OCR H240/03 2020 November — Question 6 11 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2020
SessionNovember
Marks11
PaperDownload PDF ↗
TopicImplicit equations and differentiation
TypeShow dy/dx equals given expression
DifficultyChallenging +1.2 This question requires implicit differentiation (standard A-level technique) followed by finding points where tangents are horizontal/vertical and calculating distance. Part (a) is routine differentiation. Part (b) involves solving for specific points and distance calculation, requiring careful algebra across multiple steps but no novel insights. The 8 marks and 'show detailed reasoning' indicate extended working, but the techniques are all standard Core/Pure content.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07s Parametric and implicit differentiation

In this question you must show detailed reasoning. \includegraphics{figure_6} The diagram shows the curve with equation \(4xy = 2(x^2 + 4y^2) - 9x\).
  1. Show that \(\frac{dy}{dx} = \frac{4x - 4y - 9}{4x - 16y}\). [3] At the point \(P\) on the curve the tangent to the curve is parallel to the \(y\)-axis and at the point \(Q\) on the curve the tangent to the curve is parallel to the \(x\)-axis.
  2. Show that the distance \(PQ\) is \(k\sqrt{5}\), where \(k\) is a rational number to be determined. [8]

In this question you must show detailed reasoning.

\includegraphics{figure_6}

The diagram shows the curve with equation $4xy = 2(x^2 + 4y^2) - 9x$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = \frac{4x - 4y - 9}{4x - 16y}$. [3]

At the point $P$ on the curve the tangent to the curve is parallel to the $y$-axis and at the point $Q$ on the curve the tangent to the curve is parallel to the $x$-axis.

\item Show that the distance $PQ$ is $k\sqrt{5}$, where $k$ is a rational number to be determined. [8]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2020 Q6 [11]}}