Simplifying Calculations and Proofs with Polar Coordinates

In one sentence

When the segments in a problem all start from the same point and are tied to angles, make that point the pole. A segment length is then just a radius $ρ$ , “turning through an angle” is adding a number to $θ$ , and relations between lengths become sums and differences of trigonometric expressions.

For a conic, the best pole is a focus: ellipse, parabola and hyperbola all share one equation

ρ = \frac{e p}{1 - e cos θ} (e the eccentricity, p the distance from focus to directrix) .

The reciprocals of a focal chord’s two pieces have a constant sum

Step 1 / 4

The focus as pole
Take the focus $F$ as the pole, and as polar axis the ray from $F$ perpendicular to the directrix $l$ (pointing away from it). The distance from a point of the curve to $F$ is just $ρ$ .

$ρ = \frac{e p}{1 - e cos θ}$
The two ends of the chord
Take reciprocals and add
The chord length comes for free

In rectangular coordinates the same problem means solving a system, using Vieta’s formulas, then taking square roots and simplifying. Polar coordinates skip the “solve the system” step: the angle gives the positions of $A$ and $B$ directly.

Predict first

$A B$ is a focal chord of the parabola $y^{2} = 4 x$ with $∣ F A ∣ = 3$ . What is $∣ F B ∣$ ?

Choose an answer first; then you can see the result.

Stop 1 · Length of a focal chord

How long is the chord of the parabola $y^{2} = 4 x$ through the focus with angle of inclination $\frac{π}{4}$ ?

One problem, several solutions

For the parabola $y^{2} = 4 x$ , the chord $A B$ through the focus $F$ has angle of inclination $\frac{π}{4}$ . Find $∣ A B ∣$ .

Rectangular: solve the system + Vieta

$F (1, 0)$ , and the line is $y = x - 1$ . Substituting gives $x^{2} - 6 x + 1 = 0$ , so $x_{1} + x_{2} = 6$ . By the definition of the parabola, $∣ F A ∣ = x_{1} + 1$ and $∣ F B ∣ = x_{2} + 1$ , so $∣ A B ∣ = x_{1} + x_{2} + 2 = 8$ .

CostYou must solve a system and remember “focal radius = x + 1”.

Polar: one line

$e = 1$ , $p = 2$ , $∣ A B ∣ = \frac{2 e p}{1 - e ^{2} cos ^{2} α} = \frac{4}{1 - \frac{1}{2}} = 8$ . For a parabola this is $∣ A B ∣ = \frac{2 p}{sin ^{2} α}$ .

CostYou only need to recognize $e$ and $p$ .

Stop 2 · Two perpendicular focal chords

Replace $α$ by $α + \frac{π}{2}$ , and $cos$ turns into $sin$ :

\frac{1}{∣ A B ∣} + \frac{1}{∣ C D ∣} = \frac{( 1 - e ^{2} cos ^{2} α ) + ( 1 - e ^{2} sin ^{2} α )}{2 e p} = \frac{2 - e ^{2}}{2 e p} .

Turn

α

: as one chord grows the other shrinks, and the sum of reciprocals stays the same. Now push

e

past

1

. When does this readout turn into “!”?

Once again it is “ $cos^{2} + sin^{2} = 1$ ” at work: polar coordinates turn perpendicularity into two angles that differ by $\frac{π}{2}$ , and the trigonometric identity does the elimination for us.

Stop 3 · The center as pole: two perpendicular radii

Take the center of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ as the pole and substitute $x = ρ cos θ$ , $y = ρ sin θ$ :

\frac{1}{ρ ^{2}} = \frac{cos ^{2} θ}{a ^{2}} + \frac{sin ^{2} θ}{b ^{2}} .

If $O A ⊥ O B$ , the polar angles of $A$ and $B$ differ by $\frac{π}{2}$ , and adding the two equations gives $cos^{2} + sin^{2}$ again:

\frac{1}{∣ O A ∣ ^{2}} + \frac{1}{∣ O B ∣ ^{2}} = \frac{1}{a ^{2}} + \frac{1}{b ^{2}} .

One more consequence: the distance

d

from

O

A B

satisfies

\frac{1}{d ^{2}} = \frac{1}{∣ O A ∣ ^{2}} + \frac{1}{∣ O B ∣ ^{2}}

(the altitude to the hypotenuse of a right triangle), so

d

is constant too: the line

A B

is always tangent to one fixed circle. Press the button to draw that circle.

Practice

Exercise 1Basic

For the parabola $y^{2} = 8 x$ , the chord $A B$ through the focus has angle of inclination $\frac{π}{6}$ . Find $∣ A B ∣$ .

Exercise 2Basic

$F$ is a focus of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ and $A B$ is a chord through $F$ . Find $\frac{1}{∣ F A ∣} + \frac{1}{∣ F B ∣}$ .

Exercise 3Advanced

For the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$ , two perpendicular chords $A B$ and $C D$ are drawn through the left focus. Find $\frac{1}{∣ A B ∣} + \frac{1}{∣ C D ∣}$ .

Exercise 4Challenge

Points $A$ and $B$ on the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ satisfy $O A ⊥ O B$ ( $O$ is the center). Find the minimum area of $△ O A B$ .

In one sentence

The reciprocals of a focal chord’s two pieces have a constant sum

Stop 1 · Length of a focal chord

Rectangular: solve the system + Vieta

Polar: one line

Stop 2 · Two perpendicular focal chords

Stop 3 · The center as pole: two perpendicular radii

Practice

Mistake Museum · Exhibits from this stop

Taking $p$ to be the distance from the directrix to the center

Copying the ellipse formula for a focal chord of a hyperbola, with absolute values