Polar Equations of Conics, precalculus lesson

A comet’s path around the sun is a conic with the sun at one focus, and the natural way to describe it is by how far the comet is from the sun in each direction. That is exactly what a polar equation records. Put the focus at the pole and every ellipse, parabola and hyperbola gets an equation of the same short shape, with two numbers in it that tell you the type of curve and where its directrix is.

Key terms

Focus-directrix property

Definition Let $F$ be a point, let $L$ be a line that does not pass through $F$ , and let $e > 0$ . The set of all points $P$ in the plane with

\lvert PF \rvert = e \cdot \operatorname{dist}(P, L),

where $\operatorname{dist}(P, L)$ is the perpendicular distance from $P$ to $L$ , is a conic with focus $F$ , directrix $L$ and eccentricity $e$ . It is an ellipse when $0 < e < 1$ , a parabola when $e = 1$ and a hyperbola when $e > 1$ .

In plain words Fix a point and a line. Collect every point whose distance to the point is a fixed multiple of its distance to the line. The points you collect form a conic, and the multiple decides which kind.

Example Take $F$ at the origin, the directrix $x = 3$ and $e = \frac{1}{2}$ . A point $(x, 0)$ on the $x$ -axis is on the conic when $\lvert x \rvert = \frac{1}{2}\lvert 3 - x \rvert$ . Points with $x \ge 3$ never work, since there the left side is larger. For $0 \le x < 3$ the condition reads $x = \frac{1}{2}(3 - x)$ , so $x = 1$ . For $x < 0$ it reads $-x = \frac{1}{2}(3 - x)$ , so $x = -3$ . The conic crosses the axis at $(1, 0)$ and $(-3, 0)$ . The first point is $1$ from the focus and $2$ from the directrix, the second is $3$ from the focus and $6$ from the directrix.

Focus at the pole

Definition Let $e > 0$ and $d > 0$ . The conic with focus at the pole, directrix $x = d$ and eccentricity $e$ is the graph of the polar equation

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

where $\theta$ runs over all angles with $1 + e\cos\theta \ne 0$ .

Here is why. A point $P$ with polar coordinates $(r, \theta)$ has $x = r\cos\theta$ , its distance to the pole is $\lvert r \rvert$ , and its distance to the directrix is $\lvert d - r\cos\theta \rvert$ . A point on the same side of the directrix as the focus can be named with $r > 0$ , and then the focus-directrix condition reads $r = e(d - r\cos\theta)$ . Solving for $r$ gives the formula. A hyperbola also has a branch beyond the directrix, and its points are named with a negative radius, $r < 0$ . For such a point $d - r\cos\theta$ is negative too, and the condition $\lvert r \rvert = e\lvert d - r\cos\theta \rvert$ again becomes $r = e(d - r\cos\theta)$ . Conversely, taking absolute values in $r = e(d - r\cos\theta)$ gives back the focus-directrix condition, so the graph contains no extra points.

In plain words With the focus at the pole, the distance to the focus is just $\lvert r \rvert$ , with no square root. That is why the whole conic fits into one fraction, a constant over $1 + e\cos\theta$ .

Example Use the data of the first example, $e = \frac{1}{2}$ and directrix $x = 3$ . Then $ed = \frac{3}{2}$ and

r = \frac{3/2}{1 + \frac{1}{2}\cos\theta} = \frac{3}{2 + \cos\theta}.

At $\theta = 0$ this gives $r = 1$ , the point $(1, 0)$ . At $\theta = \pi$ it gives $r = 3$ , and the polar point $(3, \pi)$ is the rectangular point $(-3, 0)$ . These are the two axis points found before.

Eccentricity

Definition The eccentricity $e$ of a conic is the constant ratio

e = \frac{\lvert PF \rvert}{\operatorname{dist}(P, L)}

from the focus-directrix property. In the polar equation of the previous section it is the coefficient of $\cos\theta$ , and you can read it off once the constant term of the denominator is $1$ . For an ellipse or a hyperbola with center to vertex distance $a$ and center to focus distance $c$ , this ratio equals $c/a$ , the eccentricity from the lessons on ellipses and hyperbolas. A circle is the limiting case $e = 0$ . It has no directrix, so it has no equation of this form, and a circle centered at the pole is simply $r = a$ .

In plain words The eccentricity says how far the conic is from round. Below $1$ the curve closes up into an ellipse, exactly $1$ gives a parabola, and above $1$ it opens into a hyperbola.

Example The ellipse $r = \frac{3}{2 + \cos\theta}$ from above has $e = \frac{1}{2}$ and directrix $x = 3$ . Wherever you stand on it, your distance to the pole is half your distance to the directrix.

$\theta$	$r$	Point $P$	$\lvert PF \rvert$	Distance to $x = 3$	Ratio
$0$	$1$	$(1, 0)$	$1$	$2$	$\frac{1}{2}$
$\frac{\pi}{3}$	$\frac{6}{5}$	$(\frac{3}{5}, \frac{3\sqrt{3}}{5})$	$\frac{6}{5}$	$\frac{12}{5}$	$\frac{1}{2}$
$\frac{\pi}{2}$	$\frac{3}{2}$	$(0, \frac{3}{2})$	$\frac{3}{2}$	$3$	$\frac{1}{2}$
$\frac{2\pi}{3}$	$2$	$(-1, \sqrt{3})$	$2$	$4$	$\frac{1}{2}$
$\pi$	$3$	$(-3, 0)$	$3$	$6$	$\frac{1}{2}$

Directrix distance

Definition The directrix distance $d > 0$ is the distance from the focus to the directrix. When the focus is at the pole and the equation is written as

r = \frac{k}{1 + e\cos\theta}, \qquad k > 0,

with constant term $1$ in the denominator, the numerator is $k = ed$ , so

d = \frac{k}{e}.

The numerator has a meaning of its own. At $\theta = \frac{\pi}{2}$ the equation gives $r = ed$ , so $ed$ is the distance from the focus to the conic measured perpendicular to the axis. In the other three forms of the next section $ed$ is again this distance, reached at the angles where the denominator equals $1$ .

In plain words First make the constant term of the denominator equal to $1$ . The numerator is then $e$ times $d$ , not $d$ itself, so divide it by $e$ to find how far the directrix is from the focus.

Example Take

r = \frac{20}{4 + 5\cos\theta}.

Dividing top and bottom by $4$ gives

r = \frac{5}{1 + \frac{5}{4}\cos\theta},

so $e = \frac{5}{4}$ and $ed = 5$ , which gives $d = 4$ . The curve is a hyperbola with directrix $x = 4$ . As a check, at $\theta = \frac{\pi}{2}$ the point is $(0, 5)$ . It is $5$ from the focus and $4$ from the line $x = 4$ , a ratio of $\frac{5}{4}$ .

Position of the directrix

Definition Let $e > 0$ and $d > 0$ and put the focus at the pole. When the directrix is parallel to a coordinate axis, it is one of the four lines at distance $d$ from the pole in the table, and each one gives its own form.

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

The axis of the conic is the line through the focus perpendicular to the directrix. In every form, once the constant term of the denominator is $1$ , the eccentricity is the size of the coefficient of $\cos\theta$ or $\sin\theta$ , and the numerator is $ed$ . A directrix can also be tilted. If the point of the directrix closest to the pole lies in the direction $\theta = \varphi$ , the equation is

r = \frac{ed}{1 + e\cos(\theta - \varphi)},

and the four rows of the table are the cases $\varphi = 0$ , $\pi$ , $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , in that order.

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

Example Find the parabola with focus at the pole and directrix $y = -4$ . Here $e = 1$ and $d = 4$ , and the directrix is horizontal and below the pole, so

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

At $\theta = \frac{3\pi}{2}$ we get $r = 2$ , the point $(0, -2)$ , halfway between the focus and the directrix. That is the vertex. At $\theta = 0$ we get $r = 4$ , the point $(4, 0)$ , which is $4$ from the focus and $4$ from the line $y = -4$ . The denominator is zero at $\theta = \frac{\pi}{2}$ , so the parabola opens upward, away from its directrix.

Vertices and axes from the polar equation

Definition For

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

the axis of the conic is the $x$ -axis, and the vertices come from $\theta = 0$ and $\theta = \pi$ . When $e \ne 1$ ,

r(0) = \frac{ed}{1 + e}, \qquad r(\pi) = \frac{ed}{1 - e},

and the vertices are the points $(r(0), 0)$ and $(-r(\pi), 0)$ in rectangular coordinates. For an ellipse $r(\pi) > 0$ , so the second vertex lies on the negative $x$ -axis. For a hyperbola $r(\pi) < 0$ , so both vertices lie on the positive $x$ -axis. In both cases the center is the midpoint of the vertices, $a$ is half the distance between them, $c$ is the distance from the center to the pole, and $e = c/a$ . For a parabola only $\theta = 0$ gives a vertex, the point $(\frac{d}{2}, 0)$ . The other three forms work the same way along their own axis, with $\theta = 0$ and $\theta = \pi$ for cosine and $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$ for sine.

In plain words Plug in the two angles that point along the axis to get the two vertices. From those two points, the center, $a$ and $c$ are plain distances on a number line.

Example Take

r = \frac{16}{5 + 3\cos\theta} = \frac{16/5}{1 + \frac{3}{5}\cos\theta}.

So $e = \frac{3}{5}$ , an ellipse, and

d = \frac{16}{5} \div \frac{3}{5} = \frac{16}{3}.

Now $r(0) = \frac{16}{8} = 2$ and $r(\pi) = \frac{16}{2} = 8$ , so the vertices are $(2, 0)$ and $(-8, 0)$ . The major axis has length $10$ , so $a = 5$ , and the center is $(-3, 0)$ . The pole is $3$ from the center, so $c = 3$ , and indeed $c/a = \frac{3}{5} = e$ . Finally $b = \sqrt{a^2 - c^2} = 4$ .

Converting to rectangular form

Definition To turn a polar conic into a rectangular equation, clear the denominator, replace $r\cos\theta$ by $x$ (or $r\sin\theta$ by $y$ ), isolate $r$ , and square using $r^2 = x^2 + y^2$ . For the form with $1 + e\cos\theta$ the result is

x^2 + y^2 = e^2(d - x)^2.

Squaring adds no new points here, because this equation says exactly that the distance to the pole is $e$ times the distance to the line $x = d$ .

In plain words Trade $r$ and $\theta$ for $x$ and $y$ , then tidy up by completing the square. The familiar rectangular form shows the center and the axes at once.

Example Continue with the ellipse $r = \frac{16}{5 + 3\cos\theta}$ . Clearing the denominator gives $5r + 3r\cos\theta = 16$ , so $5r = 16 - 3x$ . Squaring,

25(x^2 + y^2) = 256 - 96x + 9x^2,

which collects to $16x^2 + 96x + 25y^2 = 256$ . Completing the square gives $16(x + 3)^2 + 25y^2 = 400$ , or

\frac{(x + 3)^2}{25} + \frac{y^2}{16} = 1.

This is the ellipse with center $(-3, 0)$ , $a = 5$ and $b = 4$ , just as the vertices predicted.

Check yourself

Find the eccentricity, the directrix and the vertices of $r = \frac{15}{2 + 3\cos\theta}$ . What kind of conic is it?

Show the answer

Dividing top and bottom by $2$ gives

r = \frac{15/2}{1 + \frac{3}{2}\cos\theta},

so $e = \frac{3}{2}$ and $ed = \frac{15}{2}$ , which gives $d = 5$ . It is a hyperbola with directrix $x = 5$ . At $\theta = 0$ the value $r = 3$ gives the vertex $(3, 0)$ . At $\theta = \pi$ the value is $r = -15$ , which names the point $(15, 0)$ . The vertices are $(3, 0)$ and $(15, 0)$ . As a check, the center is $(9, 0)$ , so $a = 6$ and $c = 9$ , and indeed $c/a = \frac{3}{2}$ .

Write a polar equation for the conic with focus at the pole, eccentricity $\frac{2}{3}$ and directrix $y = -6$ . What kind of conic is it?

Show the answer

The directrix is horizontal and below the pole, so the denominator is $1 - e\sin\theta$ . With $ed = \frac{2}{3} \cdot 6 = 4$ ,

r = \frac{4}{1 - \frac{2}{3}\sin\theta} = \frac{12}{3 - 2\sin\theta}.

It is an ellipse, since $e < 1$ .

On the parabola $r = \frac{6}{1 - \cos\theta}$ , find the point with $\theta = \frac{\pi}{2}$ and check that it is as far from the focus as from the directrix.

Show the answer

Here $e = 1$ and $d = 6$ , and the minus sign with cosine puts the directrix at $x = -6$ . At $\theta = \frac{\pi}{2}$ the value is $r = 6$ , the point $(0, 6)$ . Its distance to the pole is $6$ and its distance to the line $x = -6$ is also $6$ .

Polar Equations of Conics

Key terms

Focus-directrix property

Focus at the pole

Eccentricity

Directrix distance

Position of the directrix

Vertices and axes from the polar equation

Converting to rectangular form

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus