Precaution: This chapter requires some knowledge in Elementar Geometr. Coordinates Sstem Locating a point: Introduction the coordinates sstem In mathematics and dail life we often need to describe the position of an object. But mostl the description is imprecise, for eample: How would ou describe the location of the gra circle? It is above the red line and on the right of the blue line. But there is a problem: this description occupies a large area, but the circle is onl a small object inside it. That means this method is not suitable to find out the actual position of an object. One solution is to tell eactl how far the object is from the referencing ones. For eample, the gra circle is 0.8 cm above the red line and 1.3 cm on the right of the blue line. This will determine a unique point on the plane and thus the object is accuratel positioned. This is an eample of coordinates sstem. In a coordinates sstem defined on a plane, a point can be defined precisel b a pair of numbers. This pair of numbers is called the coordinates of the point. In our eample above the sstem is called a rectangular coordinates sstem. This sstem is constructed b followings: 1. Create a horizontal real line. Identif the zero on it, and make the positive side points right.. Create another vertical real line that its zero matches the horizontal one. Make the positive side points up. -1 1 0 (, 1) 1 3 4 5 The horizontal real line is called the -ais and is marked with a letter. Similarl, 108
the vertical real line is called the -ais. The intersection of the two aes is called the origin, which is marker with a letter O. In this sstem if we place an object, then we can move this horizontall until reaching the -ais. The point it coincide with the -ais is called the ordinate or -coordinate of the object. Likewise, we can move the object verticall and arrive at the -ais. The point is called the abscissa or -coordinate. For the figure on the last page, the -coordinate of black circle is and the -coordinate is 1. If the - and -coordinate of a point is ζ and η respectivel, then the coordinates of this point are (ζ, η). Conversel, if we consider the coordinates sstem is the set of all points on a plane, then this set is defined as {(, ), R}. However, this is also the definition of R. Therefore, the set R and the coordinates sstem is actuall the same thing. From now on, when we sa R it means the coordinates sstem unless otherwise specified. The coordinates sstem can be divided into four parts b the two aes. Each part is called a quadrant. As shown below, the part {(, ), R + } is called quadrant I, {(, ) R, R + } is called quadrant II, etc. The aes are not included in an quadrant. II III I IV Application of the coordinates sstem* The coordinates sstem is useful in solving geometrical problems and algebraic equations. The first use is obvious. We can plug the geometric figures into the sstem directl. And for algebraic equations, we can draw curves about the equations and find the corresponding intersections in the sstem (to be discussed in the net section). 109
Line Segment Curves on the coordinates sstem Curves are one-dimensional objects that is bent from a line or line segment. Circles, lines are eamples of curves. Eample of curves In coordinates sstem, the shape of a curve can be described b collection of points. For eample, the red points below are some points that can make up a circle, and the green curve is the circle created b infinite points like that. 1 0.5 0 0.5 1 We can sa a curve in the coordinates sstems is a set of points that satisfies some predefined conditions. That means a curve C is a subset of R that is equals to {(, ) f (, ) = 0 } R for some function f: R R. Since onl f is important in how the curve is shaped, this curve is usuall just denoted f(, ) = 0. The following show two curves. The purple one is = (that is the same as = 0) and the green one is =. 110
4-0 Line segments Line segments are a special kind of curve that is not curved at all, and the length is finite. Eamples of line segments It is guaranteed that the shortest path that an object travels from one point to another is a line segment and there is onl one unique line segment for this. That means two points on the coordinates sstem are enough to fi a line segment. Problems involving line segments are usuall how long is that segment? or how steep is that segment? or what is the area of a polgon enclosed b n segments? or something like these. In this section we will go in depth answering these questions. Length of segment ( 1, 1 ) 1 (1.1) (, ) 1 In the figure there is a line segment ended with two points ( 1, 1 ) and (, ). The length of the segment is eas to find. As figured we assume the segment is the hpotenuse of a right triangle as shown. Then b Pthagoras theorem we have: ( ) ( ) 111 Length of segment = + (1.) 1 1
Eample 1a: Show that the points (, 1), (3, 8) and (5, ) forms a right triangle. Solution 1a: Let A = (, 1), B = (3, 8) and C = (5, ). B (11.), we have got: AB= BC = CA= 50 40 10 (1.3) Therefore, b the converse of Pthagoras theorem, ABC is a right triangle. Q.E.D. Slope of segment A measure of steepness is to find out the rise in one horizontal step for a segment. We define this measure to be the slope of the segment. In the following, the red segment has slope 0.5, the blue has slope 1 and green has slope 3. 3 1 However, the horizontal distance of the two end points ma not be 1. To solve this, we can shrink or enlarge the segment to have horizontal distance to be 1. If the original horizontal distance is X, then the rise will be divided b X after the dilation. Referring to the notation in (11.1), the slope of a line segment is defined as: 1 (1.4) 1 But the equation ma give negative value. B definition, a negative slope means the segment is falling : Positive / Rise Slope > 0 1 Slope < 0 Negative / Fall If the slope of a segment is 0, then it is flat, i.e., horizontal. The slope for a vertical segment is not defined, because there can never be an horizontal step. 11
Point of division Given a line segment we can alwas cut it into two different segments. The place of the cut is called the point of division. A common question is, how to determine the point of division if I want to divide the segment into ratio r : s?. This can be easil done with help of coordinate geometr. Eample 1b: Find out the formula for determining the point of division. Solution 1b: First of all, observe that the ratio does not change if we project the segment onto either ais. Thus, we ma find out the point of division of the projection on the two aes to determine the coordinates of the point of division of the segment. r : s r : s Now assume the end points of the green segment are ( 1, 1 ) and (, ), and the point of division is ( 0, 0 ). So, on the blue segment, the point 0 divides the interval [ 1, ] 0 1 r 0 r into ratio r : s. This means = and =. Solving both r+ s r+ s 1 1 r + s1 r + s1 equations give 0 =. Similarl, 0 =. Therefore: r+ s r+ s r : s r + s1 r + s1 The point, r+ s r+ s divides the segment Connecting ( 1, 1 ) and (, ) into ratio r : s. (1.5) The formula (1.5) is known as the section formula. In particular, if the point divides the segment into ratio 1:1, i.e., the point is the midpoint of the segment, then its coordinates are: +, + This is called the midpoint formula. 1 1 (1.6) Eternal division Till now we can onl cut the segment inside it into ratio r : s. But we can also cut the segment outside, and also result in ratio r : s. Consider the segment connecting the 113
AP 1 origin and A = (, 0). For a point P that OP = 3, the coordinates of P can be 3,0 4 b section formula. But P can also be (3, 0) and the ratio is still 1 : 3. Such a point that cut the segment outside it is called an eternal point of division. Likewise, the one that cut inside it is an internal point of division. To classif the eternal and internal point of division, we define the eternal one divides a negative ratio. In our eample, the P divides the segment into ratio 1 : -3. With this definition, the section formula holds for eternal point of division as well. Eample 1c: Given two points A (5, 6) and B (-, 5). If a point P (, -1) divides the segment AB into ratio r : s with r < s, find: r : s Solution 1c: 5r+ 6s Appling the section formula we have 1=. Solving this gives r : s = -7 : 6. r+ s Appl section formula again gives = -44. Eample 1d: Show that the point A (1, 1.5), B (-4, ) and C (3, 1) are not collinear. Solution 1d: Assume A, B and C are collinear. Then C must be a point of division of AB, no matter internal or eternal. So let the ratio be r : s. B section formula, for -coordinates, r+ s 1.5=, thus r : s = 1 : 1. However, for -coordinates, r+ s implies r : s = 5 :, a contradiction. Area of polgon enclosed b segments 3r 4s 1=, which r+ s Q.E.D. Area of triangle A triangle is form b three joined segments, which in turn is defined b three distinct points. Thus we can find the area of a triangle using the coordinates of the points. Eample 1e: Given three points A ( 1, 1 ), B (, ) and C ( 3, 3 ). Find the area of ABC. 114
Solution 1e: Enclose a rectangle ARQP around ABC, and without loss of generalit assume A is the one of the vertices. A ( 1, 1 ) P ( 3, 1 ) C ( 3, 3 ) R ( 1, ) B (, ) Q ( 3, ) From the figure, we have: Therefore, the area of ( )( ) AP= AR= 3 1 1 RB= PC = 1 3 1 BQ= CQ= ACP 3 3 (1.7) ( )( ) ( )( ) 3 1 3 1 3 3 =, BQC =, ( )( ) 1 1 ARB= and ARQP= 3 1 1. But area of ABC = ARQP ACP BQC ARB. Therefore, after simplifing, the area of ABC is: 1 1 + 3 3 + 3 1 1 3 (1.8) (1.8) is not eas to remember. So we define: a 1 1 n 1 n 1 k k 3 3 = = k k+ 1 k+ 1 k k= 1 ak+ 1 bk+ 1 k= 1 n b a b a b a b a b a b a b n (1.9) So the area of triangle is: 1 1 1 3 3 1 1 (1.10) 115
Eample 1f: Redo eample 1d. Solution 1f: If A, B and C are collinear then the area of ABC should be zero. But area of ABC = 3 1 1 1 1.5 4 3 1 = 0.75 0, so A, B, C are not collinear. Q.E.D. Note that when using (1.10) please ensure the points are arranged counterclockwise. Otherwise a negative value will be obtained. If ou don t know whether the points are counterclockwise or not, take an absolute value. Area of polgon A polgon can be dissected into triangles. So we can just find out the area of each triangle and sum up them to give the area of the polgon. In fact, if the vertices of the polgon is ( 1, 1 ), (, ),, ( n, n ) in counterclockwise order, then its area is: Eample 1g: Let X be = 3 quadrilateral OYPX. Solution 1g: 1 1 1 n (1.11) n 1 1 1 1,, Y be 1,, O be the origin, C be the curve 3 3 and P be a point on C in quadrant I. Find the maimum area of Let P be (, ). So 0< < 3. Then: 0 0 3 1 3 1 Area of OYPX = 1 1 0 0 1 + = + 3 1 + 7 = 6 (1.1) 116
= = + 1 7 1 7 3 7 1 7 1 589 = 1 14 196 7 589 1 = 1 196 14 = 7 589 1 196 589 336 Equalit occurs when = 1 14. Thus, the maimum area is 589 336. 117
Lines Lines in coordinates sstem Lines are curves that are straight and infinite in length. There are no end points for lines. (A line with two end points is a segment, and one is a ra.) A line can be thought as an etension of a line segment. A line and a segment As two points can fi a segment, it is also true for lines. There eists a unique line that passes through two given distinct points. Since line is a tpe of curve, we are interested in finding out the equation of a line. Equation of a line Two-point form It is given two points A and B and a line passes through them. We ma assume there is an arbitrar point P lies on the line. Then P must satisf the followings: The slope of segment AP is the same as that of AB. P is a point of division of segment AB. We will find the equation of a line using the first condition. Readers ma tr the second condition. Eample 13a: Given two points A ( 1, 1 ) and B (, ). Find the equation of the line passing though them. Solution 13a: Let P (, ) be an arbitrar point lies on the line. Then slope of AP is equal to slope of AB. Thus: 1 1 = (13.1) 1 1 This equation holds for all P on the line. Thus (13.1) is the equation for the line passing through A and B. (13.1) is known as the two-point form of a line. Eample 13b: Find the equation of the line joining (5, -1) and (, 7). 118
Solution 13b: The equation of the line is: 7 1 7 = 5 3 7 8 ( ) = ( ) 8+ 3 37= 0 (13.) (, 7) 5 8 + 3 37 = 0 0 5 (5, -1) Point-slope form Besides determining a line b two points, one can also find it out b its slope and a point on it. Using a similar approach in eample 13a, if the point is ( 1, 1 ) and the slope is m, then the equation is: 1 = m (13.3) 1 Or: 1 = m ( 1 ) (13.4) (13.4) is known as the point-slope form of a line. Slope-intercept form An intercept of a curve is a point that cut the ais. If the ais is an -ais, the intercept is called an -intercept; if the ais is -ais, it is called a -intercept. For eample, the -intercept of the line in eample 13b is 37 8 and the -intercept is 37 3. If the -intercept and slope of a line is given, we can determine that line. It is because the -intercept is just a point and we can use the point-slope form to find out the equation of the line. If the -intercept is c and the slope is m, then the equation is: 119
This is the slope-intercept form of a line. = m + c (13.5) Intercept form If we are given the two intercepts of the line, we can also tell the equation the line using two-point form. If the -intercept is b and -intercept is c, then: + = 1 (13.6) b c This is known as the intercept form of a line. General form Recall that we define a curve as set of points satisfing f(, ) = 0. If we can arrange the equation for a curve in the form = 0, then we sa that is the general form of the curve. For a line, its general form is: A + B + C = 0 (13.7) For real numbers A, B and C. B rearranging (13.7) into slope-intercept and intercept form, we have: C -intercept = A C -intercept = B A Slope= B Distance between point and line* (13.8) The distance between a point ( 0, 0 ) and a line L: A + B + C = 0 can be given b: A0 + B0 + C d =± (13.9) A + B The sign is taken opposite to that of C. If the point and the origin are ling on the same side of L, the distance is negative. Otherwise, it is positive. Usuall we just use the absolute value of (13.9) in practical use. 10
Intersecting Lines Intersection of two lines When there are two lines in a coordinates sstem, we ma ask, where will the intersect? This will be introduced here. Assume we have two lines L 1 : a + b + c = 0 and L : d + e + f = 0. If P ( 0, 0 ) is the intersection of L 1 and L then this should satisf both equations L 1 and L, i.e., a0 + b0 + c= 0 d0 + e0 + f = 0 (14.1) This is just a set of ordinar simultaneous linear equations in unknowns. Using the notation in set theor we ma write P = L 1 L. Eample 14a: Find CN : DN = 1 : and BP = CP. Area of red triangle Area of rectangle in the following figure, if AM : MB = A M B I J P D N C Solution 14a: Without loss of generalit assume the figures lies on a coordinates sstem with origin at D and A = (0, ) and C = (3, 0). It is because ratio is invariant in scaling. Therefore B = (3, ), P = (3, 1), M = (1, ) and N = (, 0). Thus: DP : = 3 DM : = (14.) AN : + = 4 Thus I = DM AN =, 3 3 and J = DP AN = 3 1,. So the area of the red 0 0 1 3 1 5 triangle is =. But the area of the rectangle is 6. Thus the answer is 5 3 4 3 6 36. 0 0 11
Angle between two lines We are also interested in finding the angle formed between two lines. But we need a lemma: Another representation of slope P X θ R PR Using the definition of slope and the above figure, we know that the slope=. But XR from definition of trigonometric function we know PR tan θ XR =. Hence we have: Slope = tan θ (14.3) The value θ is known as the inclination of the line, which is the angle between the line and the -ais. Angle between two lines Refer to the following figure, L L 1 θ β β 1 X X 1 It is eas to show that β β 1 = θ. If the slope of L 1 is m 1 and L is m, then: 1
( ) tan θ= tan β β 1 tan β1 tan β = 1 + tan β tan β m1 m = 1 + m m 1 1 (14.4) However, the last value ma be negative. If we are onl concentrated to the acute angle between the two lines, we can takes its absolute value, i.e., tan θ m m 1 m m 1 = (14.5) + When two lines parallel, the angle between is 0. Thus from (14.5), (L 1 // L ) (m 1 = m ) (14.6) When two lines are perpendicular to each other, i.e., the angle between them is 90, the denominator of its tangent is zero. Thus: (L 1 L ) (m 1 m = 1) (14.6) Eample 14b: (HKCEE 1997/A) L is the line = + 3. A line with slope m makes an angle of 45 with L. Find the value(s) of m. Solution 14b: The slope of L is. From (14.4), ( ) m tan 45 = 1 + m 1 1+ m = m 1 1+ m= m or 1+ m= m m= 3 or 1 m= 3 (14.7) 13
Locus and Parametric Equation Locus In mathematics, locus is a curve traced out b a moving point. This point usuall satisfies some conditions we want. To find out the equation of a locus, we usuall let the point be (, ) and join the other conditions to make up an equation in terms of and. Eample 15a: Find the locus of a point P such that the distant between P and (, 5) is alwas 3. Solution 15a: Let P = (, ). Hence: Is the required locus. ( ) ( ) + 3 = 3 4 + 4+ 6 + 9= 9 + 4 6 + 4= 0 (15.1) Eample 15b: Given two curves H: = 1 and C: + = 4. X is an intersection of C and H in the first quadrant. Let P be a point on C such that the line XP intersects H at Q. Find the locus of the midpoint of PQ. Solution 15b: Q M P X -4-0 4 H - C Locus of M This figure shows a brief look of the graph. Here we let M to be the midpoint of PQ. 14
Since we are going to find the locus of M, we let M = (, ). For P, Q and X, we let them to be (a, b), (c, d) and (m, n). Since X is an intersection of H and C, we have: m n = 1 m + n = 4 m, n> 0 (15.) 5 3 Thus m= and n=. If we let the equation of XP to be n = k ( m) for some constant k denoting the slope, then: ( ) ka b+ n km = 0 a + b = 4 (15.3) And: ( ) kc d+ n km = 0 c d = 1 (15.4) ( ) ( ) ( ) ( ) 5 k 1 3k 3 k 1 5 Solving (15.3) gives a=, b=. Solving (15.4) k + 1 k + 1 ( ) ( ) ( ) ( ) 5 k + 1 3k 3 k + 1 + 5k gives c=, d =. Since a + c, b + = = d, k 1 k 1 we have: 4 3 5( k + 1) 3k = 4 ( k 1) 4 3( k + 1) + 5k = 4 k 1 ( ) This is a parametric equation, which will be introduced in this section: (15.5) Parametric equation Parametric equations are equations describing a curve in the form ( ) ( ) = f t = g t or (f(t), g(t)). Here f and g are real-valued functions and t is a real variable, called the parameter of the curve. As t changes, the point (f(t), g(t)) moves. If we find the locus of this point the result will be a curve this parametric equation describing. 15
Eample 15c: Sketch the graph with = t + 5, = 5t 1 for t [-3, 3]. Solution 15c: We can plot a few points with respect to t then join them up to get a rough image of the graph. t = 3 10 t = t = 1 t = 1/ 0 4 6 8 t = 0 t = -1-10 t = - t = -3 The curve is a segment. Converting general equation into parametric equation* If the curve is of the form = f(), then we can just set = t and = f(t) to form a parametric equation. If the curve is f(, ) = 0, then we ma (I) convert it into the form = g(t) or (II) find two functions h 1 (t) and h (t) such that f(h 1 (t), h (t)) = 0. Then = h 1 (t) and = h (t). Eample 15d: Convert the curve + = 1 into parametric form. Solution 15d: Since sin t + cos t = 1 for all t, the parametric form is (sin t, cos t). Converting parametric equation into general equation To convert a parametric equation into its general form, the onl wa is to eliminate the parameter and connect and in one equation. To eliminate the parameter one can (I) epress t in terms of or then substitutes this into the other equation, or (II) find a function such that f((t), (t)) = constant. t Eample 15e: Convert t+, t into general form. 16
Solution 15e: t Let = t + and =. t Method (I): Hence t =. Substitute this into gives: = (15.6) 4 That means 4 + = 0 is the general form. Method (II): t 4 First of all, notice that =, which is simpler in epressing. If we subtract t t 4 = from it, the result will be t t t = = t. Subtract again from it t gives = 4. Rearranging this also result in 4 + = 0. 17
Circles What are circles? In coordinates geometr, circles are loci of a point P that is equidistant from another point O. The point O is called the center of the circle and the length OP is the radius. If the center of the circle is ( 0, 0 ) and the radius is r, then the equation of the circle is: ( 0 ) + ( 0 ) = r (16.1) This is called the standard form of the circle. We can epand (16.1) into: + 0 0 + ( 0 + 0 r ) = 0 (16.) To get its general form. If it is given the general form of a circle to be: + + a + b + c = 0 (16.3) Then: a b Center =, (16.4) a + b 4c Radius= So, when a + b 4c = 0, the circle reduced into a point. This is called a point circle. When a + b 4c < 0, the radius will be unreal and no graph can be drawn. Then that equation is said to be representing an imaginar circle. Eample 16a: Given a triangle ABC such that the vertices A, B, C are (, 5), (1, 7) and (-4, ) respectivel. Find: The equation of the circumcircle Γ of ABC, and The intersection (other than A) of Γ and the altitude of ABC from A. Solution 16a: The first question means to find a circle Γ such that A, B, C all lies on it. We let Γ to be + + a + b + c = 0. So: + + + + = ( ) 5 a 5b c 0 1 + 7 + a+ 7b+ c= 0 5 + 5a + b + c = 0 Solving this gives Γ: + + 3 9 + 10 = 0. (16.5) 18
For the second equation, let the altitude be L. Therefore, L BC. Since the slope of BC is 1, the slope of L is 1. B point-slope form, the equation of L is + 7 = 0. Thus the intersection satisfies: + 7= 0 + + 3 9 + 10= 0 (16.6) Hence the intersection is (-1, 8). (The other solution (, 5) should be rejected because it is just the point A.) 8 Γ L 7 B 6 5 A 4 3 C 1-6 -4-4 Tangent of circle If a curve C intersects a curve C at eactl one point somewhere and C does not appear on both sides separated b C there, C and C are tangent to each other. This is especiall important if C is a line; this will be called the tangent of C. In this section we will show how to find the tangent of a circle. Tangent of circle passing through a point on circle Let C: + + a + b + c be a circle with center O and T ( 0, 0 ) be a point on the C such that a tangent L passes through it. Thus OT L. 0 + b 0 + b Since slope of OT is =, the slope of L is + a + a 0 0 0 0 + a. Therefore, b point-slope form, the equation of L is: + b C O T L 19
( ) a+ b+ + a b + = 0 But because T is on C, we have: 0 0 0 0 0 0 + a = ( ) 0 0 0 0 + b (16.7) + = a b c (16.8) 0 0 0 0 Substitute this back to (16.7) and simplifing gives: + + L a b c 0 0 : 0 + 0 + + + = 0 (16.9) Eample 16b: Using the figure in eample 16a, let the L A, L B and L C be the three tangents of Γ passing through A, B and C respectivel. Find the length of the segment connecting L A L B and L A L C. Solution 16b: The tangent through A (L A ) is: 3 9 L : + 5+ + + 5 + 10= 0 A : 7+ 19= 0 ( ) ( ) (16.10) Similarl, the L B is + 8 = 0 and L C is + + = 0. Thus the two intersections are 7 11 11 37, and, 6 6 and the distance is 5 3. Lengths of tangent from eternal point If a point is outside a circle and tangents passes through it, there will be eactl two tangents. Nevertheless, their lengths (XT and XU in the figure) are the same. We wish to find the value of this. If the circle is ( ζ) + ( η) = r and X = ( 0, 0 ), then b Pthagoras theorem, XT + r = OX. But OX = ( 0 ζ) + ( 0 η). So: O T U X ( ) ( ) 0 0 XT = ζ + η r (16.11) But this is just the value of the equation of the circle when (, ) = ( 0, 0 ). Hence, if the circle in general form is + + a + b + c = 0, the length of tangent is: + + a + b + c (16.1) 0 0 0 0 Man circles If we are given two circles on a coordinates sstem, we can tell their relationship b the distance of their centers and their radii. If the radii of the two circles are r 1 and r 130
respectivel with r 1 > r, then: Distance of centers in Implication (r 1 + r, ) Two circles separates {r 1 + r } The are tangent to each other eternall (r 1 r, r 1 + r ) The intersect at points {r 1 r } [0, r 1 r ) The are tangent to each other internall One circle is inside the other If the two circles intersect, there will be a common chord which passes through the two intersections. If the are C 1 : + + a 1 + b 1 + c 1 = 0 and C : + + a + b + c = 0, simple calculation suggests that the equation of the common chord is: L: (a 1 a ) + (b 1 b ) + (c 1 c ) (16.13) Or an eas-to-remember representation: L: C 1 C (16.14) If C 1 and C are tangent to each other, then L reduced to one of the common tangents of them. Eample 16c: Prove that the common chords of an three intersecting circles are alwas concurrent. Solution 16c: Let the three circles be C 1 : + + a 1 + b 1 + c 1 = 0, C : + + a + b + c = 0 and C 3 : + + a 3 + b 3 + c 3 = 0. Thus the three chords are L 3 : (a 1 a ) + (b 1 b ) + (c 1 c ) = 0, L 1 : (a a 3 ) + (b b 3 ) + (c c 3 ) = 0 and L : (a 3 a 1 ) + (b 3 b 1 ) + (c 3 c 1 ) = 0. Let the intersection of L 3 and L 1 be X. Since X satisf both L 1 and L 3, it would also satisf: L 1 + L 3 : (a 1 a 3 ) + (b 1 b 3 ) + (c 1 c 3 ) = 0 (16.15) But this is just L. Therefore, X is on L 1, L and L 3. That means the three chords are concurrent. Q.E.D. 131
Parabola The quadratic polnomial in coordinates geometr If we graph the curve = on a coordinates sstem, the result would be something like this: This curve is called a parabola. In phsics, the (upside-down) parabola resembles the path of an object thrown at an angle, and it is wh this curve named like this (it means to throw across in Greek). In mathematics, a parabola is the curve = a + b + c for a 0. The following shows some eample of parabolas: 4-4 - 0 4-13
Some properties associated to parabola -Intercepts In Elementar Algebra, we learnt that the discriminant = b 4ac determines how man real roots a does a quadratic equation has. In coordinates geometr, the root of a curve is same as its -intercept. Therefore, decides how the parabola will cross the -ais. The -intercepts would be the roots of the quadratic polnomial. 4-0 4 Three parabolas with different values of. The purple one: < 0. The ellow one: = 0. The blue one: > 0. -Intercept A function will cross the -ais when = 0. B substituting this, we know that the -intercept of a parabola is the constant term of its equation, i.e., c. 0 5 10-5 -10 The parabola = ½ 5 + 1. The -intercept is 1 according to the graph. 133
Etremum and direction of opening An etremum means the maimum or minimum point of a function. B completing the square, a quadratic polnomial can be epressed as: a ( h) + k (17.1) b Where h =, k a = 4a. Because n 0 for an real number n, the maimum / minimum of (17.1) will be k depending on the sign of a. The value k can be achieved when = h. Thus, (h, k) is the etremum of a parabola. (0.5, -.75) -5-10 (5, -11.5) 0 5 10 Two parabolas: = - + 3 and = ½ 5 + 1. The etremum will be a minimum when a > 0 (Wh?). It will be a maimum when a < 0. If the etremum of a parabola is a minimum, then no points can lower than this point. Thus the parabola opens upwards. Similarl, if the etremum is a maimum, the parabola opens downwards. That means when a is positive, the parabola opens upwards. When a is negative, the parabola opens downwards. In some books the point etremum is called the verte of a parabola. Eample 17a: (HKCEE 1990/) The graph of = a + b + c is given as shown. Which of the following is/are negative: a, b or c? O 134
Solution 17a: From the graph, the -intercept is negative. Thus c is negative. Also, the parabola b opens downward, thus a is also negative. The sum of roots is positive, thus > 0. a This means b is positive. Therefore, a and c are negative. 135
Coordinates Geometric Transformation What is transformation? In realit, if ou have an object, ou can move it, rotate it, squeeze it, pull it, destro it, etc. These are transformations of the object. Transformation of something means to change the shape and/or position of something in a specific wa. In coordinates geometr, a transformation can be represented b a function τ: R R. This function maps a point to a different (or same) coordinates. Since a curve is composed of points, if we transform ever point on the original curve, the result will be another curve transformed b τ. Usuall if we transform a curve C b τ, we simpl write τ(c), which actuall means the set {τ() C}. Translation Translation is a transformation. This transformation will move ever point b a fied distance and direction. If the origin is mapped to (a, b), then the transformation is given b: T(, ) = ( + a, + b) (18.1) If the original curve it C: f(, ) = 0, the translated curve will be f( a, b) = 0. We sa the curve is translated b (a, b). - 0 4 6 Scaling - The curve + = 4 is translated b (5, 1). Scaling, also called homothet, is a transformation that makes an object bigger or smaller. To demonstrate the mechanism, imagine ou want to take a picture of something far awa using a digital camera. This thing looks too small in preview. To have a clear look of this thing, ou would zoom in. During zooming, the objects around the thing are getting farther awa (in the picture!), and the result would be the 136
whole view getting larger and larger. This is eactl how scaling works. One point is fied during scaling, and the other will move awa from / towards that point. The points that are farther apart from the fied point are moving faster. If the fied point is (a, b), and a line segment of length 1 is scaled up / down to length r, than the transformation function is: H(, ) = (r ( a) + a, r ( b) + b) (18.) Usuall the fied point is taken to be the origin. Hence: H 0 (, ) = (r, r) (18.3) If a curve C: f(, ) = 0 is transformed b H, the new curve will be + a + b f a, b = 0. The curve C is said to be scaled b a factor of r around r r (a, b). - 0 4 - -4 The curve + = 4 is scaled b a factor of around (-1, 1). If the factor is greater than 1, the figure is enlarged. If the factor is in between 0 and 1, the figure is shrunk. If the factor is negative, the figure is inverted. Reflection If ou look at a mirror, ou would find ourself behind the mirror with the direction inverted. This is a kind of reflection. In geometr, reflection is a transformation that maps ever point P to another point P with respect to a line L such that the distance between P and L is same as that of P and L and PP is perpendicular to L. 137
Eample 18a: Find the function of reflection if L is A + B + C = 0. Solution 18a: Let P = (, ), P = (, ) and the function be S: R R. Therefore: Since PP is perpendicular to L, the slope of PP is S(, ) = (, ) (18.4) B. That means: A B =. (18.5) A Now because the distance from P and P to L are the same, and P and P are on opposite sides of L (otherwise P = P, and this is not a reflection), we have: A + B + C = A B C (18.6) Using (13.9). Solving (18.5) and (18.6) for and gives: ( B A ) A( C+ B) ( A B ) B( C+ A) S(, ), A + B A + B = (18.7) If a curve f(, ) = 0 is transformed b S, the new curve is ( B A ) + A( C+ B) ( A B ) + B( C+ A) f, = 0. We sa the curve is A + B A + B reflected along L. 5-10 -5 0-5 The curve = is reflected along + + 3 = 0. The resulting curve is 9 4 + 16 16 + 63 + 96 = 0. 138
The formula (18.7) is etremel hard to memorized, so we recommend readers do not use this if possible. Rotation* Rotation is a transformation that moves a point P to another point P around a fied point O such that POP is constant. If this angle is α, we sa that P is P rotated b an angle of α about O. However, there ma be two choices of P, as shown on the right. In mathematics, we usuall measure angle in counterclockwise direction. O P Therefore, if α is positive, we take the blue P. If we want the red one, make α negative. P In coordinates geometr, rotation is evaluated as: R(, ) = (cos α ( a) sin α ( b) + a, sin α ( a) + cos α ( b) + b) (18.8) Where (a, b) is the center of rotation (the point O) and α is the angle. A curve is transformed as: ( ( ) ( ) ( ) ( ) ) f cos α a + sin α b + a, sin α a + cos α b + b = 0 (18.9) 6 4-4 - 0 The curve = is rotated about (0, -1) b 45. Invariants in transformations* In the transformations listed above, some quantities are not changed. These are the invariants of that transformation. The table below shows some invariants: 139
Invariants Translation Scaling Reflection Rotation Length and area Ratio Angle between two lines Slope Tpe / number of intersections Coordinates Points on L 140
Summar Coordinates Sstem In a coordinates sstem, a point can be defined b a pair of numbers called coordinates. The rectangular coordinates sstem is created b two perpendicular real lines, called the -ais and -ais respectivel, that are crossing at their zeros. The coordinates of a point are given b the readings on the aes. The are denoted as (, ). Coordinates sstem = {(, ), R} = R. The aes divide the coordinates sstem into four quadrants. II I III IV Line Segment {,, 0 } Curves are defined as C ( ) f ( ) = R =, where f: R R. Usuall C is just written as f(, ) = 0. Line segments are of curve that is not curved at all, and length is finite. If a segment connects ( 1, 1 ) and (, ), then: Length = ( ) + ( ) Slope= The point internall. 1 1 1 1 r + s1 r + s1, r+ s r+ s divides the segment into ratio r : s A negative ratio of division means the point divides the segment eternall. 1 1 1 1 1 1 Area of triangle=. Area of polgon=. 3 3 n n 1 1 1 1 141
Lines Lines are curves that are straight and infinite in length. There eists a unique line that passes through an two distinct points. 1 1 Two-point form of a line: = Point-slope form: 1 = m ( 1 ). Slope-intercept form: = m + c Intercept form: + = 1. b c 1 1 General form of a line: A + B + C = 0, where * Distance between a point and a line is is taken opposite to that of C. C -intercept = A C -intercept =. B A Slope= B A0 + B0 + C d =±, where the sign A + B Intersecting Lines Finding intersection P of two lines L 1, L is same as finding solution to a set of simultaneous linear equations in unknowns. We write P = L 1 L. Slope = tan θ, where θ is the inclination of a line, which is the angle between it and the -ais. Angle between two lines: tan θ m 1 =. + 1 m m m 1 (L 1 // L ) (m 1 = m ) (L 1 L ) (m 1 m = 1) Locus and Parametric Equation Locus is a curve traced out b a moving point. We let the moving point be (, ) to make up the equation of locus. Parametric equations are equations describing a curve in the form (f(t), g(t)). Circles Circles are loci of a point that is equidistant from another point. 14
If the center is ( 0, 0 ) and the radius is r, then the circle is ( 0 ) + ( 0 ) = r. This is the standard form of a circle. General form of circle: + + a + b + c = 0, where a b Center =,. a + b 4c Radius= If a curve C intersects a curve C at eactl one point somewhere and C does not appear on both sides separated b C there, C and C are tangent to each other. The tangent line of + + a + b + c = 0 at ( 0, 0 ) is 0 + 0 + 0 + 0 + a + b + c= 0. Length of the tangent from an eternal point ( 0, 0 ) is + + a + b + c. 0 0 0 0 Relationship of two circles: Distance of centers in Implication (r 1 + r, ) {r 1 + r } (r 1 r, r 1 + r ) {r 1 r } [0, r 1 r ) Common chord/tangent of two circles: L: C 1 C. Parabola Parabola is the curve = a + b + c for a 0. -intercepts of parabola are roots of that quadratic polnomial. decides how the parabola will cross the -ais. -intercept = c. b Etremum / Verte =, a 4a. 143
(a > 0) (parabola opens upwards). (a < 0) (parabola opens downwards). Coordinates Geometric Transformation Transformation means to change shape and/or position of something. Translation: T(, ) = ( + a, + b) Scaling: H(, ) = (r ( a) + a, r ( b) + b) Reflection: ( B A ) A( C+ B) ( A B ) B( C+ A) S(, ), A + B A + B = * Rotation: R(, ) = (cos α ( a) sin α ( b) + a, sin α ( a) + cos α ( b) + b). * Invariants in transformation: Invariants Translation Scaling Reflection Rotation Length and area Ratio Angle between two lines Slope Tpe / number of intersections Coordinates Points on L. 144
Self Assessment Eercises Level 1 1) Solve the followings: a) Find and epress the equation of the line passing through (cos t, sin t) and (-sin t, cos t) in terms of t. b) Show that this line is tangent to the circle + 1 = 0. [Hint: Consider the distance from the origin to the line] ) (HKCEE 1994/) In the figure, the line = m + k cuts the curve = + b + c at = α and = β. Find the value of αβ in terms of b, c, m and/or k. = + b + c = m + k α β 3) Prove the length is invariant in translation. 4) (HKMO 1996/HG Q1) In the figure, the quadratic curve = f() cuts the -ais at two points (1, 0) and (5, 0) and the -ais at point (0, -10). Find the value of p. (4, p) 1 5-10 5) (IMSC 000/FG1 Q1) Let u, v be integers such that 0 < v < u. Let A = (u, v), B be the reflection of A along the line =, C be reflection of B along the -ais, D be the reflection of C along the -ais and E be the reflection of D along the -ais. The area of the pentagon ABCDE is 451. Find u + v. Level 6) (JSMQ 1999/FG Q) A belt fits around three circles + = 1, ( 14) + = 1 and ( 9) + ( 1) = 1. Find the length of the belt. 7) (Apollonian Circle) Let A be (-1, 0) and B be (1, 0). Find the locus of P such that AP = r and epress it in terms of r. What kind of curve is it? PB 145
8) (PSMSIMC 004/S Q14c) In the figure, the radius of the circle is 1 and AP is a diameter of the circle. B is a point on the circumference and C is a point on BP such that PC : CD = 5 : 3. If M is the mid-point of AC and BM is etended to meet AP at Q, find the maimum possible area of BPQ. [Hint: Find the locus of C as B moves and show that Q is a fied point] B M C A Q P 9) (IMOPHK 004 Q13) Find the area enclosed b the graph + = + on the -plane. 10) (ISMC 003/I Q7) A piece of graph is folded once so that (0, ) is matched with (4, 0) and (7, 3) is matched with (m, n). Find m + n. 11) In the figure, P is a point on the line = 7. Find the minimum value of AP + BP and the corresponding coordinates of P. = 7 P A (5, -3) 1) Solve the followings: B (3, -6) a) Find the locus P of point X such that the distance from X to the line L: = -¼ and the point F (0, ¼) is the same. b) The circle C: + = 1 intersects P and L at four points. Find the area of the quadrilateral enclosed b these four points. c) It is given that if X ( 0, 0 ) is a point on P, the slope of the tangent to P passing through X is 0. Find the common tangent(s) of C and P. 146
Answers 1) a) dfdf 4) 6 5) 1 6) π + 4 7) ( r+ ) 1 + + 1= 0. This is a curve with center on the -ais. r 1 8) 8 11 9) π + 10) 6.8 11 10 11) 17; P= (, ) 3 3 ( 5 1 ) ( 15+ 5 ) 1) a) =. b) 16. c) =± + 5 5. 147
Chinese Translation of Important Terms Abscissa 橫坐標 Altitude 垂線 Center 圓心 Circle 圓 Circumcircle 外接圓 Collinear 共線 Common chord 公共弦 Common tangent 公切線 Concurrent 共點 Coordinates 坐標 Coordinates sstem 坐標系統 Curve 曲線 Equidistant 同距 Eternal point of division 外分點 Etremum 極值 General form 一般式 Homothet 位似變換 Hpotenuse 斜邊 Imaginar circle 虛圓 Inclination 傾角 Intercept 截距 Intercept form 截距式 Internal point of division 內分點 Invariant 不變量 Line segment 線段 Locus (pl. loci) 軌跡 Midpoint formula 中點公式 Ordinate 縱坐標 Origin 原點 Parabola 抛物線 Parameter 參數 Parametric equation 參數方程 Point circle 點圓 Point of division 分點 Point-slope form 點斜式 148
Polgon 多邊形 Project 投影 Pthagoras theorem 畢氏定理 Quadrant 象限 Quadrilateral 四邊形 Radius (pl. radii) 半徑 Ra 射線 Rectangular coordinates sstem 直角坐標系統 Reflection 反射 Right triangle 直角三角形 Rotation 旋轉 Scaling 縮放 Section formula 截點公式 Slope 斜率 Slope-intercept form 斜截式 Standard form 標準式 Tangent 相切 切線 Transformation 變換 Translation 平移 Two-point form 兩點式 Verte (pl. vertices) 頂點 -ais 軸 -coordinate 坐標 -intercept 軸截距 -ais 軸 -coordinate 坐標 -intercept 軸截距 149
References: Wikipedia, the free encclopedia. [http://en.wikipedia.org/] MathWorld A Wolfram Web Resource. [http://mathworld.wolfram.com/] Mathematics Database. [http://www.mathdb.org/] MathsWorld-001 [http://mathsworld.ath.c/] Original documents: Coordinates Sstem: Chapter 0 Before You Start Section ; Chapter Linear Equations Section 1; Chapter 8 Trigonometr Section 5. Line Segment: Chapter Linear Equations Section 1, 5, 7 and 8. Lines: Chapter Linear Equations Section and 5. Intersecting Lines: Chapter : Linear Equations Section 3 and 6; Chapter 8 Trigonometr Section 1. Locus and Parametric Equation: Chapter 13: Parametric Equation and Locus Section 1, and 3. Circles: Chapter 14: Quadratic Curve Section 10. Parabola: Chapter 3: Quadratic Equation and Polnomial Factorization Section 8. Coordinates Geometric Transformation: Chapter 14: Quadratic Curve Section 9. 150