1. Polar Form of Conic Sections
Dr. Shildneck

2. Conic Sections
In general, a conic section can be defined as the locus of points such that the distance from a point, P, to the focus and the distance from a point (P) to a fixed line not containing P (the directrix) is a constant ratio.
The constant ratio is called the eccentricity of the conic and is denoted as e.
directrix
𝑃𝐹 𝑃𝑄 =𝑒 P Q F

3. Eccentricity
The type of conic section can be determined by finding its eccentricity. The values of e for each type of conic are listed below. Note, e is always positive.

4. Polar Form of Conics Let our conic have a focus at the origin.
P (x, y) Q
F (0, 0)
Directrix
x = d
Let our conic have a focus at the origin.
Begin with the definition for e.
Rewrite as a constant multiplier.
Find PF and PQ.
Substitute.
Convert to Polar form using conversion formulas.
d - x
𝑥 2 + 𝑦 2
𝑃𝐹 𝑃𝑄 =𝑒
Begin with
Multiply by PQ
𝑃𝐹=𝑒∙𝑃𝑄
The distance PQ is the horizontal distance from d to x
𝑃𝐹=𝑒∙(d – x)
The distance PQ is the slanted distance from (x,y) to (0,0)
𝑥 2 + 𝑦 2 =𝑒∙(d – x)
Substitute 𝑟 2 for 𝑥 2 + 𝑦 2 and 𝑟𝑐𝑜𝑠𝜃 for x
𝑟 2 =𝑒∙(d – r cos𝜃)
Distribute
𝒓= 𝒆d (𝟏+𝒆 cos𝜽)
𝑟=𝑒d –𝑒r cos𝜃
Add to get r’s on one side
𝑟+𝑒r cos𝜃=𝑒d
Factor out r
𝑟(1+𝑒 cos𝜃)=𝑒d
Divide to solve for r

5. Polar Equations of Conics
The conic section with eccentricity e > 0, d > 0, and focus at the pole has the polar equation:
𝒓= 𝒆d (𝟏 + 𝒆 cos𝜽) , when the directrix is the vertical line x = d (right of the pole).
𝒓= 𝒆d (𝟏 − 𝒆 cos𝜽) , when the directrix is the vertical line x = -d (left of the pole).
𝒓= 𝒆d (𝟏 + 𝒆 sin𝜽) , when the directrix is the horizontal line y = d (above the pole).
𝒓= 𝒆d (𝟏 − 𝒆 sin𝜽) , when the directrix is the horizontal line y = −d (below the pole).

6. Examples
Identify the eccentricity, type of conic, and equation of the directrix of each polar equation. 1. 𝑟= 𝑐𝑜𝑠𝜃
Remember, we want to get the equation in the form 𝑟= 𝑒𝑑 1 +𝑒 𝑐𝑜𝑠𝜃
The key number in this problem is the one in the denominator
𝑟= 9 3(1 +0.5𝑐𝑜𝑠𝜃)
Factor
𝑟= 3(3) 3(1 +0.5𝑐𝑜𝑠𝜃)
Simplify and reduce… we want only “1+ecos(Ѳ)” in the denominator
𝑟= 𝑐𝑜𝑠𝜃
From this, we can tell that the eccentricity is e = 0.5.
We also know that ed = 3.
Since the eccentricity is e = 0.5, we know that the equation is for an ellipse. 0<𝑒<1
Next: Since ed = 3, 𝑑=3
Thus, for a conic in this form,
the directrix is the vertical line 𝒙=𝟔
𝑑= =6

7. Examples
Identify the eccentricity, type of conic, and equation of the directrix of each polar equation. 2. 𝑟= 5 4𝑠𝑖𝑛𝜃 − 2

8. Examples
Write the polar equation for each conic section. 3. 𝑒=2 with directrix: 𝑦=4
Since 𝑒=2, we know that the conic is a hyperbola
with a horizontal directrix at 4 and focus at the origin. 4
Now, a conic with a positive horizontal directrix
has the equation: 𝑟= 𝑒𝑑 1 +𝑒 𝑠𝑖𝑛𝜃
Since we know 𝑒=2 and 𝑑=4,
𝑟= 𝑒𝑑 1 +𝑒 𝑠𝑖𝑛𝜃 = 2(4) 1 +2 𝑠𝑖𝑛𝜃 = 𝑠𝑖𝑛𝜃
So, our solution is:
𝑟= 𝑠𝑖𝑛𝜃

9. Examples
Write the polar equation for each conic section. 4. 𝑒= 1 2 with vertices at (-4, 0) and (12, 0)

10. Examples
Write each polar equation in rectangular form. 5. 𝑟= 4 1 − 𝑠𝑖𝑛𝜃

11. Examples
Write each polar equation in rectangular form. 6. 𝑟= −0.6 𝑐𝑜𝑠𝜃

12. ASSIGNMENT Alternate Text (from Blog) Page 726 # 5-10 all #11-19 odd