10.5Conic Curves in Polar Form
Write conics as r = ed/(1 ± e cosθ) or r = ed/(1 ± e sinθ). Identify eccentricity e: e < 1 → ellipse, e = 1 → parabola, e > 1 → hyperbola. Graph and convert to rectangular form.
The polar form of conics unifies all three conic sections into one equation using eccentricity. This form is used in orbital mechanics — Kepler's laws of planetary motion are expressed most naturally in polar coordinates.
Essential Question
How does the unified polar form r = ed/(1±ecosθ) or r = ed/(1±esinθ) describe all conic sections, and how does eccentricity e determine which type of conic we get?
Lesson Overview
Focus-Directrix Definition in Polar Form
Place the focus at the pole. The directrix is perpendicular to the polar axis at distance d from the pole. The unified polar equations are:
- r = ed/(1 + ecosθ) — directrix right of pole (x = d); conic opens left
- r = ed/(1 − ecosθ) — directrix left of pole (x = −d); conic opens right
- r = ed/(1 + esinθ) — directrix above pole (y = d); conic opens down
- r = ed/(1 − esinθ) — directrix below pole (y = −d); conic opens up
Eccentricity determines conic type:
- 0 < e < 1 → Ellipse
- e = 1 → Parabola
- e > 1 → Hyperbola
d = distance from focus (pole) to directrix.
Finding vertices: substitute θ = 0, π, π/2, 3π/2 to find r values.
Converting: use r² = x²+y², rcosθ = x, rsinθ = y, r = √(x²+y²).
Worked Examples
Identify the conic and find key features: r = 6/(1 + 2cosθ).
Form: r = ed/(1+ecosθ); e = 2, ed = 6 → d = 3
e = 2 > 1 → Hyperbola
Directrix: x = 3 (right of pole)
Vertices: θ=0: r=6/3=2; θ=π: r=6/(1−2)=−6 (r=6, opposite direction)
Center is midpoint of vertices
Identify the conic: r = 4/(1 − sinθ).
Form: r = ed/(1−esinθ); e = 1, ed = 4 → d = 4
e = 1 → Parabola
Directrix: y = −4 (below pole)
Vertex: θ=3π/2: r=4/(1−(−1))=4/2=2; vertex at (2, 3π/2)
Write the polar equation of a parabola with focus at origin and directrix x = −3.
Directrix left of pole → form r = ed/(1−ecosθ)
e = 1 (parabola), d = 3
r = (1)(3)/(1−cosθ) = 3/(1−cosθ)
Identify and find vertices: r = 10/(5 + 4cosθ).
Divide by 5: r = 2/(1 + (4/5)cosθ)
e = 4/5 < 1 → Ellipse; ed = 2 → d = 2/(4/5) = 5/2
Vertices: θ=0: r=10/9; θ=π: r=10/1=10
Semi-major axis a = (10/9+10)/2 = (10/9+90/9)/2 = 100/18 = 50/9
Convert r = 8/(2 + 2sinθ) to rectangular form.
Divide by 2: r = 4/(1+sinθ); e=1 → parabola
r(1+sinθ) = 4 → r + rsinθ = 4
r = 4 − y (since rsinθ = y)
√(x²+y²) = 4 − y
x²+y² = (4−y)² = 16 − 8y + y²
x² = 16 − 8y → x² = −8(y−2)
Guided Practice
Identify the conic and find e and d: r = 12/(3 + 3cosθ).
Hint: Divide numerator and denominator by 3 first to get the standard form r = ed/(1+ecosθ).
Identify the conic: r = 15/(3 − 5sinθ).
Hint: Divide by 3 to get standard form. Compare e to 1.
Write the polar equation of an ellipse with e = 1/2 and directrix y = 6.
Hint: Directrix above pole → form r = ed/(1+esinθ). Substitute e=1/2, d=6.
Find the vertices of r = 9/(3 + 6sinθ) by substituting θ = π/2 and θ = 3π/2.
Hint: Divide by 3 first: r = 3/(1+2sinθ). Then substitute.
Convert r = 6/(1 + cosθ) to rectangular form.
Hint: Multiply both sides by (1+cosθ): r + rcosθ = 6. Use rcosθ = x and r = √(x²+y²).
Key Vocabulary
Polar form of a conic
r = ed/(1±ecosθ) or r = ed/(1±esinθ); focus at the pole
Pole
The origin of the polar coordinate system; coincides with the focus of the conic
Eccentricity (e)
Determines conic type: 0<e<1 ellipse, e=1 parabola, e>1 hyperbola
Directrix
The fixed line used in the focus-directrix definition; distance d from the pole
Polar axis
The reference direction (positive x-axis) in polar coordinates
Vertex (polar)
Found by substituting θ = 0, π, π/2, or 3π/2 into the polar equation
Unified polar form
A single equation form that represents all three conic types based on e
Quick Check
Interactive Practice — 5 Questions
For r = 5/(1 + cosθ), the conic is a:
For r = 12/(3 + 4cosθ), after dividing by 3, e =
r = ed/(1 − esinθ) has its directrix:
An ellipse in polar form has e =
For r = 6/(2 + sinθ), dividing by 2 gives e =
Independent Practice
Independent Practice
Identify the conic and find e and d: r = 8/(2 + 4cosθ).
Identify the conic: r = 6/(3 − 3sinθ).
Write the polar equation: parabola, focus at origin, directrix y = 4.
Find the vertices of r = 12/(4 + 8cosθ).
Convert r = 4/(1 − sinθ) to rectangular form.
Common Mistakes
Common Mistakes
Forgetting to divide by the constant to get standard form before identifying e
Always rewrite as r = ed/(1±ecosθ) with coefficient 1 in the denominator. Divide numerator and denominator by the constant term.
Confusing which form (cosθ vs sinθ) corresponds to which directrix orientation
cosθ → vertical directrix (left/right of pole); sinθ → horizontal directrix (above/below pole).
Thinking e is the coefficient of cosθ or sinθ before dividing
e is the coefficient AFTER dividing to get standard form. In r=12/(3+4cosθ), e=4/3 (not 4).
Using θ=0 and θ=π/2 to find vertices for a sinθ form
For cosθ forms, use θ=0 and θ=π for vertices. For sinθ forms, use θ=π/2 and θ=3π/2.
Math Tips
Math Tips
Always divide first: get the denominator to start with 1 before reading off e and d.
Eccentricity memory: e<1 Ellipse, e=1 Parabola, e>1 Hyperbola. Think 'EPA' — Ellipse, Parabola, Hyperbola.
The sign in the denominator tells you the directrix location: + cosθ → directrix right; − cosθ → directrix left; + sinθ → directrix above; − sinθ → directrix below.
To convert to rectangular: use r²=x²+y², rcosθ=x, rsinθ=y, and r=√(x²+y²).
Vertices are the closest and farthest points from the focus. For r=ed/(1+ecosθ): closest at θ=0, farthest at θ=π.