Polar Form of Conic Sections Dr. Shildneck

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

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.

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

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).

Examples Identify the eccentricity, type of conic, and equation of the directrix of each polar equation. 1. 𝑟= 9 3 + 1.5𝑐𝑜𝑠𝜃 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= 3 1 +0.5𝑐𝑜𝑠𝜃 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, 0.5 𝑑=3 Thus, for a conic in this form, the directrix is the vertical line 𝒙=𝟔 𝑑= 3 0.5 =6

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

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 = 8 1 +2 𝑠𝑖𝑛𝜃 So, our solution is: 𝑟= 8 1 +2 𝑠𝑖𝑛𝜃

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

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

Examples Write each polar equation in rectangular form. 6. 𝑟= 3.2 1 −0.6 𝑐𝑜𝑠𝜃

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