Chapter 13 Parametric Equation and Locus Wh the graph is so strange? Let s investigate a few points. t -5-4 -3 - -1 0 1 3 4 4 15 8 3 0-1 0 3 8 15-4.04-3.4-3.14 -.91-1.84 0 1.84.91 3.14 3.4 B plotting these points out, we get: What is Parametric Equation? Parametric Equation ( 參數方程 ) is a tpe of graph which its - and -coordinate of a point is determined b a function. A parametric equation has the form: = ( t) = ( t) (t) and (t) are the parametric functions ( 參數函數 ) of the graph, and t is called the parameter ( 參數 ). Eample: The parametric equation = t 1 and = t + sin t has the following form: 10 0 10 0 If we connect these points together, we will get that strange graph. Range of parameters 5 0 0 40 60 80 The range of parameters means the interval of t we are allowed to work on. Limiting the range usuall makes to graph not so complete. If we are onl allowed to draw < t < for our graph, it will become: -5 0-10 - 97
Converting between Parametric and Normal Equation Converting from normal to parametric equation The convert the normal equation = f() or f(, ) = 0 into parametric form, the easiest wa is to set = t. Eample: Represent = + parametricall. = t = t + t Of course, it ma be better to substitute for the others. Eample: Represent ( 3) + ( 7) = 9. We set 3 = 3 sin t, because (3 sin t) + (3 cos t) = 9, which ma reduce the compleit of the parametric equation a lot. Thus: ( t) ( ) 3sin + = 9 ( ) ( ) = 9 9sin t = 9cos t = 3cost = 3cost+ Thus, the parametric form of this equation is: = 3+ 3sin t = + 3cost 4 Converting from parametric to normal equation The main job is to eliminate the parameter t, which gives = f() or f(, ) = 0 at last. = t + t Eample: Eliminate the parameter in = t 1 Obviousl t = + 1. Substitute this into the first equation to give = ( + 1) + ( + 1). After epanding gives the normal form = + 3 +. This ma not be as eas as converting from normal to parametric. Sometimes the parameter cannot be eliminated, or the result is ver complicated. Take our strange graph in page 97 as eample. In normal form, the graph is: Another eample is the ccloid ( 擺線 ): Which in normal form the equation is: ± = 1+ + sin 1+. = t - sin t = 1 - cos t 0 π π ( ) 1 π n± cos 1 ± = The two ± signs are independent from each other, and n is an integral constant. We see that converting a parametric equation to normal can be etremel wearisome ( 厭煩 )! 0 4 6 98
What is Locus? Eample: A (3,0) and B (6, 0) are two points. A point P moves in a wa that OPB = 90, where O (0, 0) is the origin. Find the equation of the locus of the mid-point M of AP. Locus ( 軌跡 ) is a path that a point runs from one position to another position under some conditions. If we throw a ball at an angle, and record the position of the ball ever fied time period, we will get the locus of the ball: M P 0 O A B We have two variable points here, P and M. But the movement of M is relative to P, so we don t need to care about it so much. 0 4 6 8 10 The green line connecting all the position of the balls (red dots) is the locus of the ball. Equation of locus ( 軌跡方程 ) The equation of locus is the equation representing the locus. The equation of locus of the above is = 5 ( 5). Eample: Find the equation of locus made b a point P moving around (3, 4) at a distance of 5 units. Assume P = (, ). We have: ( ) ( ) 3 + 4 = 5 ( ) ( ) 3 + 4 = 5 + 6 8 = 0 It is the required equation. 0 Assume P = (, ), M = (, ). Thus, slope of OP = and slope of PB =. 6 Since OP PB, we have = 1 6 ( ) ( ) = 1 6 ( ) = 6 ( ) ( ) Also, since M is the mid-point of AP, we + 3 have =, =, i.e., = 3, =. Substitute this into (*), we have: 4 + 4 4+ 7= 0 This is the locus of M. ( ) = 6( 3) ( 3) 4 = 1 18 4 9+ 1 1-1 3 4 5 When attacking locus problems, the strateg is to let the moving point that forms the locus to be (, ). 99
Revision In this chapter, we ve learnt: 1. Parametric equations. Converting between parametric and normal equations 3. Locus P Eercise In the followings, if not specified, is the variable for -ais, and is for -ais. Use t as the parameter for parametric equations. 1. Convert the following normal equations into parametric equations: a) + 4 = 0 b) + = 9 c) 9 5 = 5 d) = 4 e) 3 + 3 = 6. Convert the following parametric equations into normal equations: a) = 6t 1, = t + 4 t b) = t+, = t c) = tan t + cot t, = tan t cot t d) = cos 3t + sin 3t, = cos t sin t 3. (HKCEE 1990) S and T are variable points on the lines = 0 and = 0 respectivel, such that ST =. Find the locus of the mid-point of ST. 4. On the coordinate plane, A (10, 0) is a fied point and P is a variable point. Find the loci (plural form of locus) of P such that the area OAP is alwas 100 units. 5. (HKCEE 1994, Modified) Given a curve P: = 8. L 1 : = m 1 + c 1 and L : = m + c are two tangents to P intersecting at point A. (See figure on the right) a) Epress c 1 in terms of m 1. b) Show that the coordinates of A are ((m 1 + m ), m 1 m ). c) If the angle between L 1 and L is θ, epress the locus of A in terms of tan θ. L A L1 100
Suggested Solutions for the Eercise 1a) = t, = t b) = 3 sin t, = 3 cos t c) = 5 sec t, = 3 tan t d) = 1/t, = 4t e) 6t 6t =, = 1+ t 1+ t a) 6 + 5 = 0 b) 4 + = 0 c) 4 = 0 d) 3 + 3 = 0 3) 4 + 5 1 = 0 4) = ± 0 5a) c 1 = - m 1 c) (tan θ ( + )) 8 = 0 101