Conics

Lessons

Senior 6 · Mathematics

Conics

5 lessons

Lesson 5 of 5

Polar coordinates and polar equations of conics

40 min

Measuring from a focus

An astronomer describing a planet's orbit does not measure from the centre of the ellipse, which is an empty point in space, but from the Sun, which is at a focus. The natural coordinates are then a distance from the Sun and an angle. These are polar coordinates, and in them every conic with a focus at the origin has one short equation. This lesson sets up polar coordinates, writes lines, circles and conics in them, and ends with the main applications of conics.

Definition

Fix a point $O$ , the pole, and a ray from it, the polar axis (usually the positive $x$ -axis). A point $P \ne O$ has polar coordinates $(r, \theta)$ , where $r = OP \gt 0$ and $\theta$ is the angle from the polar axis to $O P$ , measured anticlockwise. Example: the point $(0, 2)$ has polar coordinates $\left(2, \frac{\pi}{2}\right)$ .

The same point also has coordinates $(r, \theta + 2k\pi)$ , $k \in \mathbb{Z}$ . It is convenient to allow $r \lt 0$ as well: $(-r, \theta)$ means the point $(r, \theta + \pi)$ , on the opposite side of the pole.

Important

Conversion formulas

x = r\cos\theta , \quad y = r\sin\theta ; \qquad r = \sqrt{x^2 + y^2} , \quad \cos\theta = \frac{x}{r} , \ \sin\theta = \frac{y}{r} .

Use $\tan\theta = \frac{y}{x}$ only together with the signs of $x$ and $y$ : $\tan\theta = -1$ is true for both $\frac{3\pi}{4}$ (second quadrant) and $-\frac{\pi}{4}$ (fourth quadrant).

Worked example 1: converting coordinates

(a) Write $\left(4, \frac{2\pi}{3}\right)$ in Cartesian coordinates. (b) Write $(- 3, 3)$ and $(1, -\sqrt{3})$ in polar coordinates.

Solution. (a) $x = 4\cos\frac{2\pi}{3} = -2$ , $y = 4\sin\frac{2\pi}{3} = 2\sqrt{3}$ : the point $(-2, 2\sqrt{3})$ .

(b) $r = \sqrt{9 + 9} = 3\sqrt{2}$ ; $x \lt 0 \lt y$ , second quadrant, so $\theta = \frac{3\pi}{4}$ : $\left(3\sqrt{2}, \frac{3\pi}{4}\right)$ . For $(1, -\sqrt{3})$ : $r = 2$ , fourth quadrant, $\theta = -\frac{\pi}{3}$ (or $\frac{5\pi}{3}$ ): $\left(2, -\frac{\pi}{3}\right)$ .

Lines and circles in polar form

Curve	Polar equation	Reason
line through $O$ at angle $\alpha$	$\theta = \alpha$	every point has the same direction
line at distance $p \gt 0$ from $O$ , with foot of the perpendicular $N(p, \alpha)$	$r\cos(\theta - \alpha) = p$	the projection of $O P$ on $O N$ is always $p$
vertical line $x = a$ ; horizontal line $y = b$	$r\cos\theta = a$ ; $r\sin\theta = b$	$x = r\cos\theta$ , $y = r\sin\theta$
circle, centre $O$ , radius $a$	$r = a$	$O P = a$
circle, centre $(a, 0)$ , radius $a$ (through $O$ )	$r = 2a\cos\theta$	$x^{2} + y^{2} = 2 a x$ gives $r^2 = 2ar\cos\theta$
circle, centre $(0, a)$ , radius $a$	$r = 2a\sin\theta$	$x^{2} + y^{2} = 2 a y$
circle, centre with polar coordinates $(r_0, \theta_0)$ , radius $a$	$r^2 - 2rr_0\cos(\theta - \theta_0) + r_0^2 = a^2$	the cosine rule in triangle $O$ , centre, $P$

Worked example 2: from Cartesian to polar and back

(a) Write the line $x + y = 4$ in the form $r\cos(\theta - \alpha) = p$ . (b) Write the circle $x^{2} + y^{2} - 6 x = 0$ in polar form. (c) Identify $r = 4\sin\theta$ .

Solution. (a) $r\cos\theta + r\sin\theta = 4$ . Since $\cos\theta + \sin\theta = \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)$ , this is $r\cos\left(\theta - \frac{\pi}{4}\right) = 2\sqrt{2}$ : the line is $2\sqrt{2}$ from the pole, and the perpendicular from $O$ points at $\frac{\pi}{4}$ .

(b) $r^2 - 6r\cos\theta = 0$ , so $r = 6\cos\theta$ (the pole is included, at $\theta = \frac{\pi}{2}$ ).

(c) Multiply by $r$ : $r^2 = 4r\sin\theta$ , that is $x^{2} + y^{2} = 4 y$ , or $x^{2} + (y - 2)^{2} = 4$ : the circle with centre $(0, 2)$ and radius 2.

Important

Polar equation of a conic. Put a focus at the pole and the directrix at $x = d$ ( $d \gt 0$ ). For $P(r, \theta)$ , $P F = r$ and the distance to the directrix is $d - r\cos\theta$ . The definition $PF = e \cdot PM$ gives $r = e(d - r\cos\theta)$ , so

r = \frac{l}{1 + e\cos\theta} , \qquad l = ed .

The number $l$ is the semi-latus rectum: at $\theta = \frac{\pi}{2}$ , $r = l$ . Other positions of the directrix change only the denominator:

Directrix	$x = d$	$x = - d$	$y = d$	$y = - d$
Equation	$r = \frac{l}{1 + e\cos\theta}$	$r = \frac{l}{1 - e\cos\theta}$	$r = \frac{l}{1 + e\sin\theta}$	$r = \frac{l}{1 - e\sin\theta}$

As always, $e \lt 1$ gives an ellipse, $e = 1$ a parabola and $e \gt 1$ a hyperbola. For an ellipse, the two vertices on the axis are at $r = \frac{l}{1 + e}$ and $r = \frac{l}{1 - e}$ , so the major axis is $2a = \frac{2l}{1 - e^2}$ , that is $l = a (1 - e^{2})$ .

The ellipse

r = \frac{6}{2 + \cos\theta}

on a polar grid (circles 1 unit apart, rays every

30^\circ

), drawn from its equation, with the polar coordinates of one point.

Worked example 3: identifying a conic from its polar equation

Identify $r = \frac{10}{5 - 3\cos\theta}$ . Give its eccentricity, directrix and vertices.

Solution. Make the constant in the denominator 1 by dividing above and below by 5: $r = \frac{2}{1 - 0.6\cos\theta}$ . So $e = 0.6 \lt 1$ , an ellipse, and $l = 2$ . Then $d = \frac{l}{e} = \frac{10}{3}$ , and the minus sign with $\cos\theta$ puts the directrix at $x = -\frac{10}{3}$ .

Vertices: $\theta = 0$ gives $r = \frac{2}{0.4} = 5$ , the point $(5, 0)$ ; $\theta = \pi$ gives $r = \frac{2}{1.6} = 1.25$ , the point $(- 1.25, 0)$ . So $2 a = 6.25$ , and the centre is $(1.875, 0)$ . Check: $c = 1.875$ (the distance from the centre to the focus at $O$ ) and $\frac{c}{a} = \frac{1.875}{3.125} = 0.6 = e$ .

Worked example 4: a parabola in polar form

Identify $r = \frac{3}{1 + \sin\theta}$ and find its Cartesian equation.

Solution. $e = 1$ : a parabola with focus at $O$ and directrix $y = 3$ . Its vertex is at $\theta = \frac{\pi}{2}$ , $r = \frac{3}{2}$ , the point $\left(0, \frac{3}{2}\right)$ . In Cartesian form: $r + r\sin\theta = 3$ , so $r = 3 - y$ , and squaring, $x^{2} + y^{2} = 9 - 6 y + y^{2}$ , that is $x^2 = -6\left(y - \frac{3}{2}\right)$ : opening downwards, with $4 a = 6$ , so the focus is $\frac{3}{2}$ below the vertex, at $O$ .

Worked example 5: the orbit of Mercury

NASA gives Mercury's orbit a semi-major axis $a = 57.909 \times 10^6\,\mathrm{km}$ and eccentricity $e = 0.2056$ . Write its polar equation with the Sun at the pole and $\theta$ measured from the perihelion, and check the perihelion and aphelion distances.

Solution. $l = a(1 - e^2) = 57.909 \times (1 - 0.2056^2) \approx 55.46$ , so (in millions of km)

r = \frac{55.46}{1 + 0.2056\cos\theta} .

At $\theta = 0$ : $r = \frac{55.46}{1.2056} \approx 46.0$ ; at $\theta = \pi$ : $r = \frac{55.46}{0.7944} \approx 69.8$ . NASA's fact sheet lists $46.000$ and $69.818$ .

Where conics are used

Conic	Property used	Examples
parabola	rays parallel to the axis reflect through the focus	satellite dishes, radio telescopes, solar cookers, car headlights and torches (lamp at the focus)
parabola	path under constant gravity	a ball or a jet of water, if air resistance is ignored
ellipse	orbits under gravity, with the Sun at a focus (Kepler's first law)	planets, comets that return, artificial satellites
ellipse	rays from one focus reflect through the other	whispering galleries; lithotripters that break kidney stones
hyperbola	constant difference of distances	locating a sound or radio source from arrival times
hyperbola	shape of a strong, light surface; reflection towards the other focus	cooling towers (hyperboloids); the small second mirror of a Cassegrain telescope

Warning

Make the constant term 1 before reading $e$ . In $r = \frac{10}{5 - 3\cos\theta}$ the eccentricity is $\frac{3}{5}$ , not 3. Reading 3 would wrongly make the curve a hyperbola.

Squaring can add points. Multiplying $r = 6\cos\theta$ by $r$ adds the pole, which is already on the circle here; but always check whether a converted equation describes exactly the same set.

Tip

For any conic $r = \frac{l}{1 \pm e\cos\theta}$ , plug in $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ first. These four values give the vertices on the focal axis and the ends of the latus rectum, which is enough for a quick sketch.

The Organic Chemistry Tutor: Polar Equations of Conic Sections in Polar Coordinates (about 30 min). Watch how the eccentricity and the directrix are read from the denominator, as in worked example 3, and then used to sketch the graph.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

(a) Write $\left(6, \frac{5\pi}{6}\right)$ in Cartesian coordinates. (b) Write $(0, - 5)$ in polar coordinates.
Write in polar form: (a) $y = 3$ (b) $x^{2} + y^{2} = 16$ (c) $x - \sqrt{3}y = 6$ , in the form $r\cos(\theta - \alpha) = p$ .
Find the Cartesian equation of $r = 8\cos\theta$ , and describe the curve.
Identify $r = \frac{12}{3 + 3\cos\theta}$ , give its directrix and vertex, and find its Cartesian equation.
Identify $r = \frac{4}{1 + 2\sin\theta}$ and find its two vertices. (Careful: one of them has $r \lt 0$ .)
Write the polar equation of the conic with a focus at the pole, eccentricity $\frac{1}{2}$ and directrix $x = - 4$ .
For the ellipse $r = \frac{l}{1 + e\cos\theta}$ with $0 \lt e \lt 1$ , prove that the length of the major axis is $\frac{2l}{1 - e^2}$ .

Show answersHide answers

(a) $(-3\sqrt{3}, 3)$ (b) $\left(5, \frac{3\pi}{2}\right)$ , or $\left(5, -\frac{\pi}{2}\right)$
(a) $r\sin\theta = 3$ (b) $r = 4$ (c) $r\cos\left(\theta + \frac{\pi}{3}\right) = 3$ , that is $\alpha = -\frac{\pi}{3}$ , $p = 3$
$x^{2} + y^{2} = 8 x$ , so $(x - 4)^{2} + y^{2} = 16$ : the circle with centre $(4, 0)$ and radius 4.
$r = \frac{4}{1 + \cos\theta}$ : a parabola, directrix $x = 4$ , vertex $(2, 0)$ ; $y^{2} = - 8 (x - 2)$ .
$e = 2$ : a hyperbola with directrix $y = 2$ . $\theta = \frac{\pi}{2}$ : $r = \frac{4}{3}$ , the point $\left(0, \frac{4}{3}\right)$ ; $\theta = \frac{3\pi}{2}$ : $r = - 4$ , the point $(0, 4)$ .
$r = \frac{ed}{1 - e\cos\theta} = \frac{2}{1 - \frac{1}{2}\cos\theta} = \frac{4}{2 - \cos\theta}$
The vertices are at $\theta = 0$ and $\theta = \pi$ , on opposite sides of the pole, so $2a = \frac{l}{1 + e} + \frac{l}{1 - e} = \frac{2l}{1 - e^2}$ .

1Measuring from a focus

2Lines and circles in polar form

3Where conics are used

Measuring from a focus

Lines and circles in polar form

Where conics are used