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150 Chapter 5 Trigonometric function Find the ranges of x where the sine curve comes above the straight line y = − 1 . 2 7 11 0≦x< 6 π, 6 π < x < 2π // Q. Solve the following equations and inequalities, when 0 ≦ x < 2π. √ 1 3 (1) cos x = 2√ (2) sin x = √2 (3) sin x < − 2 (4) cos x ≧ 3 2 2 Exercise √ Solve the equation tan x = 3, when 0 ≦ x < 2π. Solution As shown in the figure, the point Y X=1 √ √ T(1, 3) is given on the straight line X = 1, 3T and the positions where the unit circle in- 1P tersects with the straight line OT are rep- H′ O H X resented with P and Q. The angle resulting −1 1 from the radius OP and OQ, when within the given range, is the value of x we are look- Q −1 ing for. By letting H(1, 0), H′(−1, 0), ∠HOP = ∠H′OQ = π 3 Thus, x= π , 4 π // 3 3 Q. Solve the following equations, when 0 ≦ x < 2π. (1) tan x = √1 (2) tan x = −1 3 The invers numbers of sin θ, cos θ, tan θ are written as cosec θ, sec θ, cot θ and called Cosecant, Secant, Cotangent respectively. cosec θ = 1 , sec θ = 1 , cot θ = 1 sin θ cos θ tan θ These are useful to express the trigonometric functions without using frac- tions. However, in this textbook we don t use these unless necessary.

§ 2 Trigonometric function 151 Exercises 2–A 1. Find the following values. (1) sin 236˚ (2) cos(−510˚) (3) tan 680˚ ( 14 ) ( 21 ) ( 11 ) − π cos π π (4) sin 3 (5) 6 (6) tan 5 2. Find the values of sin θ and tan θ, when θ is the third quadrant angle and cos θ = − 12 . 13 √ 3. Find the values of sin θ and cos θ, when tan θ = 5. 4. Prove the following equalities. (1) (1 + sin θ + cos θ)(1 + sin θ − cos θ) = 2(1 + sin θ) sin θ (2) cos2 θ − sin2 θ = 1 − tan θ 1 + 2 sin θ cos θ 1 + tan θ 5. Find the periods of the following functions and plot the graphs. (1) y = −2 sin x (2) y = 2 cos 1 x (3) y = − 1 tan x 2 (4) y = ( π ) 3 sin −x 2 6. Solve the following equations, when 0 ≦ x < 2π. (1) sin x = 1 (2) cos x = − 1 (3) tan x = −1 2 2 7. Solve the following inequalities, when 0 ≦ x < 2π. √ (2) 2 cos x − 1 < 0 (3) tan x > 1 (1) 2 sin x − 3 > 0

152 Chapter 5 Trigonometric function Exercises 2–B √ A 1. The circle O with a radius of 2 2 inter- sects with the other circle O′ with a radius of 2 at the two points A and B as shown on the 2 figure. Find the area of the overlapped por- π O′ tion of these two circles, when ∠AOB = π O 2 √3 3 2 and ∠AO′B = π . B 2 2. Find the values, when sin θ + cos θ = 2 . 3 (1) sin θ cos θ (2) sin θ − cos θ (3) sin3 θ + cos3 θ (4) sin3 θ − cos3 θ 3. Find the periods of the following functions and plot the graphs. (1) ( π) (2) y = sin(−2x + π) + 1 y = cos 2x + 3 4. For the function y = 2 sin2 x − 2 cos x + 1, answer the following questions. (1) Let t = cos x, express y with an equation of t. (2) Find the maximum and minimum values, when 0 ≦ x ≦ 5 π. 6 5. Solve the following equations. (1) cos 2x = − 1 (0 ≦ x < 2π) 2√ (2) 2 sin(π − x) − 3 = 0 (0 ≦ x < 2π) 6. Solve the following inequalities. (1) 2 sin 2x < −1 (0 ≦ x < 2π) √ (2) 2 cos(π + x) − 2 > 0 (−π ≦ x < π)

§ 3 Addition theorem and its applications 153 § 3 Addition theorem and its applications 3 1 Addition theorem On a coordinate plane, let the position where the unit circle intersects with the radiuses that make the angle α be P and the angle α + β be Q. (when Ox is an initial line). If the angle α, β and α + β are acute angles, draw a perpendicular line QH from Q to the segment OP. Then, draw another line parallel to the y-axis passing through H and let the intersection point of the line and the x-axis be L. Thus, OH = cos β, QH = sin β y HL = OH sin α = cos β sin α QM Draw a line parallel to the x-axis starting α from Q and let the intersection point of the α P line and the extension of the line HL be M. α+β H ∠QHM = ∠HOL = α β Thus, α x MH = QH cos α = sin β cos α O NL Therefore, drawing a perpendicular line QN from Q to the x-axis, QN = ML = HL + MH = cos β sin α + sin β cos α On the other hand, since QN = sin(α + β), sin(α + β) = sin α cos β + cos α sin β (1) Likewise, ON = OL − LN = OL − QM OL = OH cos α = cos β cos α, QM = QH sin α = sin β sin α From above, the following is obtained. cos(α + β) = cos α cos β − sin α sin β (2) Note (1) (2) can prove any angles of α and β.

154 Chapter 5 Trigonometric function By applying −β to (1) and (2) instead of β. (3) sin(α − β) = sin α cos(−β) + cos α sin(−β) (4) = sin α cos β − cos α sin β cos(α − β) = cos α cos(−β) − sin α sin(−β) = cos α cos β + sin α sin β These are obtained. From (1) and (2), tan(α + β) = sin(α + β) = sin α cos β + cos α sin β cos(α + β) cos α cos β − sin α sin β If the numerator and denominator of the right side of the equation are divided by cos α cos β. tan(α + β) = tan α + tan β 1 − tan α tan β This is obtained. The same again, from (3) and (4), the following equation is obtained. tan(α − β) = tan α − tan β 1 + tan α tan β Therefore, the following addition theorem is obtained.  addtion theorem  sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α ± β) = tan α ± tan β (Double sign in same order) 1 ∓ tan α tan β   Ex.1 cos 75˚= cos(45˚+ 30˚) = cos 45˚cos 30˚− sin 45˚sin 30˚ √√√ √√ 2 3 2 1 6− 2 = 2 × 2 − 2 × 2 = 4 ) ( π cos π sin π = √1 (sin θ + cos θ) sin θ + 4 = sin θ 4 + cos θ 4 2 Q. Find the values of sin 75˚, tan 75˚, sin 15˚, cos 15˚, tan 15˚. Q. ( π) with tan θ. Represent tan θ + 4

§ 3 Addition theorem and its applications 155 Exercise Find the value of sin(α + β), when α is the second quadrant angle and β is the third quadrant angle, and cos α = − 4 and sin β = − 5 . 5 13 Solution From the conditions, sin α > 0, cos β < 0, √ √ 1 ( 4 )2 3 1 − sin α = − cos 2 α = − = 55 √ √( cos β = − 1 − sin2 β = − 1 − − 5 )2 = − 12 13 13 ∴ sin(α + β) = sin α cos β + cos α sin β 3 ( 12 ) ( 4 ) ( 5 ) 16 5 − 13 − 5 − 13 65 = × + × = − // Q. Find the values of sin(α − β) and cos(α − β), when α is the second quadrant angle and β is the third quadrant angle, √ √1 5 and sin α = 3 and sin β = − 6 . Q. Find the value of tan(α + β) and the angle α + β, when 0 < α < π and 0 < β < π , and tan α = 1 and tan β = 1 . 2 2 2 3 3 2 Application of addition theorem In the addition theorem, let β α. sin 2α = sin(α + α) = sin α cos α + cos α sin α = 2 sin α cos α cos 2α = cos(α + α) = cos α cos α − sin α sin α = cos2 α − sin2 α (1) By replacing sin2 α with 1 − cos2 α in (1), cos 2α = cos2 α − (1 − cos2 α) = 2 cos2 α − 1 By replacing cos2 α with 1 − sin2 α in (1), cos 2α = (1 − sin2 α) − sin2 α = 1 − 2 sin2 α Furthermore, tan 2α = tan(α + α) = 2 tan α 1 − tan2 α

156 Chapter 5 Trigonometric function To summarize, the following double angle formulas are obtained.  double angle formula  sin 2α = 2 sin α cos α cos 2α = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α tan 2α = 2 tan α 1 − tan2 α   Exercise Find the value of sin 2α, when α is the third quadrant angle and sin α = − 2 . 3 Solution From the condition, cos α < 0. √ √ 1 ( 2 )2 √ −1 − 3 5 cos α = − sin2 α = − = − 3 ( 2 ) ( √ ) √ − 3 − 5 45 Thus, sin 2α = 2 × × 3 = // 9 Q. Find the values of sin 2α, cos 2α, tan 2α, when α is the second quadrant angle and cos α = − 4 . 5 From the double angle formula, cos 2α = 1 − 2 sin2 α, sin2 α = 1 − cos 2α (2) 2 And from cos 2α = 2 cos2 α − 1, cos2 α = 1 + cos 2α (3) 2 From (2) and (3) 1 − cos 2α 2 tan2 α = sin2 α = = 1 − cos 2α (4) cos2 α 1 + cos 2α 1 + cos 2α 2 By replacing α with α in (2), (3), (4) the following half-angle formulas 2 are obtained.  half-angle formulas  sin2 α = 1 − cos α cos2 α = 1 + cos α , tan2 α = 1 − cos α 22 22 2 1 + cos α

§ 3 Addition theorem and its applications 157 Exercise Find the value of sin π . 8 Solution Let α = π , then α = π . By using the half-angle formula, 4 2 8( ( √) √ π 1 π ) 1 2 2− 2 8 2 1 − cos 4 2 2 sin2 = = 1− = 4 √√ Since sin π > 0, sin π = 2 − 2 8 82 // Q. Find the value of cos π . 8 Q. Find the values of sin α , cos α , tan α , 2 2 2 when π < α < π and cos α = − 1 . 2 9 Let’s find the formulas transforming products of the trigonometric functions into sums or differences. From the addition theorem, sin(α + β) = sin α cos β + cos α sin β (5) sin(α − β) = sin α cos β − cos α sin β (6) From (5)+(6) and (5)−(6), sin(α + β) + sin(α − β) = 2 sin α cos β (7) sin(α + β) − sin(α − β) = 2 cos α sin β (8) The same formulas are obtained in the case of cos(α + β) and cos(α − β). By dividing both sides of these equalities by 2, the following formulas are obtained.  Formulas transforming products into sums or differences  sin α cos β = 1 {sin(α + β) + sin(α − β)} 2 cos α sin β = 1 {sin(α + β) − sin(α − β)} 2 cos α cos β = 1 {cos(α + β) + cos(α − β)} 2 sin α sin β = − 1 {cos(α + β) − cos(α − β)} 2

158 Chapter 5 Trigonometric function Ex.2 sin 2θ cos θ = 1 {sin(2θ + θ) + sin(2θ − θ)} = 1 (sin 3θ + sin θ) 2 2 cos 75˚cos 15˚= 1 {cos(75˚+ 15˚) + cos(75˚− 15˚)} = 1 2 4 Q. Transform the following equations into sums or differences. (1) cos 3θ cos 5θ (2) sin 7θ sin 2θ By applying α + β = A and α − β = B in (7), α= A + B , β= A−B 2 2 Thus, A+B A−B 2 2 sin A + sin B = 2 sin cos Likewise, the following formulas, which transforming sums or differences into products of the trigonometric functions, are obtained.  Formulas transforming sums or differences into products  sin A + sin B = 2 sin A+B cos A−B 2 2 sin A − sin B = 2 cos A+B sin A−B 2 2 cos A + cos B = 2 cos A+B cos A−B 2 2 cos A − cos B = −2 sin A+B sin A−B 2 2   Ex.3 cos 15˚− cos 75˚= −2 sin 15˚+ 75˚ sin 15˚− 75˚ 2 2 √ ( 1) = −2 sin 45˚sin(−30˚) = −2 × 2 ×− 2 √ 2 =2 2 Q. Transform the following equations into the forms of products. (1) sin 4θ + sin 2θ (2) cos 3θ + cos 5θ

§ 3 Addition theorem and its applications 159 By using the addition theorem, a sin x + b cos x (note : a and b are constant numbers) Equation of the above form can be represented by one trigonometric function. For the radius OP, when point P(a, b) is on the coordinate plane and the initial line takes the positive direction of the X-axis, let the angle that the radius OP makes be α, Y P(a,b) √ b aX OP = a2 + b2 √ a2 + b2 cos α = √ a α a2 + b2 O sin α = √ b a2 + b2 √√ Thus, a = a2 + b2 cos α b = a2 + b2 sin α From this, √√ a sin x + b cos x = a2 + b2 cos α sin x + a2 + b2 sin α cos x √ = a2 + b2(sin x cos α + cos x sin α) √ = a2 + b2 sin(x + α) Therefore, the following formulas are obtained.  Composition of trigonometric functions  √ a sin x + b cos x = a2 + b2 sin(x + α) Note : cos α = √ a sin α = √ b a2 + b2 a2 + b2   √ Ex.4 When expressing 3 sin x − cos x with a single trigonometric func- √√ tion, by letting a = 3 and b = −1, point P( 3, −1) is given on the cOoPo=rdi√na(t√e 3p)la2 n+e.12 = 2 √ Y √ cos α = 3 Oα 3 2 X Since sin α = − 1 , α = −2 sπ6in.(Txh−en,π6 2 √ 2 = −1 √ ) P( 3, −1) 3 sin x − cos x Q. Express each of the following equations with single trigonometric √ functions. (1) y = sin x + cos x (2) y = sin x − 3 cos x

160 Chapter 5 Trigonometric function Exercise √ Plot a graph of y = 3 sin x + cos x. Solution Since √(√3)2 + 12 = 2, cos α = √ and sin α = 1, α= π. 3 √( ) 2 26 = 3 sin x + cos x = 2 sin x + Y√ ∴ y π 6 √ P( 3, 1) Therefore, the graph of y = 3 sin x+cos x is the 1 2 same as the graph of y = 2 sin x but translated α O − π parallel towards the x-axis. √X 6 3 y √ 2 y = 3 sin x + cos x y = cos x √ y = 3 sin x 5 π 4 π x 6 3 O−π π 11 π 6 3 6 −2 // Q. Find the maximum and minimum values of the following function. y = 2 sin x + 3 cos x

§ 3 Addition theorem and its applications 161 Column Trigonometric function and the chord length (1) When 0 < θ < π , 2 sin θ (sine), tan θ (tangent), sec θ (secant) Each of them represent the lengths related to sine, tangent and secant (a straight line that intersects with a circle at 2 points) of a unit circle, as shown below. cot θ 1 π −θ cos θ cosec θ 2 sec θ sin θ tan θ θ 1 And, for the complementary angle π − θ, 2 cos θ (cosine), cot θ (cotangent), cosec θ (cosecant) (2) are corresponding to (1). In astronomy, from the ancient times, it was necessary to find the chord length, which subtends the center angle of a circle. By the second century AD, the table of chords was existed, giving the value of the chord for the angles ranging from 0˚ to 180˚ by increments of half a degree for a circle with radius R. This is an equivalent to the table of the value 2R sin α in 2 modern terms. The trigonometric function with general angles was invented by Leonhard Euler in 18th century after many years had been spent for this achievement.

162 Chapter 5 Trigonometric function Exercises 3–A 1. Find the values of the followings when α and β are both obtuse angles, and tan α = − 3 and cos β = − √2 . 45 (1) sin(α + β) (2) cos(α + β) (3) tan(α − β) 2. Find the values of sin 2α and cos 2α, when sin α = − √1 and π < α < 3 π. 32 3. Prove the following equations. (1) sin(α + β) = tan α + tan β sin(α − β) tan α − tan β tan ( π ) ( π ) x −x =1 (2) + tan 44 4. Prove the following triple angle formulas. (1) sin 3θ = 3 sin θ − 4 sin3 θ (2) cos 3θ = 4 cos3 θ − 3 cos θ 5. Simplify the following equations. (1) sin θ cos 3θ + sin θ cos 5θ + sin θ cos 7θ (2) sin θ sin 3θ + sin θ sin 5θ + sin θ sin 7θ 6. Express the following equations with single trigonometric functions. √ √ (1) 3 sin x + 3 cos x (2) − 3 sin x + cos x 7. Plot a graph of y = sin x − cos x and find the maximum and minimum values, and also find the value of x at that time. Note that 0 ≦ x < 2π.

162 Chapter 5 Trigonometric function Exercises 3–A 1. Find the values of the followings when α and β are both obtuse angles, and tan α = − 3 and cos β = − √2 . 45 (1) sin(α + β) (2) cos(α + β) (3) tan(α − β) 2. Find the values of sin 2α and cos 2α, when sin α = − √1 and π < α < 3 π. 32 3. Prove the following equations. (1) sin(α + β) = tan α + tan β sin(α − β) tan α − tan β tan ( π ) ( π ) x −x =1 (2) + tan 44 4. Prove the following triple angle formulas. (1) sin 3θ = 3 sin θ − 4 sin3 θ (2) cos 3θ = 4 cos3 θ − 3 cos θ 5. Simplify the following equations. (1) sin θ cos 3θ + sin θ cos 5θ + sin θ cos 7θ (2) sin θ sin 3θ + sin θ sin 5θ + sin θ sin 7θ 6. Express the following equations with single trigonometric functions. √ √ (1) 3 sin x + 3 cos x (2) − 3 sin x + cos x 7. Plot a graph of y = sin x − cos x and find the maximum and minimum values, and also find the value of x at that time. Note that 0 ≦ x < 2π.

164 6Chapter Geometry and Equations § 1 Point and Line 1 1 Distance between two points and internally dividing point Let s find the distance AB between two points A(x1, y1) and B(x2, y2) on a coordinate plane. y Let us suppose that the intersection point of a straight line parallel to the x-axis, pass- y2 B ing through A and a straight line parallel to |y2 − y1| the y-axis, passing through B is C. y1 A |x2 − x1| C Then we have, O x1 x2 x AC = |x2 − x1|, BC = |y2 − y1| From Pythagorean proposition, AB2 = AC2 + BC2 = (x2 − x1)2 + (y2 − y1)2 Therefore, the following formula is obtained.  Distance between two points  The distance AB between two points A(x1, y1) and B(x2, y2) is, √ AB = (x2 − x1)2 + (y2 − y1)2 In particular, the distance between the origin O(0, 0) and A is, OA = √ x2 + y2 1 1

Ex.1 § 1 Point and Line 165 Q. √√ Given A(0, 1) and B(−1, 2), AB = (−1 − 0)2 + (2 − 1)2 = 2 Given A(1, −2) and B(4, 3), find AB, OA and OB. Exercise Find the coordinates of the point P on the x-axis which is equidistant from two points A(1, −1) and B(4, 2). Solution Let the coordinates of the point P be (x, 0). Since AP = BP, √√ y (x − 1)2 + 12 = (x − 4)2 + (−2)2 By squaring both sides, 2B (x − 1)2 + 1 = (x − 4)2 + 4 By solving the above, 1 x O P4 x=3 −1 A // Therefore, the coordinates of the point P are, (3, 0). Q. Find the coordinates of the point on the y-axis which is equidistant from two points A(2, 3) and B(5, 2). Q. Given A(2, 3) and B(5, 2), find the coordinates of the point P on √ the x-axis that satisfies 2AP=BP. When a point P is on the segment AB and for positive numbers m and n, AP : PB= m : n y B is established. then we say the point P di- y2 vides the segment AB internally in the ra- tio m : n and the point P is the internally y P n dividing point of the segment AB. y1 Am Let’s find the coordinates (x, y) of the in- ternally dividing point P given A(x1, y1) and A′ P′ B′ x B(x2, y2). O x1 x x2 By drawing lines perpendicular to the x-axis AA′, PP′ and BB′ from the points A, P and B respectively, A′P′ : P′B′ =AP : PB= m : n

166 Chapter 7 Geometry and Equations When x1 < x2 A′P′ = x − x1, P′B′ = x2 − x Therefore, (x − x1) : (x2 − x) = m : n By solving for x, x = nx1 + mx2 m+n The formula above is also established when x1 ≧ x2 and can be obtained for y likewise. Coordinates of internally dividing point   The coordinates of the point which divide the segment joining two points (x1, y1) and (x2, y2) internally in the ratio m : n are, ( nx1 + mx2 , ny1 + my2 ) m+n m+n In particular, the coordinates of the midpoint are,  ( x1 + x2 , y1 + y2 ) 22  Q. Find the coordinates of the points P and Q which divide the seg- ment AB joining two points A(1, −5) and B(−6, 2) internally in the ratio 1 : 2 and 2 : 1 respectively, also find the coordinates of the midpoint M of the segment AB. For the given triangle ABC, let the midpoints A of the sides of BC, CA and AB be L, M and N re- spectively. Then the segments AL, BM, and CN 2 M intersect at single point where the intersection N point G divides each segment internally in the ratio 2 : 1. Let’s prove this by using the formula G 1 for the coordinates of internally dividing point. B L C Let the coordinates of three points be A(x1, y1), B(x2, y2) and C(x3, y3) respectively.

§ 1 Point and Line 167 Since the coordinates of the midpoint L of the side BC are, ( x2 + x3 , y2 + y3 ) 22 Let the point dividing AL internally in the ratio 2 : 1 be. Then, 1 × x1 + 2 × x2 + x3 = x1 + x2 + x3 2 x= 2+1 3 1 × y1 + 2 × y2 + y3 = y1 + y2 + y3 2 y= 2+1 3 Likewise, since the coordinates of the midpoint M of the side CA are, ( x3 + x1 , y3 + y1 ) 22 By solving for the coordinates of the point dividing BM internally in the ratio 2 : 1, ( x1 + x2 + x3 , y1 + y2 + y3 ) 33 We find the above coincides with the coordinates of G and the same thing can be said about CN. Therefore, the straight lines AL, BM and CN intersect at G where G divides each segment internally in the ratio 2 : 1. G is called the barycenter of △ABC. Barycenter of a triangle   The coordinates of the barycenter G of △ABC with three vertices of A(x1, y1), B(x2, y2) and C(x3, y3) are,  ( x1 + x2 + x3 , y1 + y2 + y3 ) 33  Q. Find the coordinates of the barycenter of the triangle with three vertices of (2, 3), (3, 5) and ( 4, 4). Q. When three vertices are at A(1, 5), B(6, 1) and C(x, y) and the barycenter of △ABC is at the origin, find the values for x and y.

168 Chapter 7 Geometry and Equations 1 2 Equation of a straight line Suppose you are given a geometric figure on a coordinate plane. When the coordinates of an arbitrary point on the figure are (x, y), the relational expression established between x and y is called the equation of the figure. Let’s think about various equations of straight lines. First let us summarize the graph of a linear function and a straight line parallel to the y-axis, which we learned previously on P.72.  Equation of a straight line (1)  The equation of a straight line with a slope of m and an intercept of n is, y = mx + n. When m = 0, y = n which represents a straight line parallel to the x-axis. The equation of a straight line parallel to the y-axis, passing through the point (k, 0) is, x=k   ℓ y y (x, y) x=k m n 1 n y=n O x Ok x Let’s find the equation of the straight line ℓ with a slope of m, passing through the point A(x1, y1). Let (x, y) be the coordinates of an arbitrary y ℓ point P different from A on the line ℓ. y P Since the slope of AP is y − y1 , y − y1 x − x1 y1 A x − x1 y − y1 x − x1 =m The above is established. Therefore, O x1 xx y − y1 = m(x − x1) Obviously, the equation above is also established for the point A.

§ 1 Point and Line 169  Equation of a straight line (2)  The equation of a straight line with a slope of m, passing through the point (x1, y1) is, y − y1 = m(x − x1)   Q. Find the equations of the following straight lines. (1) The straight line with a slope of 3, passing through the point (2, 0). (2) The straight line passing through the point (−2, 3) and making 60˚ angle with the x-axis. Let s find the equation of the straight line ℓ passing through two points A(x1, y1) and B(x2, y2). y ℓ When x1 =\\ x2, let the slope of the straight y2 B line ℓ be m. y2 − y1 m= y2 − y1 y1 A x2 − x1 x2 − x1 Since the straight line ℓ passes through the x2 x point A, O x1 y − y1 = y2 − y1 (x − x1) x2 − x1 Then, when x1 = x2, since ℓ is parallel to the y-axis, the equation is x = x1.  Equation of a straight line (3)  The equation of a straight line passing through two points (x1, y1) and (x2, y2) is, y − y1 = y2 − y1 (x − x1) When x1 =\\ x2, x2 − x1 When x1 = x2, x = x1   Q. Find the equation of a straight line passing through each of the following two points. (1) (3, 2), (5, 10) (2) (2, 0), (0, 3) (3) (1, −3), (−7, 4) (4) (−3, −3), (−3, 4)

170 Chapter 7 Geometry and Equations The equations of the straight lines y = mx+n and x = k can be transformed as follows. 1 · x + 0 · y + (−1)k = 0 mx + (−1)y + n = 0, The same thing can be said about the others and every straight line can be represented by a linear equation for x and y.  Equation of a straight line (4)  ax + by + c = 0 (a =\\ 0 or b =\\ 0)   Ex.2 (1) 2x − y + 1 = 0 is the equation of the straight y (2) (1) line with a slope of 2 and an intercept of 1. (2) 5x − 4 = 0 is the equation of the straight 1 line parallel to the y-axis, passing through − 1 4 (4 ) 2 5 x 5 0 the point , on the x-axis. O Q. Draw the straight lines which represented by the following equa- tions. (1) 3x − 4y + 5 = 0 (2) 2x + 5 = 0 (3) 3y − 4 = 0 1 3 Relation between two straight lines y m ℓ m′ ℓ′ Let’s look into the relation between two straight lines. n ℓ : y = mx + n 1 x ℓ′ : y = m′x + n′ n′ First, from the definition of the slope and intercept, 1 we find that, O When m = m′ and n =\\ n′, ℓ and ℓ′ are parallel to each other. When m = m′ and n = n′, ℓ and ℓ′ coincide with each other. Next, let’s find the condition for ℓ and ℓ′ to be perpendicular when m =\\ 0 and m′ =\\ 0. If the straight lines ℓ and ℓ′ are translated parallel to pass through the origin, the equations of the straight lines will be represented as follows. y = mx, y = m′x

§ 1 Point and Line 171 Now let us take the points P(1, m)andQ(1, m′) on these straight lines. If ℓ and ℓ′ are perpendicular to each other, from ℓ′ y ℓ y = mx Pythagorean proposition, P(1, m) OP2+OQ2 = PQ2 That is, (12 + m2) + (12 + m′2) = (m − m′)2 O 1x From the above, mm′ = −1 If mm′ = −1 is established, then we will find Q(1, m′) that the two straight lines are perpendicular to y = m′x each other.  Condition for two lines to be parallel and perpendicular  For two straight lines y = mx + n and y = m′x + n′, Condition for two lines to be parallel or coincide m = m′ Condition for two lines to be perpendicular mm′ = −1   Exercise Find the equations of the straight lines parallel and perpen- dicular respectively to the given straight line 2x + 3y − 6 = 0, passing through the point (2, 4). Solution Since the slope of the given y 3x − 2y + 2 = 0 (2, 4) straight line is − 2 , 3 2x + 3y − 16 = 0 x the equation of the straight line paral- 2x + 3y − 6 = 0 lel to the given straight line is, // y − 4 = − 2 (x − 2) 3 ∴ 2x + 3y − 16 = 0 O Also the equation of the straight line perpendicular to the given straight line is, y−4= 3 (x − 2) 2 ∴ 3x − 2y + 2 = 0

172 Chapter 7 Geometry and Equations Q. Find the equation of a straight line that satisfies each of the follow- ing conditions. (1) The straight line parallel to the given straight line y = x, passing through the point (5, 3). (2) The straight line perpendicular to the given straight line 2x+4y+5 = 0, passing through the point (3, −1). (3) The straight line perpendicular to the given straight line x + 1 = 0, passing through the point (−3, 2). (4) The straight line parallel to the y-axis, passing through the point (−2, 1). Q. Find the equation of the perpendicular bisector of the segment AB given A(2, 3) and B(5, −2).

§ 1 Point and Line 173 Exercises 1–A 1. Find the coordinates of the point which is equidistant from three points (0, 6), (6, −2) and (7, 5). 2. The point which divides the segment joining two points (a b) and (1, 5) internally in the ratio 2 : 1 is (−1, 1). Find the values for a and b. 3. Find the values for a and b such that the coordinates of the barycenter of the triangle is (−3, 1) with three vertices of (3, 4), (−1, −5) and (a, b). 4. Find the equation of a straight line for each of the followings. (1) The perpendicular bisector of the segment AB given A(2, 3) and B(4, 5). (2) The straight line perpendicular to the given straight line passing through two points (−1, 3) and (3, −2), passing through the point (0, 5). (3) The straight line parallel to the given straight line 2x + 5y − 7 = 0, passing through the intersection point of two straight lines 4x − 3y + 5 = 0 and x + 2y − 7 = 0. 5. Find the equation of the straight line passing through the point which divides the segment joining two points (12, −1) and (4, 3) internally in the ratio 3 : 1 and perpendicular to the segment. 6. Find the value for the constant k such that three straight lines 2x − 3y = 8, x − 4y = 9 and kx + y = 3 intersect at single point. 7. Each of the following conditions is established for two straight lines ax + by + c = 0 and a′x + b′y + c′ = 0. Prove each of them when b =\\ 0 and b′ =\\ 0. (1) The condition for two straight lines to be parallel or perpendicular is, ab′ = a′b (2) The condition for two straight lines to be perpendicular is, aa′ + bb′ = 0

174 Chapter 7 Geometry and Equations Exercises 1–B 1. Find the value for the constant m such that two straight lines x+my+m−3 = 0 and mx + (m + 2)y − 2 = 0 are parallel to each other. 2. With respect to the given straight line y = 2x + 1, find the coordinates of the point symmetric to the point (3, 4). 3. For the straight line intersecting perpendicular to two parallel straight lines 3x + 4y + 5 = 0 and 3x + 4y − 6 = 0, find the length of the segment cut by two parallel straight lines. 4. For △ABC with vertices of A(x1, y1), B(x2, y2) and C(x3, y3), let the points which divide the sides BC, CA and AB of △ABC internally in the ratio m : n be P, Q and R respectively. Prove that the barycenter of △ABC coincides with that of △PQR. Note that m > 0 and n > 0. 5. For the straight line passing through the origin and perpendicular to the given straight line ℓ : ax + by + c = 0, let the y point where the straight line intersects ℓ be H. ℓ Answer the following questions. H (1) Find the coordinates of the point H. O x (2) Prove that the distance OH between O and ℓ is given by the following equation. OH = √ |c| a2 + b2 6. Prove that the distance d between the point (x1, y1) and the straight line ax + by + c = 0 is given by the following equation. d = |ax√1 + by1 + c| a2 + b2

§ 2 Quadratic curve 175 § 2 Quadratic curve 2 1 Equation of a circle In geometry, when a point P moves on a plane satisfying a given condition, we call it the locus of a point P. A circle is the y locus of points such that the distance from a P(x, y) fixed point C has a constant value of r. Cr The fixed point C is called the center of a circle and r is called the radius. O x Let s find the equation of a circle. Let the center of a circle be C(a, b), the radius be r and the point on the circumference be P(x, y). Since CP = r, √ (x − a)2 + (y − b)2 = r By squaring both sides, (1) (x − a)2 + (y − b)2 = r2 Equation of a circle   The equation of a circle with the center point (a, b) and a radius of r is, (x − a)2 + (y − b)2 = r2 In particular, the equation of a circle with the center at the origin and a radius of r is, x2 + y2 = r2   Ex.1 The equation of a circle with the cen- y ter point (3, −2) and a radius of 4 is, 3 x (x − 3)2 + (y + 2)2 = 42 O4 Q. Find the equations of the following circles. −2 C (1) The circle with the center (−1, 2) and a radius of 2. (2) The circle which has the origin and the point (8, 6) as the endpoints of a diameter. (3) The circle which has two points (2, 7) and (−8, 1) as the endpoints of a diameter.

176 Chapter 7 Geometry and Equations Let us expand the equation (1). Then we have, x2 + y2 − 2ax − 2by + a2 + b2 − r2 = 0 Therefore, the equation of a circle can also be written as, x2 + y2 + Ax + By + C = 0 (where A, B, C are constants) (2) Ex.2 Let us transform the equation x2 + y2 − 2x + 8y − 10 = 0, x2 − 2x + 1 + y2 + 8y + 16 = 10 + 1 + 16 Therefore, (x − 1)2 + (y + 4)2 = 27 [ = (3√3)2] 27 √ This represents the circle with the center (1, −4) and a radius of 3 3. Note The equation (2) does not always represent a circle. For example, Let us transform x2 + y2 − 2x + 8y + 18 = 0. Then we have, (x − 1)2 + (y + 4)2 = −1. Therefore, there s no point that satisfies x2 + y2 − 2x + 8y + 18 = 0. Q. Find the center and radius for each of the following circles repre- sented by given equations. (1) x2 + y2 + 4x − 6y + 4 = 0 (2) (x − 1)(x − 3) + (y − 5)(y − 7) = 7 Exercise Find the equation of the circle passing through the points P(−2, −1), Q(2, −3) and R(0, 3). Solution Let us suppose that the required equation of the circle is, x2 + y2 + Ax + By + C = 0 y R Since the circle passes through the points P(−2, −1), Q(2, −3) and R(0, 3), 1 x 4 + 1 − 2A − B + C = 0 4 + 9 + 2A − 3B + C = 0 O1 P 9 + 3B + C = 0 Q By solving the simultaneous equation, A = −2, B = 0 and C = −9. ∴ x2 + y2 − 2x − 9 = 0 that is, (x − 1)2 + y2 = 10 //

§ 2 Quadratic curve 177 Q. Find the equations of the following circles. (1) The circle passing through three points at (1, 1), (2, −1) and (3, 2). (2) The circle with the center on the y-axis, passing through two points (−1, 1) and (3, 5). Exercise Find the locus of a point P that satisfies AP : BP = 2 : 1 given two points A(−3, 0) and B(3 0). Solution Let the coordinates of a point P be (x, y). Then we have, √ AP = (x + 3)2 + y2 y √ BP = (x − 3)2 + y2 P Since AP=2BP A 1 x √√ O 1B (x + 3)2 + y2 = 2 (x − 3)2 + y2 Let us simplify by squaring both sides. Then we have, (x − 5)2 + y2 = 16 Therefore, the required locus is the circle with the center (5, 0) and a radius of 4. // Q. Given A(2, 0) and B(0, 1), find the locus of a point P that satisfies each of the following conditions. (1) AP : BP = 3 : 2 (2) AP2 + BP2 = 4 In general, when an equation is given by a quadratic equation for x and y, the curve is called a quadratic curve. Since a circle can be represented by the equation (2), it is a quadratic curve. The parabola y = ax2 + bx + c is also a quadratic curve. As for a fractional function, (3) y = cx + d (a =\\ 0). ax + b Since it can be transformed into axy + by − cx − d = 0, the curve represented by (3) is also a quadratic curve.

178 Chapter 7 Geometry and Equations 2 2 Various types of quadratic curves ⋄Ellipse Given a and b are positive constants, when the circle x2+y2 = a2 is stretched or compressed towards y-axis by a factor of b , it shows the following curves. a a>b a<b yy a bB P Q bB Q a P A′ O H A x A′ A x −a a −a O H a −b B′ a a −b B′ The curve formed in this way is called an ellipse. Let the coordinates of a point P on the circle be (x, y) and the coordinates of a point Q on the ellipse be (X, Y ). x2 + y2 = a2 (1) (2) X =x x=X (3) b where, a a b Y = y y= Y The above is established. By substituting (2) for (1), X2 + a2 Y 2 = a2 b2 X2 + Y 2 = 1 a2 b2 Therefore, the equation of an ellipse is represented as follows. x2 + y2 = 1 a2 b2 (3) is called the standard form of the equation of an ellipse. The intersection points of the ellipse (3) and the x-axis are A(a, 0) and A′(−a, 0). The intersection points with the y-axis are B(0, b) and B′(0, −b). A, A′, B and B′ are each called the vertex of the ellipse and the origin O is called the center.

§ 2 Quadratic curve 179 When a > b, an ellipse becomes a horizontally long ellipse and when a < b, an ellipse becomes a vertically long ellipse. For AA′ and BB′, the longer one is called the major axis and the shorter one is called the minor axis. An ellipse is symmetric with respect to the major axis and the minor axis. For the ellipse (3), √ When a > b, F(c, 0), F′(−c, 0), ( Note c = a2 − b2) When a < b, F(0, c), F′(0, −c), √ ( Note c = b2 − a2) From the above, find two points F and F′. Then it follows that for a point P on the ellipse, the value of PF + PF′ can be proved to be constant. F and F′ are each called the focus of the ellipse. An ellipse is the locus of points such that the sum of the distances from two foci is constant. a>b a<b y y bB P bB cF P A′ F′ O F A x A′ A x −a −c c a −a O a −b B′ −c F′ −b B′ Note Although PF + PF′ has a constant value, when a > b, it becomes, AF + AF′ = (a − c) + (a + c) = 2a √ BF + BF′ = 2 c2 + b2 = 2a From the above, we find PF + PF′ = 2a. Likewise, when a < b, PF + PF′ = 2b. It is known that when a curved mirror is placed along an ellipse, all the light emitted from one focus is concentrated in the other focus.

180 Chapter 7 Geometry and Equations Equation of an ellipse   The equation of an ellipse x2 + y2 = 1 (a > 0, b > 0) When a > b, a2 b2 Foci F(c, 0), F′(−c, 0) √ (Note c = a2 − b2) The length of major axis is 2a, the length of minor axis is 2b When a < b, √ Foci F(0, c), F′(0, −c) (Note c = b2 − a2) The length of major axis is 2b, the length of minor axis is 2a   Ex.3 Given x2 + y2 = 1, 9 4√ √ a = 3, b = 2, c = 32 − 22 = 5 √√ Therefore, the foci of the ellipse are F( 5, 0) and F′(− 5, 0). The lengths of major and minor axes are 6 and 4 respectively. Ex.4 Given the foci at (0, 3) and (0, −3) and the length of the major axis 12, for a vertically long ellipse, c = 3, b = 6. Also from c2 = b2 − a2, a2 = b2 − c2 = 62 − 32 = 27 Therefore, the equation of the ellipse is x2 + y2 = 1 27 36 Q. Find the equation of the ellipse with four vertices of (7, 0), (−7, 0), (0, 4) and (0, −4), also find the coordinates of the foci. Q. Find the equation of the ellipse with two foci at (2, 0) and (−2, 0) and the length of the minor axis 4, also draw a rough sketch of the ellipse. Q. Find the equation of the ellipse with two foci at (0, 4) and (0, −4) and the length of major axis 10, also draw a rough sketch of the ellipse. Q. Find the coordinates of the foci and the lengths of the major and minor axes for each of the following equations, also draw a rough sketch for each of them. (2) 4x2 + y2 = 4 (3) 4x2 + 9y2 = 4 (1) x2 + y2 = 1 36 25

§ 2 Quadratic curve 181 ⋄Hyperbola Given a and b are positive constants, x2 − y2 = 1 (4) a2 b2 (5) or x2 − y2 = −1 a2 b2 The geometric figure represented by the equation above is called a hyperbola. (4) and (5) are called the standard form of the equation of a hyperbola. The rough sketches of the hyperbolas (4) and (5) are shown below. x2 − y2 =1 x2 − y2 = −1 a2 b2 a2 b2 y y A′ A x bB x −a O a O −b B′ Let s look into the properties of the hyperbola (4). First of all, (4) is symmetric with respect to the x-axis and y-axis. The intersection points with the x-axis are A(a, 0) and A′(−a, 0) and there’s no common point with the y-axis. A and A′ are each called the vertex, the origin O is called the center and AA′ is called the principal axis. √ For the hyperbola (4), given c = a2 + b2, y F( c, 0) and F′(−c, 0) are each called the focus. Then it follows that, for an arbitrary P point P, the value of |PF − PF′| can be proved F′ F x to be always 2a. That is, a hyperbola is the −c −a O ac locus of points such that the difference of the distances from two foci is constant.

182 Chapter 7 Geometry and Equations A hyperbola has asymptotes and this is shown as follows. Given x > 0 and y > 0, by transforming (4), y y= b x ( )( ) O a x y x y =1 a − b a + b x − y = 1 y b x a b b a x + y a ax Since the value of the denominator on the right side gets infinitely larger when a point (x, y) on y y= b x a the curve gets infinitely far away from the origin, b F′ A′ x y 1 ( b ) −c −a O a b b a y − = x − −b AF x ac the value of the above gets infinitely closer to 0. That is, the straight line y = b x is the asymp- y = − b x a a tote of a hyperbola. Likewise, for other quadrants, the asymptotes y = ± b x can be found. a  Equation of a hyperbola (1)  The equation of a hyperbola x2 − y2 = 1 a2 b2 The coordinates of the foci are, √ F(c, 0) F′(−c, 0) Note c = a2 + b2 The equation of the asymptotes is, y = ± b x a   Ex.5 Given x2 − y2 = 1, y y= 2x 3 32 22 √ √ 2 a = 3, b = 2, c = 32 + 22 = 13 The foci are F′ F x √√ √ √ − 13 −3 O 3 13 F( 13, 0) and F′(− 13, 0) −2 2 The asymptotes are y = ± 3 x y=−2x 3

§ 2 Quadratic curve 183 Ex.6 Given the foci at (3, 0) and (−3, 0) and the length of the principal axis 4, c = 3, a = 2, b2 = c2 − a2 = 32 − 22 = 5 The equation of the hyperbola is x2 − y2 = 1 4 5√ 5 and the equation of the asymptotes is y = ± 2 x. Q. Find the coordinates of the foci and the equation of the asymptotes for the hyperbola x2 − y2 = 1. 32 42 Q. Find the equation of the hyperbola with the foci at (6, 0) and (−6, 0), passing through two points (3, 0) and (−3, 0), also find the equa- tion of the asymptotes. Q. Find the coordinates of the foci and the equation of the asymptotes for each of the following hyperbolas, also draw a rough sketch for each of them. (1) x2 − y2 = 1 (2) x2 − y2 = 1 (3) 4x2 − 9y2 = 4 16 9 For the hyperbola (5), the intersection points y with the y-axis are B(0, b) and B′(0, −b) and there’s no common point with the x-axis. B F y= bx and B′ are each called the vertex, the origin bB a O is called the center and BB′ is called the principal axis. −a a x −b B′ Also, the foci are on the y-axis and there are − b two asymptotes just like (4). F′ a Then, the above can be summarized as follows. y = x  Equation of a hyperbola (2)  The equation of a hyperbola x2 − y2 = −1 a2 b2 The coordinates of the foci are, √ F(0, c) F′(0, −c) Note c = a2 + b2 The equation of the asymptotes is, y = ± b x a   Q. Find the coordinates of the foci and the equation of the asymptotes for each of the following hyperbolas, also draw a rough sketch for each of them. (1) x2 − y2 = −1 (2) x2 − y2 = −1 (3) x2 − y2 = −1 49 16 9

184 Chapter 7 Geometry and Equations ⋄Parabola For a nonzero constant p, y2 = 4px (6) The geometric figure represented by the equation above is called a parabola and (6) is called the standard form of the equation of a parabola. The parabola (6) is symmetric with respect to ℓy the x-axis. The x-axis is called the axis of the H P(x, y) parabola and the origin O is the vertex. The point F(p, 0) is called the focus of the F x parabola and the straight line x = −p is called −p O p the directrix of the parabola. When the perpendicular line PH is drawn from a given point P on the parabola to the directrix, PF = PH is established. That is, a parabola is the locus of all points P such that the length of the perpendicular line PH equals to that of the segment PF.  Equation of a parabola  The equation of a parabola y2 = 4px The directrix ℓ is, x = −p The focus is, F(p, 0)   (7) y x2 = 4py The geometric figure represented by the equa- tion above is also a parabola. Since (7) is p P x transformed into y = 1 x2, it is a quadratic F H 4p O −p function which we learned on Chapter 3. The ℓ directrix of (7) is y = −p and the focus is F(0, p). Q. Find the equation of the parabola with the coordinates of the focus at (2, 0) and the straight line x = −2 as the directrix, also draw a rough sketch of the parabola. Q. Find the coordinates of the focus and the equation of the directrix for each of the following parabolas. (1) y2 = 16x (2) y2 = −4x (3) x2 = y (4) x2 = −2y

§ 2 Quadratic curve 185 2 3 A tangent to a quadratic curve The coordinates of the common point of a quadratic curve C and a straight line ℓ can be found by solving the simultaneous equations so that the equations of C and ℓ are simultaneously satisfied. Especially when a quadratic equation for x and y has a double root, C and ℓ have single point in common. This common point is called a point of contact and ℓ is called a tangent to C. Exercise Find the tangent to the ellipse x2 + y2 = 1 and with a slope of 1. 35 Solution Let the required tangent be y = x + k. y√ y=x+2 2 By solving simultaneous equations of the √ ellipse and he tangent, 5 x2 + y2 = 1 (1) √ 35 3 y = x+k (2) √ Ox −3 By substituting (2) for (1), x2 + (x + k)2 = 1 √ √ 35 −5 y=x−2 2 By simplifying the above, 8x2 + 6kx + (3k2 − 15) = 0 Since the condition of tangency is, D = 36k2 − 32(3k2 − 15) = 0 From this, // √ ∴ k = ±2 2 Q. Find the value for the constant k such that the straight line y = −x + k is the tangent to the parabola y2 = 4x. Q. Find the tangent to the hyperbola x2 − y2 = −1 and with a slope of 2. 9

186 Chapter 7 Geometry and Equations Exercise Prove that the equation of the tangent to the point A(x0, y0) on the circle x2 + y2 = r2 is, x0x + y0y = r2 Proof The center of the circle is O(0, 0) and the radius is r. Let a given point on the tangent be P(x, y). y Since the tangent is perpendicular to OA, A(x0, y0) from Pythagorean proposition, r x OP2 = AP2 + OA2 O P(x, y) Where OA = r, x2 + y2 = (x − x0)2 + (y − y0)2 + r2 By expanding and simplifying the above, 2x0x + 2y0y = x2 + y2 + r2 = 2r2 0 0 Therefore, x0x + y0y = r2 // Q. Find the equation of a tangent to each of the following points on the circle x2 + y2 = 25. (1) (3, 4) (2) (−4, 3) (3) (−5, 0) A circle which is inscribed in three sides of a tri- A angle is called the inscribed circle of a triangle Inscribed circle and the center of an inscribed circle is called the incenter of a triangle. The incenter of a triangle incenter O C is the intersection point of three angle bisectors r of a triangle. B Q. When the radius of an inscribed circle is r and the lengths of three sides of △ABC are a, b and c, prove that the area S of △ABC is, S= 1 (a + b + c)r 2 Q. Find the area of the triangle with three sides whose length are 13, 14 and 15 by using Heron’s formula, also find the radius of an inscribed circle of the triangle.

§ 2 Quadratic curve 187 2 4 Inequality and domain When an inequality involving variables for x and y is given, the range of points (x, y) that satisfies the inequality is called the domain represented by the inequality. Inequality (1) y > ax + b Let’s find the domain represented by the inequality above. Straight line (2) y = ax + b The coordinate plane is divided into two areas by the straight line, the area above the line and the area below the line. If the point P(x1, y1) is in the domain represented by the inequality (1), y1 > ax1 + b (3) y The above is established. y1 P Since the point Q(x1, ax1 + b) is on the ax1 + b y = ax + b straight line, from (3), we find the point Q P(x1, y1) is in the area above the straight O x1 x line. Therefore, y > ax + b represents the area above the straight line (2). Likewise, y < ax + b represents the area below the straight line (2). Ex.7 Let us find the domain represented by the inequality x + 2y − 2 ≦ 0. By transforming the inequality, y y ≦ − 1 x + 1 1 2 Therefore, the domain represented by the inequality is the straight line y = − 1 x+1 O 2x 2 and the area below the line, as shown in the shaded area in the figure. Note that the boundary is included.

188 Chapter 7 Geometry and Equations Likewise, for the curve y = f (x), the domain represented by y > f (x) is the area above the curve y = f (x), the domain represented by y < f (x) is the area below the curve y = f (x). Q. Show the domains represented by the following inequalities. (1) y < 2x − 1 (2) 3x + 2y > 6 (3) y < x2 + 4x (4) x2 − 2x − y − 3 ≦ 0 Exercise Show the domain represented by the inequality x2 + y2 ≦ 1. Solution The equation x2 + y2 = 1 represents the circle with the center at the origin O and a radius of 1. y Let us take the point P(x, y) inside the circle 1 or on the circumference. Since OP ≦ 1, −1 O x √ 1 x2 + y2 ≦ 1 From the above, x2 + y2 ≦ 1 −1 Therefore, this satisfies the condition. Then let us take the point P(x, y) outside the circle. √ Since OP > 1, x2 + y2 > 1 From the above, x2 + y2 > 1 We find this does not satisfy the inequality. Therefore, the required domain is the area inside the circle and on the circumference as shown in the figure. Note As in the case with a circle, the areas inside and outside the ellipse x2 + y2 = 1 are represented with the inequalities a2 b2 x2 + y2 < 1 and x2 + y2 > 1 respectively. a2 b2 a2 b2 Q. Show the domains represented by the following inequalities. (1) x2 + y2 > 4 (2) x2 − 2x + y2 ≦ 0 (3) x2 + y2 < 1 94

§ 2 Quadratic curve 189 The range of points (x, y) that satisfies multiple inequalities at the same time is called the domain represented by the simultaneous inequalities. Exercise Show the domain represented by the following simultaneous inequalities. x+y > 1 (1) x2 + y2 < 4 (2) Solution From (1), y > −x + 1 y Therefore, the domain represented by (1) is 2 the area above the straight line y = −x + 1. 1 Then, the area represented by (2) is the area inside the circle with the center at the origin −2 O 1 x 2 and a radius of 2. The required domain is the −2 overlapped area between the two domains. Note that the boundary is not included. // Q. Show the domains represented by the following simultaneous in- equalities. x2 + y2 > 4 x−y+2≧0 (1) (2) x2 + y2 < 9 x2 − y < 0 Exercise When x and y satisfy the following inequalities, find the maximum value for x + y. y ≦ −2x + 5, x + 2y − 6 ≦ 0, x ≧ 0, y ≧ 0 Solution The domain represented by the y 6x simultaneous inequalities is the shaded area 5 in the figure. Note that the boundary is in- cluded. Let us take x + y = k. O y = −x + k

190 Chapter 7 Geometry and Equations This represents the straight line with a slope of −1 and an intercept of k. For the straight line, the point where the in- tercept k has its maximum while having the shaded area and the common point is the intersection point P, where the straight line crosses the inter- section point P of the two straight lines. y = −2x + 5, x 2y −6 =0 ( 4 7 ) finding the coordinates of the intersection 3 3 . By point P, we have, , Therefore, x + y has its maximum given x = 4 and y = 7 . 33 The maximum value is 11 . // 3 Q. When x and y satisfy the simultaneous inequalities, x + 2y − 6 ≦ 0, 2x + y − 6 ≦ 0, x ≧ 0, y ≧ 0 find the maximum value for each of the following expressions. (1) x + y (2) 4x + y Column Quadratic curve (Conic section) Let’s look into the properties of the focus F of a parabola. Let us take the point P(x1, y1) on the parabola y2 = 4px. √ PF= √(x1 − p)2 + y2 y ℓ 1 = (x1 − p)2 + 4px1 H θ √ M P(x1, y1) = (x1 + p)2 = x1 + p (x1 + p > 0) −p O p F x Since PH= x1 + p, PF=PH is established. The tangent to the point P on the parabola 2y1y = 4px + y2 is the straight 1 line ℓ. The above can be found because the equation obtained by eliminating 4px from two equations, where 2y1y = y2 + y2 , hmaisdapodinout bMle(r0o,oyt1y)=ayn1d. 1 Since the straight line ℓ passes through the 2 the slopes of ℓ and HF are 2p and − y1 respectively, we find that the straight y1 2p

§ 2 Quadratic curve 191 line ℓ is the perpendicular bisector of HF. Therefore, by letting the angle between the straight line HP and ℓ be θ, we have ∠FPM= θ. That means, when a curved mirror is placed along the parabola, the light emitted from the focus F is reflected parallel to the principal axis. The following figures show the cases of an ellipse and a hyperbola, yy P P Fx F′ F x O F′ O There are upper and lower cones obtained by rotating the straight line y = mx about the y-axis. If the cones are cut through with a plane not passing through the origin, the conic sections will be an ellipse (1), a parabola (2) or a hyperbola (3) depending on the angle of the cut. From this, a quadratic curve is also called a conic section. It is said that conic sections have been studied systematically in Greece since around the fourth century B.C. y y = mx (3) (3) (1) x (1) (2) (2)

192 Chapter 7 Geometry and Equations Exercises 2–A 1. Find the coordinates of the center and the radius of a circle represented by each of the following equations. (1) x2 + y2 + 4x − 6y + 5 = 0 (2) x2 + y2 − 2y − 4 = 0 2. Find the equations of the following circles. (1) The circle with the coordinates of the center at (−1, 2), passing through the point (3 5). (2) The circle with the center on the straight line y = x, passing through two points (0, 0) and (1, 2). 3. Given two points A(−2, 0) and B(2, 0), find the equation of the locus of a point P that satisfies each of the following conditions, also draw the locus for each of them. (1) AP2 + BP2 = 10 (2) AP2 − BP2 = 1 4. Draw the rough sketches for the following quadratic curves. (1) x2 + y2 = 1 (2) 5y2 − 4x = 0 (3) x2 − y2 = 1 43 95 5. Find the equations and draw the rough sketches for the following quadratic curves. √√ (1) The ellipse with the foci at (0, 5) and (0, − 5), and the length of the major axis 8. (2) The hyperbola with the asymptotes y = ± 3 x, 2 passing through the point (2, 0). 6. Find the value for the constant k such that the straight line y = x + k is tangent to the hyperbola 4x2 − y2 = −4, also find the coordinates of the point of contact. 7. Show the domain represented by each of the following inequalities.   x2 + y2 < 4  x2 + y ≦ 0 (1)  2x + y + 2 > 0 (2)  x − y − 2 ≦ 0 8. When x and y satisfy the following inequalities, find the maximum and min- imum values for 2x + y. y ≦ x + 2, y ≧ −x + 2, y ≧ 4x − 1

§ 2 Quadratic curve 193 Exercises 2–B 1. Find the value for the constant m such that the straight line y = mx is tangent to the circle x2 + y2 − 4x − 6y + 12 = 0, also find the range of the value for m such that they have no common point. 2. Prove that the equation of the tangent to the point (x1, y1) on the circle (x − a)2 + (y − b)2 = r2 is, (x1 − a)(x − a) + (y1 − b)(y − b) = r2 In addition, by using the above, find the equation of the tangent to the point (3, 6) on the circle (x + 2)2 + (y − 3)2 = 34. 3. Find the number of common points of the ellipse 3x2 + 4y2 = 12 and the straight line y = mx + 2. 4. Given two points A and B on the parabola y2 = 4px, △OAB is a equilateral triangle. Find the area of △OAB. 5. For the segment AB with a length of 3, when both ends of A and B move along the x-axis and y-axis respectively, let the point dividing the segment AB internally in the ratio 1 : 2 be the point P(x, y). y Answer the following questions. B (1) Let the coordinates of A and B be (a, 0), (0, b), O P(x, y) respectively. Find the relation between a and b. x (2) Express x and y in terms of a and b. A (3) Find the equation of the locus of a point P and draw the locus. 6. Prove that two tangents drawn from a point on the directrix to the parabola x2 = 4py are perpendicular to each other. 7. When x and y satisfy the following simultaneous inequalities, find the maximum and minimum values for 2x − 3y. y2 ≦ 4x, x + y ≦ 3

194 Chapter 7 Number of cases and sequence of numbers § 1 Number of cases 1 1 Number of cases The roads connecting four points P, Q, R and S are shown below. Let s check how many possible ways to travel from P to S there are without passing the same point twice. PQ R S S S There are three ways to travel from P to R, and R S R S for each way, there are four ways to travel from R P R S to S. Therefore, there are S S 3 × 4 = 12 (possible ways) (1) Q S S A tree diagram of the possible ways to travel from S S P to S is shown on the right, and is calculated as above (1). S In general, for the two events A and B, if there are m possible ways for A to happen, and for each way, if there are n possible ways for B to happen, the number of cases for both A and B to happen is mn. This is called Product rule.

§ 1 Number of cases 195 The previous example can also be considered as follows. (i) When traveling without passing through Q. P S There is only one way to travel from P to S R S R without passing through Q, and there are S four possible ways to travel from R to S. Therefore, 1 × 4 = 4 (possible ways) S (ii) When traveling through Q. S R S PQ S S There are two ways to travel from Q to R, S and there are four possible ways to travel from R to S. R Therefore, 2 × 4 = 8 (possible ways) S S At this time, the way to travel from P to S would be either (i) or (ii). Therefore, the number of possible ways is 4 + 8 = 12 (possible ways) In general, if there are m possible ways for A to happen and n possible ways for B to happen, when they don t happen at the same time, the number of cases for either A or B to happen is m + n. This is called Sum rule. The product rule and the sum rule are established likewise, in the case of three or more events. Exercise How many divisors does 360 have ? Solution The prime factorization of 360 is 360 = 23 × 32 × 51 Therefore, the divisors of 360 are expressed with one of the possible number arrangements. 2p 3q 5r (p = 0, 1, 2, 3, q = 0, 1, 2, r = 0, 1) Here, there are 4, 3, 2 possible choices of numbers for p, q, r respectively. Therefore, the number of divisors that we are looking for is, from the product rule, 4 × 3 × 2 = 24 (divisors) // Q. How many divisors does 300 have? Q. Find the numbers of divisors for the following formulas. (1) (x + 1)4(x + 2)2(x + 3)5 (2) x4 − x2

196 Chapter 7 Number of cases and sequence of numbers Exercise How many pairs (x, y) of positive integers are there, which satisfy 5x + y ≦ 20 ? Solution From 5x + y ≦ 20, y ≦ 20 − 5x When x = 1, 1 ≦ y ≦ 15. Thus 15 (pairs) When x = 2, 1 ≦ y ≦ 10. Thus 10 (pairs) When x = 3, 1 ≦ y ≦ 5. Thus 5 (pairs) The number that we are looking for is, from the sum rule, 15 + 10 + 5 = 30 (pairs). Q. How many pairs (x, y) of positive integers are there, which satisfy 3x + y ≦ 12 ? Q. How many sets (x, y, z) of positive integers are there, which satisfy 2x + y + z = 8 ? Q. When three dice are thrown (large, medium, small) at the same time, how many possible cases are there, which the number of large die is equal to the sum of medium and small dice. b c d 1 2 Permutation acb d Choose three letters from a choice of four letters a, b, c, d b c and d, then, line them up as shown below. Let’s check how ac d a many different ways to line them up there are. b c d abc abd acb acd bac bad d a c There are four possible choices for the first letter, and for a b d a each of them, there are three possible choices for the second c b d letter, and two possible choices for the third letter. d a b Therefore, the total number of different ways to line them a b c up is, from the product rule, a dbc 4 × 3 × 2 = 24 (different ways) c a b

§ 1 Number of cases 197 In general, when taking r out of n choices and lining them up, it is called permutation of r from n choices and denoted as nPr. Similar to the previous example, you will see the following formula is estab- lished. Ex.1 Pn r = n(n − 1)(n − 2) (n − r + 1) Q. P4 3 = 4 · 3 · 2 = 24, P5 5 = 5 · 4 · 3 · 2 · 1 = 120 Find the following values. (1) P6 3 (2) P5 1 (3) P7 4 (4) P6 6 The permutations of n from n choice are Pn n = n(n − 1)(n − 2) 3 · 2 · 1 The product of all natural numbers from 1 to n (right side of formula) is called factorial and denoted as n!. Ex.2 2! = 2 · 1 = 2, 5! = 5 · 4 · 3 · 2 · 1 = 120 10! = 3628800, 15! = 1307674368000 6! = 6 ·⧹5 ·⧹4 ·⧹3 ·⧹2 ·⧹1 = 6 5! ⧹5 ·⧹4 ·⧹3 ·⧹2 ·⧹1 Q. Find the following values. (1) 4! (2) 7! (3) 9! (4) n! 7! (n − 1)! From the definition, Pn n = n! When r < n, Pn r = n(n − 1)(n − 2) (n − r + 1) = n(n − 1)(n − 2) (n − r + 1) × (n − r)! = n! (n − r)! (n − r)! Therefore, the following formula is obtained.    Pn r = n(n − 1)(n − 2) (n − r + 1) = n! (n − r)!

198 Chapter 7 Number of cases and sequence of numbers Suppose r = n, in the formula. Left side = Pn n = n!, Right side = n! Here, define as 0! 0! = 1, so that the formula is established, when r = n. Exercise Form seven-digit integers using integers 0, 1, 2, 3, 4, 5, and 6. Note that the integers are not allowed to be used twice. Answer the following questions. (1) How many seven-digit integers can be formed ? (2) How many seven-digit even integers can be formed ? Solution (1) There are six possible integers for the first digit excluding 0, and for each of them, there are six possible integers for the rest of digits. Therefore, from the product rule, 6 × P6 6 = 6 × 6! = 4320 (2) If the seventh digit is 0, you just need to arrange the rest of numbers. P6 6 = 6! = 720 If the seventh digit is either 2, 4 or 6, you can think likewise to (1), 5 × 5P5. Thus, from the product rule, 3 × 5 × P5 5 = 1800 Therefore, from the sum rule, 720 + 1800 = 2520 // Q. When five letters a, b, c, d, and e are arranged in a line, how many possible arrangements are there in which the both ends of the line are vowels? Q. Three people A, B and C stay at room 1 to 7 separately. Answer the following questions. (1) How many different arrangements of rooms are there for these three people? (2) How many different arrangements are there for these three people to stay in the rooms with consecutive numbers?

§ 1 Number of cases 199 When forming pairs of letters using three letters a, b, and c, a . . . aa how many possible letter arrangements are there? a b . . . ab c . . . ac Note that the letters are allowed to be used more than once. a . . . ba There are three possible choices for the first letter, and for b b . . . bb c . . . bc each of them, there are three possible choices for the second a . . . ca letter as shown in the figure. Therefore, the total number of c b . . . cb letter arrangements is, from the product rule, c . . . cc 3 × 3 = 32 = 9 In general, when taking r out of n choices, with repetition, and lining them up, it is called repeated permutation of r from n choices. The following formula is established regarding repeated permutation.  Formula of repeated permutation  The total number of repeated permutations, where taking r out of n choices and lining them up, is nr.   Ex.3 When integers 1, 2, 3, 4, and 5 are arranged with repetition to form three-digit integers, the total number of three-digit integers is 53 = 125 Q. When four dice are thrown, how many different arrangements are there? Q. When three people play rock-paper-scissors, how many different arrangements are there? Q. When eight numbers are lined up using only 0 and 1 with repetition, how many possible arrangements are there ? Note The minimum unit of data represented by 0 and 1 is called a bit. Also, 8 bits are called 1 byte.