SPS SPS SM Pure 2020 February — Question 4 8 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2020
SessionFebruary
Marks8
TopicParametric curves and Cartesian conversion
TypeConvert to Cartesian (tan/sec/cot/cosec identities)
DifficultyModerate -0.8 This is a straightforward parametric equations question with standard techniques: part (a) is routine trigonometric identity manipulation, part (b) uses the given identity to eliminate the parameter (yielding x² - y² = a²), and part (c) requires finding dy/dx using the chain rule and substituting θ = π/3. All steps are textbook exercises with no novel problem-solving required, making it easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07s Parametric and implicit differentiation

4
  1. Using \(\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1\), show that \(\tan ^ { 2 } \theta + 1 \equiv \sec ^ { 2 } \theta\).
    A curve is given parametrically by $$x = a \sec \theta , \quad y = a \tan \theta$$ where \(a\) is a constant.
  2. Find a Cartesian equation of the curve.
  3. Determine an equation of the tangent to the curve at the point \(\theta = \frac { \pi } { 3 }\), giving your answer in exact form.
    [0pt] [5]

4
\begin{enumerate}[label=(\alph*)]
\item Using $\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1$, show that $\tan ^ { 2 } \theta + 1 \equiv \sec ^ { 2 } \theta$.\\

A curve is given parametrically by

$$x = a \sec \theta , \quad y = a \tan \theta$$

where $a$ is a constant.
\item Find a Cartesian equation of the curve.
\item Determine an equation of the tangent to the curve at the point $\theta = \frac { \pi } { 3 }$, giving your answer in exact form.\\[0pt]
[5]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2020 Q4 [8]}}