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topics_11_6_12_10_13_1

# topics_11_6_12_10_13_1 - Topics 11.6 12.10 13.1 11.6 Conic...

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Topics 11.6; 12.10; 13.1 11.6. Conic Sections in Polar Coordinates The conics are parabola, ellipse, and hyperbola. Parabola can be defined in terms of a focus and a directrix. Ellipse and hyperbola can be defined in terms of two foci. Here we give a unified approach to all these 3 types of conic sections in terms of only a Focus and a Directrix. We shall place the Focus at the Origin and then, the Conic Sections will have simple Polar Equations that are convenient to describe motions of planets, satellites, etc. Theorem. Let F be a fixed point (called FOCUS ) and l be a fixed line (called DIRECTRIX and let e be a fixed positive number (called ECCENTRICITY ). The set of all point P on the plane such that | PF | | P l | = e is a conic section and the points on each conic section can be described in this way. Moreover, if e < 1 then the conic is an ellipse ; if e = 1, then the conic is a parabola ; if e > 1 then the conic is a hyperbola. Proof. Without any restriction we place the focus F at origin and take as directrix the line x = d that is parallel to the y - axis and d units to the right of the y -axis. 1

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| PF | = r ; | Pl | = d - r cos( θ ); | PF | | Pl | = r d - r cos( θ ) = e r = e ( d - r cos( θ )) r = ed 1 + e cos( θ ) the polar equation of the conic . 2
Now our next considerations will be based on r = e ( d - r cos( θ )) . To take a Cartesian equation of the conic we square both sides: x 2 + y 2 = e 2 ( d 2 - 2 dx + x 2 ) (1 - e 2 ) x 2 + 2 de 2 x + y 2 = e 2 d 2 . Case 1: e=1. 2 dx + y 2 = d 2 - 2 d x - d 2 ! = y 2 and this is a parabola with a focus at the origin and a directrix x = d . Case 2: e < 1. We take (1 - e 2 ) x 2 + 2 de 2 x + y 2 = e 2 d 2 divide both sides by (1 - e 2 ) and after completing the square with respect to x we obtain: x + e 2 d 1 - e 2 2 + y 2 1 - e 2 = e 2 d 2 (1 - e 2 ) 2 ( * ) and this is an ellipse ( x - h ) 2 a 2 + y 2 b 2 = 1 centered at ( h, 0) = ( - e 2 d 1 - e 2 , 0) with a 2 = e 2 d 2 / (1 - e 2 ) 2 and b 2 = e 2 d 2 / (1 - e 2 ). The foci of the ellipse must be at a distance c from its center, where c 2 = a 2 - b 2 = e 4 d 2 / (1 - e 2 ) 2 . Hence, c = e 2 d 1 - e 2 and e = c a and the focus defined in the Theorem means the same as focus defined in Section 1.5.

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