Polar Conics, calculus lesson

A planet’s orbit is an ellipse with the sun at one focus, not at the center. Measuring from the sun is natural, and that is exactly what polar coordinates do. With the focus at the pole, every ellipse, parabola and hyperbola has an equation of the same short form, and one number in it tells you which of the three you have.

Key terms

Focus and directrix

Definition Let $F$ be a fixed point (the focus), $\ell$ a fixed line not through $F$ (the directrix) and $e > 0$ a fixed number (the eccentricity). The set of all points $P$ in the plane with

\frac{\lvert PF \rvert}{\lvert P\ell \rvert} = e,

where $\lvert P\ell \rvert$ is the distance from $P$ to $\ell$ , is a conic section. It is an ellipse if $e < 1$ , a parabola if $e = 1$ and a hyperbola if $e > 1$ .

In plain words Pick a point and a line. Ask for all points whose distance to the point is a fixed multiple of their distance to the line. The multiple decides the shape.

Example With $F = (0, 0)$ , directrix $x = 4$ and $e = 1$ , a point $(x, y)$ with $x < 4$ satisfies $\sqrt{x^2 + y^2} = 4 - x$ . Squaring gives $y^2 = 16 - 8x$ , a parabola opening to the left with vertex $(2, 0)$ , halfway between focus and directrix.

Eccentricity

Definition The eccentricity $e$ is the constant ratio in the focus directrix definition. For an ellipse with semimajor axis $a$ , semiminor axis $b$ and focal distance $c = \sqrt{a^2 - b^2}$ from the center, $e = c/a$ . For a hyperbola with semitransverse axis $a$ , semiconjugate axis $b$ and $c = \sqrt{a^2 + b^2}$ , also $e = c/a$ . A circle is the limiting case $e = 0$ , which has no directrix.

In plain words Eccentricity measures how far a conic is from round. Near $0$ the ellipse is almost a circle, near $1$ it is long and thin.

Example An ellipse with $a = 5$ and $b = 3$ has $c = \sqrt{25 - 9} = 4$ , so $e = \frac{4}{5} = 0.8$ .

Polar equation of a conic with focus at the pole

Definition Put the focus at the pole and let the directrix be the line $x = d$ with $d > 0$ . Then a conic with eccentricity $e > 0$ has the polar equation

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

Here is why. For a point $P$ with polar coordinates $(r, \theta)$ , $r \ge 0$ , on the same side of the directrix as the focus, $\lvert PF \rvert = r$ and $\lvert P\ell \rvert = d - r\cos\theta$ . The condition $r = e(d - r\cos\theta)$ solves to the formula above. For a hyperbola there is also a branch on the far side of the directrix. A point there with polar coordinates $(r, \theta)$ , $r > 0$ , has $\lvert P\ell \rvert = r\cos\theta - d$ , so

r = \frac{ed}{e\cos\theta - 1}.

Its other name $(-r, \theta + \pi)$ satisfies the formula above, because

\frac{ed}{1 + e\cos(\theta + \pi)} = \frac{ed}{1 - e\cos\theta} = -r.

So the far branch is drawn by values of $\theta$ that make $r$ negative.

In plain words Any conic with a focus at the pole has the shape constant over one plus $e$ times a cosine. Read off $e$ from the denominator after making its constant term $1$ .

Example Dividing top and bottom by $5$ gives

r = \frac{10}{5 + 5\cos\theta} = \frac{2}{1 + \cos\theta},

so $e = 1$ and $ed = 2$ gives $d = 2$ . It is a parabola with focus at the pole and directrix $x = 2$ .

Orientation of the directrix

Definition With the focus at the pole, $e > 0$ and $d > 0$ , the four positions of the directrix give these equations.

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

In plain words Cosine means a vertical directrix, sine means a horizontal one. A plus sign puts the directrix on the positive side, a minus sign on the negative side.

Example Take

r = \frac{4}{1 + \sin\theta}.

Here $e = 1$ and $d = 4$ , so the directrix is $y = 4$ . Clearing the denominator gives $r + r\sin\theta = 4$ , so $r = 4 - y$ and $x^2 + y^2 = (4 - y)^2$ . This simplifies to

y = 2 - \frac{x^2}{8},

a parabola opening downward away from its directrix.

Ellipse

Definition Let $d > 0$ . When $0 < e < 1$ , the polar equation

r = \frac{ed}{1 + e\cos\theta}

describes an ellipse with one focus at the pole and major axis along the $x$ -axis. Its vertices are at $\theta = 0$ and $\theta = \pi$ , and in terms of the semimajor axis $a$ ,

ed = a(1 - e^2), \qquad r_{\min} = a(1 - e), \qquad r_{\max} = a(1 + e).

In plain words The denominator is never zero, so $r$ stays finite and the curve closes up. The nearest and farthest points from the focus lie on the major axis.

Example Dividing top and bottom by $3$ gives

r = \frac{10}{3 - 2\cos\theta} = \frac{10/3}{1 - \frac{2}{3}\cos\theta},

so $e = \frac{2}{3}$ , $ed = \frac{10}{3}$ gives $d = 5$ , and the directrix is $x = -5$ . The vertices are $r = 10$ at $\theta = 0$ , the point $(10, 0)$ , and $r = 2$ at $\theta = \pi$ , the point $(-2, 0)$ . So $2a = 12$ , $a = 6$ , the center is $(4, 0)$ and $c = 4$ . Indeed $c/a = 2/3$ , and $b = \sqrt{36 - 16} = 2\sqrt{5}$ .

Parabola

Definition Let $d > 0$ . When $e = 1$ , the equation

r = \frac{d}{1 + \cos\theta}

describes a parabola with focus at the pole and directrix $x = d$ . It is defined for $\theta \neq \pi$ , and $r \to \infty$ as $\theta \to \pi$ , so the parabola opens toward the negative $x$ -axis. Its vertex is at $\theta = 0$ , the point $(\frac{d}{2}, 0)$ .

In plain words Each point is exactly as far from the focus as from the directrix. The curve runs off to infinity in the one direction where the denominator becomes zero.

Example Dividing by $2$ gives

r = \frac{6}{2 + 2\cos\theta} = \frac{3}{1 + \cos\theta},

so $d = 3$ . The vertex is $(\frac{3}{2}, 0)$ . At $\theta = \frac{\pi}{2}$ we get $r = 3$ , the point $(0, 3)$ , which is $3$ from the focus and $3$ from the line $x = 3$ , as it should be.

Hyperbola

Definition Let $d > 0$ . When $e > 1$ , the equation

r = \frac{ed}{1 + e\cos\theta}

describes a hyperbola with one focus at the pole. The denominator is zero when $\cos\theta = -1/e$ , and the asymptotes are parallel to the rays at those two angles. They pass through the center of the hyperbola, not through the pole. Angles with $1 + e\cos\theta > 0$ give the branch around the pole, and angles with $1 + e\cos\theta < 0$ give negative $r$ and draw the other branch.

In plain words Now the denominator does cross zero. Near those angles $r$ blows up and the curve heads off along the asymptotes, and the negative values of $r$ draw the second branch.

Example Take

r = \frac{6}{1 + 2\cos\theta}.

Here $e = 2$ and $ed = 6$ , so $d = 3$ . At $\theta = 0$ , $r = 2$ , the point $(2, 0)$ . At $\theta = \pi$ , $r = -6$ , which is the point $(6, 0)$ . So the vertices are $(2, 0)$ and $(6, 0)$ , the center is $(4, 0)$ , $a = 2$ , $c = 4$ and $c/a = 2$ . The denominator vanishes at $\theta = \frac{2\pi}{3}$ and $\frac{4\pi}{3}$ , and the asymptotes through the center have slopes $\pm\sqrt{3}$ .

Check yourself

Identify the conic $r = \frac{8}{4 + 4\sin\theta}$ and its directrix.

Show the answer

Dividing top and bottom by $4$ gives

r = \frac{2}{1 + \sin\theta},

so $e = 1$ and $d = 2$ . It is a parabola with focus at the pole and directrix $y = 2$ .

Write a polar equation of the ellipse with focus at the pole, eccentricity $\frac{1}{2}$ and directrix $x = 4$ .

Show the answer

With $e = \frac{1}{2}$ and $d = 4$ the first row of the table gives

r = \frac{\frac{1}{2} \cdot 4}{1 + \frac{1}{2}\cos\theta} = \frac{4}{2 + \cos\theta}.

Find $e$ and the directrix of $r = \frac{12}{2 - 6\cos\theta}$ . What kind of conic is it?

Show the answer

Dividing top and bottom by $2$ gives

r = \frac{6}{1 - 3\cos\theta},

so $e = 3$ and $ed = 6$ gives $d = 2$ . The directrix is $x = -2$ and the conic is a hyperbola.

Polar Conics

Key terms

Focus and directrix

Eccentricity

Polar equation of a conic with focus at the pole

Orientation of the directrix

Ellipse

Parabola

Hyperbola

Check yourself

How this lesson connects

Builds on