| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.8 This question requires implicit differentiation (standard technique) but then asks students to find vertical tangent points by setting the denominator to zero—a conceptual step beyond routine exercises. The algebra involves solving a cubic equation after substitution, requiring careful manipulation and exact form answers. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{10-2x}{3y^2-12}\) | A1 | \(\frac{dy}{dx} = \frac{10-2x}{3y^2-12}\) with no errors |
| M1 | For a valid attempt at making \(\frac{dy}{dx}\) the subject, with two terms in \(\frac{dy}{dx}\) coming from \(x^2\) and \(2y\). | |
| A1 | Fully correct derivative \(2x + 3y^2\frac{dy}{dx} - 10 - 12\frac{dy}{dx} = 0\) | |
| M1 | either \(y^3 \rightarrow Ay^2\frac{dy}{dx}\) or \(12y \rightarrow 12\frac{dy}{dx}\) |
| Answer | Marks |
|---|---|
| \(Q(5 + \sqrt{46}, 2)\) | A1 |
| M1 | substitute \(y = 2\) in \(x^2 + y^3 - 10x - 12y - 5 = 0\) and creating a 3 term quadratic and solving |
| A1 | \(y = 2\) |
| M1 | Deduces that \(3y^2 - 12 = 0\) and proceed to find the value of \(y\) |
### Part a:
$\frac{dy}{dx} = \frac{10-2x}{3y^2-12}$ | A1 | $\frac{dy}{dx} = \frac{10-2x}{3y^2-12}$ with no errors
| M1 | For a valid attempt at making $\frac{dy}{dx}$ the subject, with two terms in $\frac{dy}{dx}$ coming from $x^2$ and $2y$.
| A1 | Fully correct derivative $2x + 3y^2\frac{dy}{dx} - 10 - 12\frac{dy}{dx} = 0$
| M1 | either $y^3 \rightarrow Ay^2\frac{dy}{dx}$ or $12y \rightarrow 12\frac{dy}{dx}$
### Part b:
$Q(5 + \sqrt{46}, 2)$ | A1 |
| M1 | substitute $y = 2$ in $x^2 + y^3 - 10x - 12y - 5 = 0$ and creating a 3 term quadratic and solving
| A1 | $y = 2$
| M1 | Deduces that $3y^2 - 12 = 0$ and proceed to find the value of $y$9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{63d85737-99d4-4916-a479-fe44f77b1505-16_871_1017_267_548}
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\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve with equation $x ^ { 2 } + y ^ { 3 } - 10 x - 12 y - 5 = 0$ a. Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 10 - 2 x } { 3 y ^ { 2 } - 12 }$
At each of the points $P$ and $Q$ the tangent to the curve is parallel to the $y$-axis.\\
b. Find the exact coordinates of $Q$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q9 [8]}}