OCR FP2 2008 January — Question 4 8 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.8 This is a Further Maths polar coordinates question requiring integration of a non-standard polar curve (involving sec θ) and conversion to Cartesian form. Part (i) needs careful application of the polar area formula with the given integral result, while part (ii) requires algebraic manipulation of r = 1 + 2sec θ using x = r cos θ. Both parts demand solid technique beyond standard A-level, but follow established methods without requiring novel insight.
Spec4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve

4 The equation of a curve, in polar coordinates, is $$r = 1 + 2 \sec \theta , \quad \text { for } - \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$$
  1. Find the exact area of the region bounded by the curve and the lines \(\theta = 0\) and \(\theta = \frac { 1 } { 6 } \pi\). [The result \(\int \sec \theta \mathrm { d } \theta = \ln | \sec \theta + \tan \theta |\) may be assumed.]
  2. Show that a cartesian equation of the curve is \(( x - 2 ) \sqrt { x ^ { 2 } + y ^ { 2 } } = x\).

4 The equation of a curve, in polar coordinates, is

$$r = 1 + 2 \sec \theta , \quad \text { for } - \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$$

(i) Find the exact area of the region bounded by the curve and the lines $\theta = 0$ and $\theta = \frac { 1 } { 6 } \pi$. [The result $\int \sec \theta \mathrm { d } \theta = \ln | \sec \theta + \tan \theta |$ may be assumed.]\\
(ii) Show that a cartesian equation of the curve is $( x - 2 ) \sqrt { x ^ { 2 } + y ^ { 2 } } = x$.

\hfill \mbox{\textit{OCR FP2 2008 Q4 [8]}}