FM June 2023 p13 q05
4217
(a) Show that the curve with Cartesian equation \(x^2 - y^2 = a\), where \(a\) is a positive constant, has polar equation \(r^2 = a \sec 2\theta\).
The curve \(C\) has polar equation \(r^2 = a \sec 2\theta\), where \(a\) is a positive constant, for \(0 \leq \theta < \frac{1}{4}\pi\).
(b) Sketch \(C\) and state the minimum distance of \(C\) from the pole.
Solution
(a) Start with the Cartesian equation \(x^2 - y^2 = a\).
In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\).
Substitute these into the equation: \((r \cos \theta)^2 - (r \sin \theta)^2 = a\).
This simplifies to \(r^2 (\cos^2 \theta - \sin^2 \theta) = a\).
Using the double angle identity, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we have \(r^2 \cos 2\theta = a\).
Thus, \(r^2 = a \sec 2\theta\).
(b) The sketch shows the curve starting at \(\theta = 0\) and increasing, with the minimum distance from the pole being \(\sqrt{a}\).
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