AS & A Level · Practice questions · Polar and Cartesian coordinates

Change the coordinates, keep the curve.

Six original Cambridge-style questions on polar and Cartesian coordinates: converting points, finding polar equations from Cartesian ones and the reverse, and using the convention \(r \ge 0\).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Substitute, simplify, check \(r \ge 0\).

Points: \(x = r\cos\theta\), \(y = r\sin\theta\); back again \(r = \sqrt{x^{2} + y^{2}}\) and \(\tan\theta = \dfrac{y}{x}\).
Quadrant check: sketch the point and place \(\theta\) in its quadrant, in the range the question asks for.
Cartesian to polar: replace \(x\), \(y\) and \(x^{2} + y^{2}\), then make \(r\) the subject.
Polar to Cartesian: multiply by \(r\) or \(r^{2}\) to build \(r\cos\theta\), \(r\sin\theta\) and \(r^{2}\); isolate any lone \(r\) before squaring.
\(r \ge 0\): values of \(\theta\) that make \(r\) negative give no points.

Recall

[2 marks]

Find the Cartesian coordinates of the point with polar coordinates \(\left(5, \dfrac{7\pi}{6}\right)\), giving exact values.

\(x = 5\cos\dfrac{7\pi}{6} = 5 \times \left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{5\sqrt{3}}{2}\). ✓
\(y = 5\sin\dfrac{7\pi}{6} = 5 \times \left(-\dfrac{1}{2}\right) = -\dfrac{5}{2}\). The point is \(\left(-\dfrac{5\sqrt{3}}{2}, -\dfrac{5}{2}\right)\), in the third quadrant as expected. ✓

Calculation

[3 marks]

Find polar coordinates of the point \((3, -4)\), giving \(\theta\) in radians to 3 significant figures, (a) with \(-\pi < \theta \le \pi\), (b) with \(0 \le \theta < 2\pi\).

\(r = \sqrt{9 + 16} = 5\). ✓
The point is in the fourth quadrant and \(\tan^{-1}\dfrac{4}{3} = 0.927\), so (a) \(\theta = -0.927\): \((5, -0.927)\). ✓
(b) Add \(2\pi\): \(\theta = 2\pi - 0.927 = 5.36\), giving \((5, 5.36)\). ✓

Calculation

[3 marks]

Show that the polar equation of the curve \(x^{2} - y^{2} = 4\) can be written \(r^{2} = 4\sec 2\theta\). State the values of \(\theta\) in \(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\) for which the curve has points.

\(r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta = 4\), so \(r^{2}(\cos^{2}\theta - \sin^{2}\theta) = 4\). ✓
\(\cos^{2}\theta - \sin^{2}\theta = \cos 2\theta\), so \(r^{2} = \dfrac{4}{\cos 2\theta} = 4\sec 2\theta\). ✓
Points need \(r^{2} > 0\), so \(\cos 2\theta > 0\): \(-\dfrac{\pi}{4} < \theta < \dfrac{\pi}{4}\). ✓

Calculation

[3 marks]

Find the Cartesian equation of the curve \(r = 4\sin\theta - 2\cos\theta\) and hence state the centre and radius of the circle.

Multiply by \(r\): \(r^{2} = 4r\sin\theta - 2r\cos\theta\), so \(x^{2} + y^{2} = 4y - 2x\). ✓
Complete the squares: \((x + 1)^{2} + (y - 2)^{2} = 1 + 4 = 5\). ✓
Centre \((-1, 2)\), radius \(\sqrt{5}\). ✓

Analysis

[4 marks]

A curve has polar equation \(r = \dfrac{3}{1 + \cos\theta}\), \(-\pi < \theta < \pi\).

(a) Show that its Cartesian equation is \(y^{2} = 9 - 6x\).

(b) Find the polar coordinates of the point of the curve on the initial line.

(a) \(r + r\cos\theta = 3\), so \(r + x = 3\) and \(r = 3 - x\). ✓
Square: \(x^{2} + y^{2} = 9 - 6x + x^{2}\), so \(y^{2} = 9 - 6x\). ✓
(b) On the initial line \(\theta = 0\): \(r = \dfrac{3}{2}\), giving \(\left(\dfrac{3}{2}, 0\right)\). ✓
Check with (a): \(y = 0\) gives \(x = \dfrac{3}{2}\), the vertex of the parabola. ✓

Analysis

[5 marks]

The curve \(C\) has polar equation \(r^{2} = 4\cos 2\theta\), with \(r \ge 0\).

(a) Find the set of values of \(\theta\) in \(-\pi < \theta \le \pi\) for which \(C\) has points.

(b) Show that the Cartesian equation of \(C\) is \((x^{2} + y^{2})^{2} = 4(x^{2} - y^{2})\).

(a) Need \(\cos 2\theta \ge 0\), with \(-2\pi < 2\theta \le 2\pi\). ✓
So \(-\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{4}\), \(\dfrac{3\pi}{4} \le \theta \le \pi\) or \(-\pi < \theta \le -\dfrac{3\pi}{4}\). ✓
(b) \(\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta\); multiply by \(r^{2}\): \(r^{4} = 4(r^{2}\cos^{2}\theta - r^{2}\sin^{2}\theta)\), so \((x^{2} + y^{2})^{2} = 4(x^{2} - y^{2})\). ✓
(c) Put \(y = 0\): \(x^{4} = 4x^{2}\), so \(x^{2}(x^{2} - 4) = 0\). ✓
The points are \((0, 0)\) (the pole), \((2, 0)\) and \((-2, 0)\), from \(\theta = 0\) and \(\theta = \pi\) with \(r = 2\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.