Coordinate Geometry. = k2 e 2. 1 e + x. 1 e. ke ) 2. We now write = a, and shift the origin to the point (a, 0). Referred to

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Bernard Johnston
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1 Coodinate Geomet Conic sections These ae pane cuves which can be descibed as the intesection of a cone with panes oiented in vaious diections. It can be demonstated that the ocus of a point which moves so that its distance fom a fixed point (the focus) is a constant mutipe (e - the eccenticit) of its distance fom a fixed staight ine (the diectix) is a conic section. If e < 1 we obtain an eipse. If e = 1 we obtain a paaboa. If e > 1 we obtain a hpeboa. See Scientific Ameican Septembe 1977-Mathematica games section p4. Catesian equation Take as the x-axis a ine pependicua to the diectix passing though the focus. Take the oigin to be whee the conic cuts the axis between the focus and diectix. Fom the definition of a conic SP = e P M + (x ek) = e (x + k) + ekx + e k = e + e kx + e k + x ( 1 e ) ke(1 + e)x = 0 If we have a paaboa whee e = 1 then the equation educes to = 4kx. If e 1 we wite the equation in the fom ( 1 e + x ke ) = k e 1 e (1 e) ke We now wite = a, and shift the oigin to the point (a, 0). Refeed to 1 e these new axes the equation becomes a + a (1 e ) = 1 The focus becomes the point ( ae, 0) and the diectix the ine x = a e. Notice that the equation is unchanged if x is epaced be x, so that thee is a second focus at x = (ae, 0) and a second diectix at x = a. e Fo an eipse e < 1 and we wite b = a (1 e ) so the equation becomes a + b = 1. 1

2 Fo a hpeboa e > 1 and we wite b = a (e 1) so the equation becomes a b = 1. Foca distance popeties Eipse (e < 1) Fom the definition S 1 P + S P = ep M 1 + ep M = e(p M 1 + P M ) = em 1 M = e a e = a So the sum of the foca distances is constant. Hpeboa (e > 1) Fom the definition S P S 1 P = ep M ep M 1 = e(p M P M 1 ) = e a e = a Simia S 1 Q S Q = a The Paaboic Mio Suppose a a of ight comes in paae to the x-axis and is efected in a diection equa incined to the tangent. We pove that it passes though the focus. Let the paaboa have equation = 4kx, so S = (k, 0), P = (x, ) d dx = 4k so d dx = k thus tan α 1 = k. Now tan α 3 = x k and α = α 3 α 1 So tan α = tan(α 3 α 1 ) = tan α 3 tan α tan α 3 tan α 1 = k (veif)

3 so α 1 = α So a paae beam of ight wi be efected though the focus. Paametic equations Because a cuve is one-dimensiona we can abe the points b means of a singe ea vaiabe, as in the foowing exampes. Taditiona the ette t is used as the paamete, anaogous with the cuve being taced out in time. Exampes i) x = a + t, = b + mt epesents the staight ine though (a, b) with sope m. ii) x = a cos t, = a sin t epesents the cice of adius a cented at (0,0). We use cos + sin = 1, t coesponds to an ange and so θ is sometimes used. iii) x = a cos t, = b sin t epesents the eipse x a + b epesents an ange but not the ange fom O to P. iv) to paameteise a hpeboa we need to find x a = f(t), that f(t) g(t) = 1. Thee ae sevea possibiities a) x a = 1 ( t + 1 ) t b = 1 ( t 1 ) t = 1 again t b = g(t) so b) x = a sec t = b tan t x c) a = 1 ( e t + e t) = cosh t b = 1 ( e t e t) = sinh t These ae caed hpeboic functions. v) to paameteise the paaboa = 4kx we use x = kt, = kt As t inceases this induces a diection on the cuve. The cuve descibed in the opposite diection can be paameteised b x = kt = kt. 3

4 We egad these two as diffeent cuves (with the same set of points). It is impotant to distinguish the diection in man appications. Poa equation of a conic We want to find the poa equation of a conic with the oigin as focus. + = e(x + k(e + 1)) (1) P S = ep M Conveting to poas gives = e cos θ + ek(e + 1) notice that fom (1) ek(e + 1) is the -vaue when x = 0 Wite = ek(e + 1). The ength = P P. P P is caed the atus ectum. is the semi-atus ectum. Thus we can wite the conic as = 1 e cos θ Note that otations ae eas in poa co-odinates, so the equation = 1 e cos(θ α) is a conic having its axis at an ange α with the initia ine. Notice that when α = π the equation becomes = 1 + e cos θ In the case of an eipse o hpeboa this is equivaent to using the othe focus as an oigin. Notice that if e > 1 we can sometimes have < 0. Athough we noma insist on > 0 in poas, in intepeting poa equations it is often convenient to aow < 0, meaning measued in the othe diection though O. e.g. 1 = 1 cos θ when θ = 0 this gives 1 = 1, = 1. We pot θ = 0, = 1 as the point ( 1, 0). 4

5 When cos θ = 3 4, (θ 41 ) this gives 1 = 1, = 41-5

Chapter 10 Sample Exam

Chapter 10 Sample Exam Chapte Sample Exam Poblems maked with an asteisk (*) ae paticulaly challenging and should be given caeful consideation.. Conside the paametic cuve x (t) =e t, y (t) =e t, t (a) Compute the length of the