-solving, elevating this above average difficulty. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dx}{dt} = 2t - 1\), \(\frac{dy}{dt} = \frac{4x(1-t) - 4x(x-1)}{(1-t)^2} = \frac{4}{(1-t)^2}\) | B1 M1 |
| \(\frac{dy}{dx} = \frac{4}{(2t-1)(1-t)^2}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(t = -1, x = 2, y = -2\), grad \(= -\frac{1}{3}\) | M1 |
| \(\therefore y + 2 = -\frac{1}{3}(x-2)\) | M1 |
| \(3y + 6 = -x + 2\) | |
| \(x + 3y + 4 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t(t-1) + 3 \times \frac{4t}{1-t} + 4 = 0\) | M1 | |
| \(-t(t-1)^2 + 12t + 4(1-t) = 0\) | ||
| \(t^3 - 2t^2 - 7t - 4 = 0\) | A1 | |
| \(t = -1\) is a solution \(\therefore (t+1)\) is a factor | M1 | |
| \((t+1)(t^2 - 3t - 4) = 0\) | M1 | |
| \((t+1)(t+1)(t-4) = 0\) | A1 | |
| \(t = -1\) (at P) or \(t = 4\) \(\therefore Q(12, -\frac{16}{5})\) | M1 A1 | (14) |
**(a)**
$\frac{dx}{dt} = 2t - 1$, $\frac{dy}{dt} = \frac{4x(1-t) - 4x(x-1)}{(1-t)^2} = \frac{4}{(1-t)^2}$ | B1 M1 |
$\frac{dy}{dx} = \frac{4}{(2t-1)(1-t)^2}$ | M1 A1 |
**(b)**
$t = -1, x = 2, y = -2$, grad $= -\frac{1}{3}$ | M1 |
$\therefore y + 2 = -\frac{1}{3}(x-2)$ | M1 |
$3y + 6 = -x + 2$ | |
$x + 3y + 4 = 0$ | A1 |
**(c)**
$t(t-1) + 3 \times \frac{4t}{1-t} + 4 = 0$ | M1 |
$-t(t-1)^2 + 12t + 4(1-t) = 0$ | |
$t^3 - 2t^2 - 7t - 4 = 0$ | A1 |
$t = -1$ is a solution $\therefore (t+1)$ is a factor | M1 |
$(t+1)(t^2 - 3t - 4) = 0$ | M1 |
$(t+1)(t+1)(t-4) = 0$ | A1 |
$t = -1$ (at P) or $t = 4$ $\therefore Q(12, -\frac{16}{5})$ | M1 A1 | (14)A curve has parametric equations
$$x = t(t - 1), \quad y = \frac{4t}{1-t}, \quad t \neq 1.$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. [4]
\end{enumerate}
The point $P$ on the curve has parameter $t = -1$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the tangent to the curve at $P$ has the equation
$$x + 3y + 4 = 0.$$ [3]
\end{enumerate}
The tangent to the curve at $P$ meets the curve again at the point $Q$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the coordinates of $Q$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [14]}}