1121 APPENDIX A Polar Form of Conic Sections We can verify that r = e p 1 + e cos u does indeed satisfy the definition of a conic section. Consider Figure 1, where the directrix is vertical and is p units (where p 70) to the right of the focus F10, 0°2. If we let P1r, u2 be a point on the graph, then the distance between P and the directrix is found as follows. PP′ = p - x = p - r cos u x = r cos u = ` p - a e p 1 + e cos ub cos u ` Use the equation for r. = ` p11 + e cos u2 - e p cos u 1 + e cos u ` Write with a common denominator. = ` p + e p cos u - e p cos u 1 + e cos u ` Distributive property PP′ = ` p 1 + e cos u ` Simplify. Because r = e p 1 + e cos u , we multiply each side by 1 e . r e = p 1 + e cos u We can substitute r e for the expression in the absolute value bars for PP′. PP′ = ` p 1 + e cos u ` = ` r e ` = r e = r e Appendices Equations and Graphs Until now we have worked with equations of conic sections in rectangular form. If the focus of a conic section is at the pole, the polar form of its equation is r = e p 1 te # ƒ1U2 , where ƒ is either the sine or cosine function. A Polar Form of Conic Sections ■ Equations and Graphs ■ Conversion from Polar to Rectangular Form Polar Forms of Conic Sections A polar equation of the form r = e p 1 te cos U or r = e p 1 te sin U has a conic section as its graph. The eccentricity is e (where e 70), and p is the distance between the pole (focus) and the directrix. F(0, 0°) 90° r cos u r y x p 0° P(r, u) P9 Directrix u Figure 1 1121
RkJQdWJsaXNoZXIy NjM5ODQ=