Unit 3: Pole, Polar, and Polar Coordinates
Table of Contents
Definition of Pole and Polar
The concept of pole and polar relates a point to a line with respect to a given conic.
- Polar: The **polar** of a point P with respect to a conic is the locus of the points of intersection of tangents drawn at the extremities (ends) of all possible chords of the conic that pass through P. This locus is always a straight line.
- Pole: The point P is called the **pole** of that straight line (the polar).
Property: If the polar of point P passes through point Q, then the polar of point Q passes through point P. Such points are called conjugate points, and their polars are called conjugate lines.
Equation of the Polar of a Point
The equation of the polar of a point P(x₁, y₁) with respect to any second-degree conic is given by the "tangent equation" T = 0.
The rule for finding T = 0 is to make the following substitutions in the conic's equation:
x²→x·x₁y²→y·y₁2xy→x·y₁ + y·x₁2x→x + x₁2y→y + y₁c→c(no change)
- If the point
P(x₁, y₁)is ON the conic, the equationT = 0gives the TANGENT at that point. - If the point
P(x₁, y₁)is OUTSIDE the conic, the equationT = 0gives the CHORD OF CONTACT of tangents drawn from P. - For any point P,
T = 0gives the POLAR of P.
Table of Polar Equations for Pole (x₁, y₁):
| Conic | Equation | Equation of Polar (T = 0) |
|---|---|---|
| Circle | x² + y² = a² | x·x₁ + y·y₁ = a² |
| Parabola | y² = 4ax | y·y₁ = 2a(x + x₁) |
| Ellipse | x²/a² + y²/b² = 1 | x·x₁/a² + y·y₁/b² = 1 |
| Hyperbola | x²/a² - y²/b² = 1 | x·x₁/a² - y·y₁/b² = 1 |
Determination of the Pole of a Straight Line
This is the reverse problem: given a line, find its pole.
Method:
Let the given conic be (for example) the ellipse x²/a² + y²/b² = 1.
Let the given line be Lx + My + N = 0.
We want to find the pole P(x₁, y₁).
- Assume the pole is
(x₁, y₁). - Write the equation of its polar:
x·x₁/a² + y·y₁/b² - 1 = 0 - Compare this polar with the given line. Since they represent the same line, their coefficients must be proportional.
(x₁/a²) / L = (y₁/b²) / M = (-1) / N - Solve for
x₁andy₁:(x₁/a²) / L = -1 / N=>x₁ = -a²L / N(y₁/b²) / M = -1 / N=>y₁ = -b²M / N
So, the pole is (-a²L/N, -b²M/N).
Example:
Question: Find the pole of the line x + 2y + 3 = 0 w.r.t. the parabola y² = 4x.
Solution:
Here a = 1. The parabola is y² = 4x.
1. Let the pole be (x₁, y₁).
2. The polar is y·y₁ = 2a(x + x₁) => y·y₁ = 2(x + x₁)
Rearranging: 2x - y·y₁ + 2x₁ = 0
3. The given line is x + 2y + 3 = 0.
4. Compare coefficients:
2 / 1 = (-y₁) / 2 = (2x₁) / 3
5. Solve:
2 / 1 = (-y₁) / 2 => 4 = -y₁ => y₁ = -4
2 / 1 = (2x₁) / 3 => 6 = 2x₁ => x₁ = 3
The pole is (3, -4).
Polar Equation of a Conic
Definition and Standard Form
A conic section can be defined as the locus of a point P moving in a plane such that the ratio of its distance from a fixed point S (the focus) to its perpendicular distance from a fixed straight line (the directrix) is a constant e (the eccentricity).
To get the polar equation, we place the focus S at the pole (origin). We let the directrix be a line perpendicular to the polar axis (the x-axis) at a distance p from the pole.
The standard form of a conic, with the focus at the pole and the directrix perpendicular to the initial line, is:
l/r = 1 + e cos(θ)
Where:
r, θare the polar coordinates of a point on the conic.eis the eccentricity.- If
e = 1, it's a Parabola. - If
e < 1, it's an Ellipse. - If
e > 1, it's a Hyperbola.
- If
lis the semi-latus rectum (the focal chord perpendicular to the axis, halved).
Variations of the equation:
| Equation | Directrix |
|---|---|
l/r = 1 + e cos(θ) | To the left of the focus (e.g., x = -p) |
l/r = 1 - e cos(θ) | To the right of the focus (e.g., x = p) |
l/r = 1 + e sin(θ) | Below the focus (e.g., y = -p) |
l/r = 1 - e sin(θ) | Above the focus (e.g., y = p) |
Equation of a Chord
Let the conic be l/r = 1 + e cos(θ).
The equation of the chord joining two points on the conic with vectorial angles α and β is:
l/r = e cos(θ) + sec((β - α)/2) * cos(θ - (α + β)/2)
A simpler form, often used, is for a chord joining points with angles (α - β) and (α + β):
l/r = e cos(θ) + sec(β) * cos(θ - α)
Equation of a Tangent
To find the equation of the tangent at a point with vectorial angle α, we take the chord equation and let the two points approach each other (i.e., set β = 0 in the second form).
The equation of the tangent at the point with vectorial angle α on the conic l/r = 1 + e cos(θ) is:
l/r = e cos(θ) + cos(θ - α)
Unit 3: Exam Quick Tips
- The T=0 Rule is Universal: Memorize the substitution rule (
x² → x·x₁, etc.). It gives the Tangent (if point is on conic) and the Polar (for any point). - Finding the Pole: Always use the "compare coefficients" method. It's systematic and works for all conics.
- Polar Conic Equation:
l/r = 1 + e cos(θ). Know this form.e = 1is Parabola.e < 1is Ellipse.e > 1is Hyperbola.
- Polar Tangent Equation:
l/r = e cos(θ) + cos(θ - α). This is a very common formula to state or use.