1. MTH 253 Calculus (Other Topics) Chapter 10 – Conic Sections and Polar Coordinates Section 10.8 – Conic Sections in Polar Coordinates Copyright © 2009 by Ron Wallace, all rights reserved.

2. Conics - Reminder  The cross product term (Bxy) can be removed by rotation of axis where

3. Conics - Reminder  Nothing: AC > 0 & D 2 /(4A) + E 2 /(4C) – F < 0 C = E = 0 & D 2 – 4AF < 0 A = D = 0 & E 2 – 4CF < 0  Point: AC > 0 & F = D 2 /(4A) + E 2 /(4C)  Line(s): A = 0, C = 0, & D and/or E ≠ 0 C = E = 0 & D 2 – 4AF ≥ 0 A = D = 0 & E 2 – 4CF ≥ 0 A > 0, C < 0, & F = D 2 /(4A) – E 2 /(4C) A 0, & F = –D 2 /(4A) + E 2 /(4C) NOTE: These are known as the “degenerate” cases.

4. Conics - Reminder  Parabola: A = 0 & C ≠ 0 A ≠ 0 & C = 0  Circle: A = C ≠ 0  Ellipse: AC > 0 & A ≠ C  Hyperbola: AC < 0 NOTE: These assume non-degenerate cases.

5. Conics in Polar Coordinates  Some applications that use conics, especially astronomy, work better with polar coordinates.

6. Lines in Polar Coordinates  Lines through the pole (i.e. origin)

7. Lines in Polar Coordinates  Lines NOT through the pole Using the right triangle … NOTE: The blue line is perpendicular to the red line.

8. Converting Polar Lines to Cartesian Lines

9. Converting Cartesian Lines to Polar Lines 1.Find the point of intersection of these two lines. 2.Convert that point into polar coordinates: (r 0, 0 ) 3.Give the polar equation …

10. Circles in Polar Coordinates Using the law of cosines … If the circle passes through the pole, r 0 = a …

11. Circles in Polar Coordinates Circles through the pole with the center on the x-axis. Circles through the pole with the center on the y-axis.

12. Reminder: The Focus-Directrix Property of Conics  Given a point F (focus) a line not containing F (directrix) a constant e (eccentricity)  A conic is the set of all points P where PF = e · PD e=1  parabola 0<e<1  ellipse e>1  hyperbola F P D

13. Polar Equations of Conics  For polar equations of conics focus at the pole (i.e. origin) directrix a vertical line: x = k > 0  PF = r  PD = k – rcos  Therefore, since PF = e · PD … r = e(k - rcos)  Solving for r … PF = e · PD e=1  parabola 0 1  hyperbola F P D k (r,)

14. Examples …  Describe the graphs of the following equations (type of conic, directrix, intercepts, vertices) PF = e · PD e=1  parabola 0 1  hyperbola F P D k (r,)

15. Polar Equations of Conics  Other orientations … Directrix: x = –k Directrix: y = k Directrix: y = –k

16. More Examples  Describe the graphs of the following equations (type of conic, directrix, intercepts, vertices)