The concept of pole and polar relates a point to a line with respect to a given conic.
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.
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)P(x₁, y₁) is ON the conic, the equation T = 0 gives the TANGENT at that point.P(x₁, y₁) is OUTSIDE the conic, the equation T = 0 gives the CHORD OF CONTACT of tangents drawn from P.T = 0 gives 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 |
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).
(x₁, y₁).x·x₁/a² + y·y₁/b² - 1 = 0
(x₁/a²) / L = (y₁/b²) / M = (-1) / N
x₁ and y₁:
(x₁/a²) / L = -1 / N => x₁ = -a²L / N(y₁/b²) / M = -1 / N => y₁ = -b²M / NSo, the pole is (-a²L/N, -b²M/N).
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).
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, θ are the polar coordinates of a point on the conic.e is the eccentricity.
e = 1, it's a Parabola.e < 1, it's an Ellipse.e > 1, it's a Hyperbola.l is 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) |
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)
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)
x² → x·x₁, etc.). It gives the Tangent (if point is on conic) and the Polar (for any point).l/r = 1 + e cos(θ). Know this form.
e = 1 is Parabola.e < 1 is Ellipse.e > 1 is Hyperbola.l/r = e cos(θ) + cos(θ - α). This is a very common formula to state or use.