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P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 CHAPTER 1 A toolbox Objectives To revise the properties of sine ,cosine and tangent To revise methods for solving right-angled triangles To revise the sine rule and cosine rule To revise basic triangle, parallel line and circle geometry To revise arithmetic and geometric sequences To revise arithmetic and geometric series To revise infinite geometric series To revise cartesian equations for circles To sketch graphs of ellipses from the general cartesian relation (x−h)2 a2 + (y−k)2 b2 =1 To sketch graphs of hyperbolas from the general cartesian relation (x−h)2 a2 − (y−k)2 b2 =1 To consider asymptotic behaviour of hyperbolas To work with parametric equations for circles, ellipses and hyperbolas The first six sections of this chapter revise areas for which knowledge is required in this course, and which are referred to in the Specialist Mathematics Study Design. The final section introduces cartesian and parametric equations for ellipses and hyperbolas. 1.1 Circular functions Defining sine, cosine and tangent The unit circle is a circle of radius one with centre at the origin. It is the graph of the relation x2+ y2= 1. (0, 1) (1, 0) (0, _1) (_1, 0) 0 x y 1

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 2 Essential Specialist Mathematics Sine and cosine may be defined for any angle through the unit circle. 0 x y θ° P(cos ( θ°), sin ( θ°)) For the angle of ◦, a point Pon the unit circle is defined as illustrated opposite. The angle is measured in an anticlockwise direction from the positive direction of the xaxis. cos( ◦) is defined as the x-coordinate of the point P and sin( ◦) is defined as the y-coordinate of P. A calculator gives approximate values for these coordinates where the angle is given. 0 x y 135 ° (_0.7071, 0.7071) 0 x y (0.8660, 0.5) 30° x y 100 ° (_0.1736, 0.9848) 0 sin 30 ◦= 0.5 (exact value) cos 30 ◦= √3 2 ≈ 0.8660 sin 135 ◦= 1√2≈ 0.7071 cos 135 ◦= −1√2≈− 0.7071 cos 100 ◦≈− 0.1736 sin 100 ◦≈ 0.9848 tan( ◦) is defined by tan( ◦)= sin( ◦) cos( ◦). The value of tan( ◦) can be illustrated geometrically through the unit circle. x y T(1, tan ( θ°)) P' O P sin ( θ°) = PP' T' θ° By considering similar triangles OPP and OTT ,itcan be seen that TT  OT = PP  OP  i.e. TT = sin( ◦) cos( ◦)= tan( ◦) B' B 1 C' θ° O C For a right-angled triangle OBC , a similar triangle OB Ccan be constructed that lies in the unit circle. By the definition, OC = cos( ◦) and CB= sin( ◦). The scale factor is the length OB . Hence BC = OB sin( ◦) and OC = OB cos( ◦). This implies BC OB = sin( ◦) and OC OB = cos( ◦)

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 3 hypotenuse B adjacent C O θ° opposite This gives the ratio definition of sine and cosine for a right-angled triangle. The naming of sides with respect to an angle ◦is as shown. sin ◦= opp hyp  opposite hypotenuse  cos ◦= adj hyp  adjacent hypotenuse  tan ◦= opp adj opposite adjacent  unit1 1 1 P A 1c _1 _1 y O Definition of a radian In moving around the circle a distance of 1 unit from Ato P the angle POA is defined. The measure of this angle is 1 radian. One radian (written 1 c) is the angle subtended at the centre of the unit circle by an arc of length 1 unit. Note: Angles formed by moving anticlockwise around the circumference of the unit circle are defined as positive . Those formed by moving in a clockwise direction are said to be negative . Degrees and radians The angle, in radians, swept out in one revolution of a circle is 2 c ∴ 2c= 360 ◦ ∴ c= 180 ◦ ∴ 1c= 180 ◦  or 1 ◦= c 180 Henceforth the cmay be omitted. Any angle is assumed to be measured in radians unless otherwise indicated. The following table displays the conversions of some special angles from degrees to radians. Angles in degrees 0 30 45 60 90 180 360 Angles in radians 0  6  4  3  2  2

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 4 Essential Specialist Mathematics Some values for the trigonometric functions are given in the following table. xin radians sin x cos x tan x 0010  6 1 2 √3 2 √3 3  4 √2 2 √2 2 1  3 √3 2 1 2 √3  2 1 0 undefined The graphs of sine and cosine As sin x= sin( x+ 2n),n∈Z, the function is periodic and the period is 2 . A sketch of the graph of f:R→ R, f(x)= sin xis shown opposite. The amplitude is 1. y x −π π 2π −π π 2 3π −1 1 0 f(x) = sin x 2 2 A sketch of the graph of f:R→ R, f(x)= cos xis shown opposite. The period of the function is 2 . The amplitude is 1. y x −π π 2π −1 1 0 f(x) = cos x −π 2 π 2 3π 2 For y= acos( nx) and y= asin( nx) a> 0, n> 0 Period = 2 n , amplitude = a, range = [−a,a] Symmetry properties for sine and cosine From the graph of the functions or from the unit circle definitions, the following results may be obtained. sin( − )= sin  sin( + )=− sin  cos( − )=− cos  cos( + )=− cos  sin(2 − )=− sin  sin( −)=− sin  cos(2 − )= cos  cos( −)= cos  sin( + 2n)= sin  cos( + 2n)= cos for n∈Z sin  2−   = cos  cos  2−   = sin 

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 5 Example 1 a Change 135 ◦into radians. b Change 1.5 cinto degrees, correct to two decimal places. c Find the exact value of: isin c ii cos 7 4 c Solution a 135 ◦= 135 ×  180 c = 3 4 c Note that angles in radians which are expressed in terms of are left in that form. b 1.5c= 1.5× 180  ◦ = 85 .94 ◦, correct to two decimal places. c isin c= 0 ii cos 7 4 c = cos 7 4 − 2 c = cos − 4 c = cos c 4 = √2 2 Example 2 Find the exact value of: a sin 150 ◦ b cos( −585 ◦) Solution a sin 150 ◦= sin(180 ◦− 150 ◦) = sin 30 ◦ = 0.5 b cos( −585 ◦)= cos 585 ◦ = cos(585 ◦− 360 ◦) = cos(225 ◦) =− cos 45 ◦ =− √2 2 Example 3 Find the exact value of: a sin 11  6  b cos −45  6  Solution a sin 11  6  = sin  2− 11  6  =− sin  6 =− 1 2 b cos −45  6  = cos −712×  = cos  − 2  = 0

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 6 Essential Specialist Mathematics The Pythagorean identity For any value of  cos 2+ sin 2= 1 Example 4 If sin x◦= 0.3, 0 < x< 90, find: a cos x◦ b tan x◦ Solution a sin 2x◦+ cos 2x◦= 1 0.09 + cos 2x◦= 1 ∴ cos 2x◦= 0.91 cos x◦=± √0.91 as 0 < x< 90 ,cos x◦= √0.91 = \b 91 100 = √91 10 b tan x◦= sin x cos x = 0.3 √0.91 = 3√91 = 3√91 91 Solution of equations If a trigonometric equation has a solution, then it will have a corresponding solution in each ‘cycle’ of its domain. Such equations are solved by using the symmetry of the graphs to obtain solutions within one ‘cycle’ of the function. Other solutions may be obtained by adding multiples of the period to these solutions. y x a b cd 1 _1 π 2π 0 Example 5 The graph of y= f(x) where f:R→ R, f(x)= sin x,x∈[0, 2 ] is shown. Find the other xvalue which has the same yvalue as each of the pronumerals marked. Solution For x= a, the value is − a. For x= b, the other value is − b. For x= c, the other value is 2 − (c− )= 3− c. For x= d, the other value is + (2− d)= 3− d. Example 6 Solve the equation sin  2x+  3  = 1 2 for x∈[0, 2 ].

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 7 Solution Let = 2x+  3 Note 0\b x\b 2 ⇔ 0\b 2x\b 4 ⇔  3 \b 2x+  3 \b 13  3 ⇔  3 \b \b 13  3 Therefore solving the equation sin  2x+  3  = 1 2for x∈[0, 2 ] is achieved by first solving the equation sin( )= 1 2for  3 \b \b 13  3 Consider sin = 1 2 ∴ =  6 or 5 6 or 2 +  6 or 2 + 5 6 or 4 +  6 or 4 + 5 6 or ... The solutions  6 and 29  6 are not required as they lie outside the restricted domain for . ∴For  3 \b \b 13  3 = 5 6 or 13  6 or 17  6 or 25  6 ∴ 2x+ 2 6 = 5 6 or 13  6 or 17  6 or 25  6 ∴ 2x= 3 6 or 11  6 or 15  6 or 23  6 ∴ x=  4 or 11  12 or 5 4 or 23  12 Using a graphics calculator A graphics calculator can be used to find a numerical solution to the equation sin  2x+  3  = 12by plotting the graphs of y= sin  2x+  3  and y= 12and considering the points of intersection. It can be seen from the graph for [0, 2 ] that there are four solutions. The xcoordinates of the points of intersection are found through using 5:intersect from the CALC menu. It is sometimes possible to find the exact value by calculating X /. The corresponding exact value is 11  12 .

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 8 Essential Specialist Mathematics Using a CAS calculator A CAS calculator can be used to solve this equation. The syntax is solve(sin(2 x+ /3) = 1/2, x)|0< = xand x< = 2. The result is shown. Sketch graphs The graphs of functions defined by rules of the form f(x)= asin( nx + ε)+ band f(x)= acos( nx + ε)+ bcan be obtained from the graphs of sin xand cos xby transformations. Example 7 Sketch the graph of h:[0,2]→ R,h(x)= 3 cos  2x+  3  + 1. Solution h(x)= 3 cos  2  x+  6  + 1 The transformations from the graph of y= cos xare a dilation from the yaxis of factor 12 a dilation from the x-axis of factor 3 a translation of  6 in the negative direction of the xaxis a translation of 1 in the positive direction of the yaxis The graph with the dilations applied is as shown below. y x 4 0 –3 y = 3 cos (2 x) π 2π π 4 π 2 3π 4 5π 4 3π 4 7π 4 The translation  6 in the negative direction of the xaxis is then applied.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 9 3 6 32 3 2 _π 12 π 3 2π 2π, 3 4π 6 5π 6 π 12 7π 12 13π 12 19π 12 25π 6 11π _3 x y y = 3 cos 2 x + 0 The final translation is applied and the graph is given for the required domain. 2 2π, 3 4π 6 5π 3 π 6 11π 5 2 5 3 0 _2 x y The x-axis intercepts are found by solving the equation. 3 cos  2x+  3  + 1= 0 i.e. cos  2x+  3  =− 1 3 The graph of tan A sketch of the graph of f:R\{(2n+ 1) 2;n∈Z}→ R,f()= tan is shown below. y 5π 2π 3π π π _π 3π 2 2 _π 2 2 0 θ Note: =−  2, 2,3 2 and 5 2 are asymptotes. Observations from the graph The graph repeats itself every units, i.e. the period of tan is . Range of tan is R. The vertical asymptotes have equation = (2k+ 1) 2 where k∈Z.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 10 Essential Specialist Mathematics Using a graphics calculator If the normal procedure is followed and y= tan xis entered into the Y = menu and then plotted by pressing ZOOM 7, we get the graph as shown here. While it might appear that the calculator has drawn in asymptotes, it hasn’t. Instead, it has simply joined up the last two points plotted either side of the point where the asymptote lies and joined them up. Note that, since tan x= sin x cos x, the domain of tan must exclude all real numbers where cos xis zero, i.e. all odd integral multiples of  2. The graph shows vertical asymptotes at x= (2n+ 1) 2 . This function has a period of as tan x= tan( x+ n),n∈Z. The concept of amplitude is not applicable here. Symmetry properties for tan From the definition of tan, the following results are obtained: tan( − )=− tan  tan( + )= tan  tan(2 − )=− tan  tan( −)=− tan  Example 8 Find the exact values of: a tan (330 ◦) b tan 4 3  Solution a tan (330 ◦)= tan(360 − 30) ◦ =− tan (30) ◦ =− √3 3 b tan 4 3  = tan  +  3  = tan  3  = √3 Solutions of equations involving tan The procedure here is similar to that used for solving equations involving sin and cos, except that only one solution needs to be selected then all other solutions are or 180 ◦apart. Example 9 Solve the equations: a tan x=− 1 for x∈[0, 4 ] b tan(2 x− )= √3 for x∈[−,] Solution a tan x=− 1 Now tan 3 4 =− 1.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 11 Therefore x= 3 4 or 3 4 +  or 3 4 + 2 or 3 4 + 3. Hence x= 3 4 or 7 4 or 11  4 or 15  4 . b tan(2 x− )= √3. Therefore −2 \b 2x\b 2and thus −3 \b 2x− \b and −3 \b \b . In order to solve tan(2 x− )= √3 first solve tan = √3. =  3 or  3−  or  3− 2 or  3− 3 ∴ =  3 or −2 3 or −5 3 or −8 3 and as = 2x−  2x−  =  3 or −2 3 or −5 3 or −8 3 Therefore 2 x= 4 3 or  3 or −2 3 or −5 3 And x= 2 3 or  6 or − 3 or −5 6 . Exercise 1A 1 a Change the following angles from degrees to exact values in radians: i 720 ◦ ii 540 ◦ iii −450 ◦ iv 15 ◦ v −10 ◦ vi −315 ◦ b Change the following angles from radians to degrees: i 5 4 c ii −2 3 c iii 7 12 c iv −11  6 c v 13  9 c vi −11  12 c 2 Perform the correct conversion on each of the following, giving the answer correct to two decimal places. a Convert from degrees to radians: i 7◦ ii −100 ◦ iii −25 ◦ iv 51 ◦ v 206 ◦ vi −410 ◦ b Convert from radians to degrees: i 1.7 c ii −0.87 c iii 2.8 c iv 0.1 c v −3c vi −8.9 c 3 Find the exact value of each of the following: a sin 2 3  b cos 3 4  c cos  − 3  d cos 5 4  e cos 9 4  f sin 11  3  g cos 31  6  h cos 29  6  i sin  −23  6  4 Find the exact value of each of the following: a sin(135 ◦) b cos( −300 ◦) c sin(480 ◦) d cos(240 ◦) e sin( −225 ◦) f sin(420 ◦)

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 12 Essential Specialist Mathematics 5 If sin( x◦)= 0.5 and 90 < x< 180, find: a cos( x◦) b tan( x◦) 6 If cos( x◦)=− 0.7 and 180 < x< 270, find: a sin( x◦) b tan( x◦) 7 If sin( x)=− 0.5 and  < x< 3 2 , find: a cos( x) b tan( x) 8 If sin( x)=− 0.3 and 3 2 < x< 2, find: a cos( x) b tan( x) 9 Solve each of the following for x∈[0, 2 ]: a sin x=− √3 2 b sin (2 x)= √3 2 c 2 cos (2 x)=− 1 d sin  x+  3  =− 1 2 e 2 cos  2  x+  3  =− 1 f 2 sin  2x+  3  =− √3 10 Find the exact values of each of the following: a tan 5 4  b tan  −2 3  c tan  −29  6  d tan(240 ◦) 11 If tan x= 1 4and \b x\b 3 2 , find the exact value of: a sin x b cos x c tan( −x) d tan( − x) 12 If tan x=− √3 2 and  2 \b x\b , find the exact value of: a sin x b cos x c tan( −x) d tan( x− ) 13 Solve each of the following for x∈[0, 2 ]: a tan x=− √3 b tan  3x−  6  = √3 3 c 2 tan x 2  + 2= 0 d 3 tan  2+ 2x  =− 3 14 Sketch the graphs of each of the following for the stated domain: a f(x)= sin 2 x,x∈[0, 2 ] b f(x)= cos  x+  3  ,x∈ − 3 , c f(x)= cos  2  x+  3  ,x∈[0,] d f(x)= 2 sin(3 x)+ 1, x∈[0, ] e f(x)= 2 sin  x−  4  + √3,x∈[0,2] 15 Sketch the graphs of each of the following for x∈[0, ], clearly labelling all intercepts with the axes and all asymptotes: a f(x)= tan(2 x) b f(x)= tan  x−  3  c f(x)= 2 tan  2x+  3  d f(x)= 2 tan  2x+  3  − 2

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 13 1.2 Solving right-angled triangles Pythagoras’ theorem This well-known theorem is applicable to right-angled triangles and will be stated here without proof: (hyp) 2= (opp) 2+ (adj) 2 A B C 5 6 x° Example 10 In triangle ABC ,∠ABC = 90 ◦and ∠CAB = x◦, AB = 6cmand BC = 5cm. Find: a AC b the trigonometric ratios related to x◦ c x Solution a By Pythagoras’ theorem, AC 2= 52+ 62= 61 ∴ AC = √61 cm b sin x◦= 5√61 cos x◦= 6√61 tan x◦= 5 6 c tan x◦= 5 6 ∴ x= 39 .81 (correct to two decimal places) Exercise 1B 1 Find the trigonometric ratios tan x◦, cos x◦and sin x◦for each of the following triangles: a b c x° 5 8 x° 5 7 9 7 x° 2 Find the exact value of ain each of the following triangles: a b c 12 a 30 ° 45 ° 6 a a 5 30 ° 30 °

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 14 Essential Specialist Mathematics 3 Find the exact value of the pronumerals for each of the following: a 5 1 a b 1 1 1 2 a b c c 3 3 h a 1 d 1 c a b d 30° 45° 4 a Find the values of a,y,z,wand x. 30° 45° x w y a z° 1 b Hence deduce exact values for sin(15 ◦), cos(15 ◦) and tan(15 ◦). c Find the exact values for sin(75 ◦), cos(75 ◦) and tan(75 ◦). 1.3 The sine and cosine rules The sine rule In section 1.2, methods for finding unknown lengths and angles for right-angled triangles were discussed. This section discusses methods for finding unknown quantities in non-right-angled triangles. The sine rule is used to find unknown quantities in a triangle when one of the following situations arises: one side and two angles are given two sides and a non-included angle are given In the first case a unique triangle is defined, but for the second it is possible for two triangles to exist. Labelling convention The following convention is followed in the remainder of this chapter. Interior angles are denoted by upper-case letters, and the length of the side opposite an angle is denoted by the corresponding lower-case letter. A a b c B C The magnitude of angle BAC is denoted by A. The length of side BC is denoted by a.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 15 The sine rule states that for triangle ABC A a b c B C a sin A = b sin B = c sin C A proof will only be given for the acute-angled triangle case. The proof for obtuse-angled triangles is similar. Proof In triangle AC D , sin A= h b ∴ h= bsin A In triangle BCD , sin B= h a ∴ h= asin B ∴ asin B = bsin A i.e. a sin A = b sin B Similarly, starting with a perpendicular from Ato BC would give b sin B = c sin C a b C D A B h A B C c 10 cm 70 ° 31 ° Example 11 Use the sine rule to find the length of AB . Solution c sin 31 ◦= 10 sin 70 ◦ ∴ c= 10 × sin 31 ◦ sin 70 ◦ ∴ c= 5.4809 ... The length of AB is 5.48 cm correct to two decimal places. Example 12 Z X Y 25 ° 6 cm 5 cm Use the sine rule to find the magnitude of angle XZY ,given that Y= 25 ◦,y= 5 and z= 6.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 16 Essential Specialist Mathematics 25 ° 30 .47 ° 149 .53 ° 6 cm 5 cm 5 cm X Y Z1 Z2 Solution 5 sin 25 ◦= 6 sin Z ∴ sin Z 6 = sin 25 ◦ 5 ∴ sin Z= 6× sin 25 ◦ 5 = 0.5071 ... ∴ Z= sin −1(0.5071 ... ) ∴ Z= 30 .4736 ... or 180 − 30 .4736 ... ∴ Z= 30 .47 ◦or Z= 149 .53 ◦correct to two decimal places. Remember: sin(180 − )◦= sin ◦ There are two solutions for the equation sin Z= 0.507 1... Note: When using the sine rule in the situation where two sides and a non-included angle are given, the possibility of two such triangles existing must be considered. Existence can be checked through the sum of the given angle and the calculated angle not exceeding 180 ◦. The cosine rule The cosine rule is used to find unknown quantities in a triangle when one of the following situations arises: two sides and an included angle are given three sides are given The cosine rule states that for triangle ABC A a b c B C a2= b2+ c2− 2bc cos Aor, equivalently, cos A= b2+ c2− a2 2bc The symmetrical results also hold, i.e. b2= a2+ c2− 2ac cos B c2= a2+ b2− 2ab cos C The result will be proved for acute-angled triangles. The proof for obtuse-angled triangles is similar. Proof A B C D a b x h c In triangle AC D b2= x2+ h2(Pythagoras’ theorem) cos A= x b and therefore x= bcos A In triangle BCD a2= (c− x)2+ h2(Pythagoras’ theorem)

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 17 Expanding gives a2= c2− 2cx + x2+ h2 = c2− 2cx + b2 (as x2+ h2= b2) ∴ a2= b2+ c2− 2bc cos A (as x= bcos A) Example 13 For triangle ABC , find the length of AB in centimetres correct to two decimal places. Solution c2= 52+ 10 2− 2× 5× 10 cos 67 ◦ = 85 .9268 ... ∴ c≈ 9.269 The length of AB is 9.27 cm correct to two decimal places. 10 cm 5 cm A C B c 67 ° Example 14 Find the magnitude of angle ABC for triangle ABC correct to two decimal places. B C A 15 cm 12 cm 6 cm Solution cos B = a2+ c2− b2 2ac = 12 2+ 62− 15 2 2× 12 × 6 =− 0.3125 ∴ B = (108 .2099 ... )◦ i.e. B ≈ 108 .21 ◦correct to two decimal places. The magnitude of angle ABC is 108 ◦12 36 (to the nearest second). Example 15 In ABC ,∠CAB = 82 ◦,AC = 12 cm, AB = 15 cm. Find, correct to two decimal places: a BC b ∠AC B 15 cm 12 cm a cm 82 ° A B C

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 18 Essential Specialist Mathematics Solution a BC is found by applying the cosine rule: a2= b2+ c2− 2bc cos A = 12 2+ 15 2− 2× 12 × 15 cos 82 ◦ = 144 + 225 − 360 × cos 82 ◦ = 318 .897 ... BC = a= 17.86 cm, correct to two decimal places. b ∠AC B is found by applying the sine rule pair: a sin A = c sin C ∴ sin C = csin A a = 15 × sin 82 ◦ 17 .86 ∴ ∠AC B = 56.28 ◦, correct to two decimal places. Note: 123.72 ◦is also a solution to this equation but it is discarded as a possible answer as it is inconsistent with the information already given. Exercise 1C 1 In triangle ABC ,∠BAC = 73 ◦,∠AC B = 55 ◦and AB = 10 cm. Find, correct to two decimal places: a BC b AC 2 In triangle ABC ,∠ABC = 58 ◦,AB = 6.5 cm and BC = 8 cm. Find, correct to two decimal places: a AC b ∠BCA 3 The adjacent sides of a parallelogram are 9 cm and 11 cm. One of its angles is 67 ◦. Find the length of the longer diagonal, correct to two decimal places. 4 In ABC ,∠AC B = 34 ◦,AC = 8.5 cm and AB = 5.6 cm. Find, correct to two decimal places: a the two possible values of ∠ABC (one acute and one obtuse) b BC in each case 5 In ABC ,∠ABC = 35 ◦,AB = 10 cm and BC = 4.7 cm. Find, correct to two decimal places: a AC b ∠AC B 6 In ABC ,∠ABC = 45 ◦,∠AC B = 60 ◦and AC = 12 cm. Find AB .

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 19 7 In PQR ,∠QPR = 60 ◦,PQ = 2cmand PR = 3 cm. Find QR . 8 In ABC ,∠ABC has magnitude 40 ◦,AC = 20 cm and AB = 18 cm. Find the distance BC correct to 2 decimal places. 9 In ABC ,∠AC B has magnitude 30 ◦,AC = 10 cm and AB = 8 cm. Find the distance BC using the cosine rule. 10 In ABC ,AB = 5 cm, BC = 12 cm and AC = 10 cm. Find: a the magnitude of ∠ABC , correct to two decimal places b the magnitude of ∠BAC , correct to two decimal places 1.4 Geometry prerequisites In the Specialist Mathematics study design it is stated that students should be familiar with several geometric results and be able to apply them in examples. These results have been proved in earlier years’ study. In this section they are listed. The sum of the interior angles of a triangle is 180 ◦. a+ b+ c= 180 b° a° c° The sum of the exterior angles of a convex polygon is 360 ◦. a° e° d° c° b° a+ b+ c+ d+ e= 360 x+ y+ z+ w = 360 a+ b+ c+ d+ e+ f= 360 x° y° w° z° b° a° c° d° e° f° a° c° b° d° Corresponding angles of lines cut by a transversal are equal if, and only if, the lines are parallel. a= band c= d aand bare corresponding angles cand dare corresponding angles

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 20 Essential Specialist Mathematics Opposite angles of a parallelogram are equal, and opposite sides are equal in length. a° c° b° d° D A B C AB = DC A D = BC c= dand a= b a° b° X B C A The base angles of an isosceles triangle are equal. (AB = AC ) a= b The line joining the vertex to the midpoint of the base of an isosceles triangle is perpendicular to the base. The perpendicular bisector of the base of an isosceles triangle passes through the opposite vertex. The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference. x= 2y A B y° x° The angle in a semicircle is a right angle. Angles in the same segment of a circle are equal. x= y= z x° y°z° A B The sum of the opposite angles of a cyclic quadrilateral is 180 ◦. x° y° a° w° z° x + y = 180 z + w = 180 a = x An exterior angle of a cyclic quadrilateral and the interior opposite angle are equal.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 21 A trangent to a circle is perpendicular to the radius at the point of contact. The two tangents to a circle from an exterior point are equal in length. XA = XB A B O X An angle between a tangent to a circle and a chord through the point of contact is equal to the angle in the alternate segment. x= y x° y° Example 16 Find the magnitude of each of the following angles: a ∠ABC b ∠ADC c ∠CBD d ∠OCD e ∠BAD 93° 85° 60° O A D B C Solution a ∠ABC = 93 ◦ (vertically opposite) b ∠ADC = 87 ◦ (opposite angle of a cyclic quadrilateral) c ∠COB = 85 ◦ (vertically opposite) ∠CBD = [180 − (60 + 85)] ◦= 35 ◦ (angles of a triangle, CBO ) d ∠CAD = 35 ◦ (angle subtended by the arc CD ) ∠ADC = 87 ◦ (From b) ∠OCD = [180 − (87 + 35)] ◦= 58 ◦ (angles of a triangle,  CAD ) e ∠BAD = [180 − (60 + 58)] ◦= 62 ◦ (opposite angles of a cyclic quadrilateral) Exercise 1D 1 Find the value of a,y,zand x. 68 ° 150 ° x° z° a° y°

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 22 Essential Specialist Mathematics 2 Find the magnitude of each of the following: a ∠RTW b ∠TSW c ∠TRS d ∠RW T 105 ° 62 ° 37 ° R S T W 3 Find the value of a,band c.AB is a tangent to the circle at C. 50 ° 40 ° a° b°c° AB C 4 ABCD is a square and ABX is an equilateral triangle. Find the magnitude of: a ∠DXC b ∠XDC X C B A D 5 Find the values of a,b,c,dand e. 69°47° 105 °c° a° b° d°e° X Y Z W 6 Find xin terms of a,band c. a° b° c° x° 7 Find the values of xand y, given that O is the center of the circle. x° y° O 40° 8 Find the values of a,b,cand d. 70° 50° a° b° c° d° 9 Find the values of xand y. 40 ° y° x° A B X 10 O is the centre of the circle. Find the values of xand y. A B C O 20° 50° x° y°

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 23 1.5 Sequences and series The following are examples of sequences of numbers: a 1,3,5,7,9... b 0.1, 0.11, 0.111, 0.111 1... c 13,19,127,181 ... d 10, 7, 4, 1, −2... e 0.6, 1.7, 2.8, 3. 9... Note that each sequence is a set of numbers, with order being important. For some sequences of numbers a rule can be found connecting any number to the preceding number. For example: for sequence A, a rule is: add 2 for sequence C, a rule is: multiply by 13 for sequence D, a rule is: subtract 3 for sequence E, a rule is: add 1.1 The numbers of a sequence are called terms . The nth term of a sequence is denoted by the symbol tn. So the first term is t1, the 12th term is t12 and so on. A sequence can be defined by specifying a rule which enables each subsequent term to be found using the previous term. In this case, the rule specified is called an iterative rule or a difference equation . For example: sequence A can be defined by t1= 1, tn= tn−1+ 2 sequence C can be defined by t1= 13,tn= 13tn−1 Example 17 Use the difference equation to find the first four terms of the sequence t1= 3, tn= tn−1+ 5 Solution t1= 3 t2= t1+ 5= 8 t3= t2+ 5= 13 t4= t3+ 5= 18 The first four terms are 3, 8, 13, 18. Example 18 Find the difference equation for the following sequence. 9,−3,1,−13... Solution −3=− 13× 9i .e. t2=− 13t1 1=− 13×− 3i .e. t3=− 13t2 ∴ tn=− 13tn−1,t1= 9

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 24 Essential Specialist Mathematics Alternatively, a sequence can be defined by a rule that is stated in terms of n. For example: tn= 2n defines the sequence t1= 2, t2= 4, t3= 6, t4= 8... tn= 2n−1 defines the sequence t1= 1, t2= 2, t3= 4, t4= 8... Example 19 Find the first four terms of the sequence defined by the rule tn= 2n+ 3. Solution t1= 2× 1+ 3= 5 t2= 2× 2+ 3= 7 t3= 2× 3+ 3= 9 t4= 2× 4+ 3= 11 The first four terms are 5, 7, 9, 11. Arithmetic sequences A sequence in which each successive term is found by adding a constant value to the previous term is called an arithmetic sequence . For example, 2, 5, 8, 1 1...isan arithmetic sequence. An arithmetic sequence can be defined by a difference equation of the form tn= tn−1+ d where dis a constant . If the first term of an artithmetic sequence t1= athen the nth term of the sequence can also be described by the rule: tn= a+ (n− 1) d where a= t1and d= tn− tn−1 dis the common difference . Example 20 Find the tenth term of the arithmetic sequence −4, −1,2,5... Solution a=− 4,d= 3,n= 10 tn= a+ (n− 1) d t10 =− 4+ (10 − 1)3 = 23 Arithmetic series The sum of the terms in a sequence is called a series . If the sequence in question is arithmetic, the series is called an arithmetic series . The symbol Snis used to denote the sum of nterms of a sequence, i.e. Sn= a+ (a+ d)+ (a+ 2d)+ ··· + (a+ (n− 1) d) If this sum is written in reverse order, then Sn= (a+ (n− 1) d)+ (a+ (n− 2) d)+ ··· + (a+ d)+ a

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 25 Adding these two expression together gives 2 Sn= n[2a+ (n− 1) d] ∴ Sn= n 2[2a+ (n− 1) d] and since the last term l= tn= a+ (n− 1) d Sn= n 2(a+ l) Geometric sequences A sequence in which each successive term is found by multiplying the previous term by a fixed value is called a geometric sequence . For example, 2, 6, 18, 5 4...isa geometric sequence. A geometric sequence can be defined by an iterative equation of the form tn= rtn−1, where ris constant. If the first term of a geometric sequence t1= a, then the nth term of the seqeunce can also be described by the rule tn= ar n−1 where r= tn tn−1 ris called the common ratio . Example 21 Calculate the tenth term of the sequence 2, 6, 18 ,... Solution a= 2,r= 3,n= 10 tn= ar n−1 t10 = 2× 3(10−1)= 39 366 Geometric series The sum of the terms in a geometric sequence is called a geometric series . An expression for Sn, the sum of nterms, of a geometric sequence can be found using a similar method to that used in the development of a formula for an arithmetic series. Let Sn= a+ ar + ar 2+ ··· + ar n−1 ... 1 Then rS n= ar + ar 2+ ar 3+ ··· + ar n ... 2 Subtract 1 from 2 rS n− Sn= ar n− a ∴ Sn(r− 1) = a(rn− 1) and Sn= a(rn− 1) r− 1

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 26 Essential Specialist Mathematics For values of rsuch that −1< r< 1, it is often more convenient to use the alternative formula Sn= a(1 − rn) 1− r which is obtained by subtracting 2 from 1 above. Example 22 Find the sum of the first nine terms of the geometric sequence 13,19,127,181 ... Solution a= 13,r= 13,n= 9 ∴ S9= 13 13 9− 1  13− 1 = −12 13 9− 1  ≈ 12(0.999 949) ≈ 0.499 975 (to 6 decimal places) Infinite geometric series If the common ratio of a geometric sequence has a magnitude less than 1, i.e. −1< r< 1, then each successive term of the sequence is closer to zero. e.g., 4 ,2,1,12,14... When the terms of the sequence are added, the corresponding series a+ ar + ar 2+ ··· + arn−1will approach a limiting value, i.e. as n→∞ ,Sn→ a limiting value. Such a series is called convergent . In example 22 above, it was found that for the sequence 13,19,127,181 ... the sum of the first nine terms, S9, was 0.499975. For the same sequence, S20 = 0.499 999 999 9 ≈ 0.5 So even for a relatively small value of n(20), the sum approaches the limiting value of 0.5 very quickly. Given that Sn= a(1 − rn) 1− r ⇒ Sn= a 1− r− ar n 1− r as n→∞ ,rn→ 0 and hence ar n 1− r→ 0 It follows then that the limit as n→∞ of Snis a 1− r So S∞ = a 1− r This is also referred to as ‘the sum to infinity’ of the series.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 27 Example 23 Find the sum to infinity of the series 1 + 12+ 14+ 18+ ... Solution r= 12,a= 1 ∴ S∞ = 1 1− 12 = 2 Using a graphics calculator Example 24 Graph the terms of the geometric sequence defined by: an= 512(0 .5)n−1for n= 1,2... Solution Step 1 Set your calculator to sequence mode. a Enter the MODE menu; move cursor down to line 4 and across to cover Seq . b Press ENTER to select sequence mode. Step 2 Enter expression defining sequence. Press Y= and enter: a set nMin = 1 b u(n)= 512(0.5) ˆ ( n− 1) (Note: nis obtained by pressing XT N ) c u(nMin) = 512 Step 3 Set viewing window. Press WINDOW and enter the values as shown. Step 4 Plot. Press GRAPH . Note that terms of the sequence can be viewed in the TABLE window.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 28 Essential Specialist Mathematics Example 25 a Generate the first 10 terms in the arithmetic sequence whose nth term is 8+ 4(n− 1) and store them in list L1 . b Find the sum of these first 10 terms. c This arithmetic sequence can also be generated by the recursive form, tn= tn−1+ 4, with t1= 8. Generate the first 10 terms in this way Solution a Select Seq mode from the MODE menu. In the Y= window enter nMin = 1, u(n) = 8+ 4(n−1), and u(nMin) = 8 Press STAT and select 1:Edit from the EDIT menu. Take the cursor to the top of L2(actually over L2) and press ENTER . In the entry line enter L2= “8 + 4(L 1− 1)” and press ENTER . The result is as shown. b Take the cursor to L3and enter L3= “cumSum(L 2)” and press ENTER . The result is as shown. cumSum is obtained by selecting 6: cumSum from the OPS submenu of LIST . c In the Y= window enter, nMin = 1, u(n) = u(n− 1) + 4and u(nMin) = 8. Note that uis obtained by pressing 2ND 7 and nis obtained by pressing XT N. The sequence can be seen in the TABLE screen. Using a CAS calculator The method is very similar to that discussed above. Press MODE and use the right arrow from Graph to reveal the Graph menu. Select 4:SEQUENCE and press ENTER .

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 29 In the Y= screen enter u1(n) = u1(n −1) + 4 and ui1 = 8. The sequence can be seen in the TABLE window. Exercise 1E 1 A difference equation has rule tn+1= 3tn− 1, t1= 6. Find t2and t3. Use a graphics calculator to find t8. 2 A difference equation has rule yn+1= 2yn+ 6, y1= 5. Find y2and y3. Use a graphics calculator to find y10 and to plot a graph showing the first ten values. 3 The Fibonacci sequence is given by the difference equation tn+2= tn+1+ tnwhere t1= t2= 1. Find the first ten terms of the Fibonacci sequence. 4 Find the sum of the first ten terms of an arithmetic sequence with first term 3 and common difference 4. 5 Find the sum to infinity of 1 − 13+ 19− 127 + ... 6 The first, second and third terms of a geometric sequence are x+ 5, xand x− 4 respectively. Find: a x b the common ratio c the difference between the sum to infinity and the sum of the first ten terms 7 Find the sum of the first eight terms of a geometric sequence with first term 6 and common ratio −3. 8 Find the sum to infinity of the geometric sequence a, a√2,a 2, a 2√2... in terms of a. 9 Consider the sum Sn= 1+ x 2+ x2 4 + ··· + xn−1 2n−1 a Calculate S10 when x= 1.5. b i Find the possible values of xfor which the infinite sum exists. Denote this sum by S. ii Find the values of xfor which S= 2S10. 10 a Find an expression for the infinite geometric sum 1 + sin + sin 2+ ... b Find the values of for which the infinite geometric sum is 2. 1.6 Circles For a circle with centre the origin and radius r, if the point with coordinates ( x,y)isonthe circle then x2+ y2= r2.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 30 Essential Specialist Mathematics The converse is also true, i.e. a point with coordinates ( x,y) such that x2+ y2= r2lies on the circle with centre the origin and radius r. Applying Pythagoras’ theorem to triangle OA P yields r2= OP 2= x2+ y2 x y O A r P(x,y) In general, the following result holds: The circle with centre ( h,k) and radius ris the graph of the equation (x− h)2+ (y− k)2= r2 This graph is obtained from the graph of x2+ y2= r2by the translation defined by (x,y)→ (x+ h,y+ k) Example 26 Sketch the graph of the circle with centre at ( −2, 5) and radius 2, and state the cartesian equation for this circle. 7 5 3 _4 _2 0 x y Solution The equation is (x+ 2)2+ (y− 5)2= 4 which may also be written as x2+ y2+ 4x− 10 y+ 25 = 0 Note: The equation x2+ y2+ 4x− 10 y+ 25 = 0 can be ‘unsimplified’ by completing the square. x2+ y2+ 4x− 10 y+ 25 = 0 implies x2+ 4x+ 4+ y2− 10 y+ 25 + 25 = 29 i.e. (x+ 2)2+ (y− 5)2= 4 This suggests a general form of the equation of a circle. x2+ y2+ Dx + Ey + F = 0 Completing the square gives x2+ Dx + D2 4 + y2+ Ey + E2 4 + F = D2+ E2 4 i.e.  x+ D 2 2 +  y+ E 2 2 = D2+ E2− 4F 4

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 31 IfD2+ E2− 4F> 0, then the equation represents a circle with centre −D 2 ,−E 2  and radius \b D2+ E2− 4F 4 . IfD2+ E2− 4F= 0, then the equation represents one point −D 2 ,−E 2  . IfD2+ E2− 4F< 0, then the equation has no graphical representation in the cartesian plane. Example 27 Sketch the graph of x2+ y2+ 4x+ 6y−12 = 0. State the coordinates of the centre and the radius. Solution x2+ y2+ 4x+ 6y− 12 = 0 ∴ x2+ 4x+ 4+ y2+ 6y+ 9− 12 = 13 i.e. (x+ 2)2+ (y+ 3)2= 25 The circle has centre ( −2, −3) and radius 5. √21 −3 + √21 −3 − −6 2 0 (−2, −3) x y Example 28 Sketch a graph of the region of the plane such that x2+ y2< 9 and x≥ 1. Solution x = 1 y _3 _3 3 3 x required region 0 Exercise 1F 1 Find the equations of the circles with the following centres and radii: a centre (2, 3); radius 1 b centre ( −3, 4); radius 5 c centre (0, −5); radius 5 d centre (3, 0); radius √2 2 Find the radii and the coordinates of the centres of the circles with the following equations: a x2+ y2+ 4x− 6y+ 12 = 0 b x2+ y2− 2x− 4y+ 1= 0 c x2+ y2− 3x= 0 d x2+ y2+ 4x− 10 y+ 25 = 0

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 32 Essential Specialist Mathematics 3 Sketch the graphs of each of the following: a 2x2+ 2y2+ x+ y= 0 b x2+ y2+ 3x− 4y= 6 c x2+ y2+ 8x− 10 y+ 16 = 0 d x2+ y2− 8x− 10 y+ 16 = 0 e 2x2+ 2y2− 8x+ 5y+ 10 = 0 f 3x2+ 3y2+ 6x− 9y= 100 4 Sketch the graphs of the regions of the plane specified by the following: a x2+ y2\b 16 b x2+ y2≥ 9 c (x− 2)2+ (y− 2)2< 4 d (x− 3)2+ (y+ 2)2> 16 e x2+ y2\b 16 and x\b 2 f x2+ y2\b 9 and y≥− 1 5 The points (8, 4) and (2, 2) are the ends of a diameter of a circle. Find the coordinates of the centre and the radius of the circle. 6 Find the equation of the circle, centre (2, −3), which touches the xaxis. 7 Find the equation of the circle which passes through (3, 1), (8, 2) and (2, 6). 8 Find the radii and coordinates of the centre of the circles with equations 4x2+ 4y2− 60 x− 76 y+ 536 = 0 and x2+ y2− 10 x− 14 y+ 49 = 0, and find the coordinates of the points of intersection of the two curves. 9 Find the coordinates of the points of intersection of the circle with equation x2+ y2= 25 and the line with equation: a y= x b y= 2x 1.7 Ellipses and hyperbolas Ellipses The curve with equation x2 a2+ y2 b2= 1 is an ellipse with centre the origin, x-axis intercepts (−a, 0) and ( a,0)and y-axis intercepts (0, −b) and (0, b). Ifa> bthe ellipse will appear as shown in the diagram on the left. Ifb> athe ellipse is as shown in the diagram on the right. Ifb= a, the equation is that of a circle with centre the origin and radius a. A' A B' B _b _a b a 0 y x A' A –b B′ b –aa B 0 y x x2 a2+ y2 b2= 1;a> b AA is the major axis BB is the minor axis x2 a2+ y2 b2= 1;b> a AA is the minor axis BB is the major axis

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 33 The general cartesian form is as given below. The curve with equation (x− h)2 a2 + (y− k)2 b2 = 1 is an ellipse with centre ( h,k). It is obtained by a translation of the ellipse x2 a2+ y2 b2= 1. The translation is ( x,y)→ (x+ h,y+ k). (h, k + b ) (h, k _ b) (h + a , k) (h _ a, k)( h, k) x y 0 The ellipse with equation x2 a2+ y2 b2= 1 can be obtained by applying the following dilations to the circle with equation x2+ y2= 1: a dilation of factor afrom the yaxis, i.e. ( x,y)→ (ax,y) a dilation of factor bfrom the xaxis, i.e. ( x,y)→ (x,by) The result is the transformation ( x,y)→ (ax,by). x x x yy y 1 –1 11 –1 –1 00 0 (x, y) ( ax, y) (x, y) ( x, by ) –a –a –b b a a Example 29 Sketch the graph of each of the following. Give the axes intercepts and the coordinates of the centre. a x2 9 + y2 4 = 1 b x2 4 + y2 9 = 1 c (x− 2)2 9 + (y− 3)2 16 = 1 d 3x2+ 24 x+ y2+ 36 = 0

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 34 Essential Specialist Mathematics Solution a 3 2 –3 –2 0 x y b 3 2 _2 –3 0 x y Centre (0, 0) Axes intercepts ( ±3, 0) and (0, ±2) Centre (0, 0) Axes intercepts ( ±2, 0) and (0, ±3) c Centre is at (2, 3) When x= 0 4 9+ (y− 3)2 16 = 1 ∴ (y− 3)2 16 = 5 9 ∴ (y− 3)2= 16 × 5 9 ∴ y= 3± 4√5 3 When y= 0 (x− 2)2 9 + 9 16 = 1 (x− 2)2 9 = 7 16 (x− 2)2= 9× 7 16 x= 2± 3√7 4 √ (2, 7) (5, 3) (2, –1) (2, 3) (–1, 3) 534 3 + √534 3 _ √743 2 _ √743 2 + 0 x y d 3x2+ 24 x+ y2+ 36 = 0 Completing the square yields 3[x2+ 8x+ 16] + y2+ 36 − 48 = 0 i.e. 3(x+ 4)2+ y2= 12 (x+ 4)2 4 + y2 12 = 1 ∴Centre ( −4, 0) Axes intercepts ( −6, 0) and ( −2, 0) √3) (_6, 0) ( _4, 0) ( _2, 0) 0 x y (_4, 2 √3) (_4, _2

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 35 Defining an ellipse In the previous section a circle was defined as a set of points which are all a constant distance from a given point (the centre). An ellipse can be defined in a similar way. Consider the set of all points Psuch that PF 1+ PF 2is equal to a constant kwith k> 2m, and the coordinates of F1and F2are ( m, 0) and ( −m, 0) respectively. We can show that the equation describing this set of points is x2 a2+ y2 a2− m2= 1 where k= 2a. F1 F2 P2 P3 P1 x y O P1F1+ P1F2= P2F1+ P2F2 = P3F1+ P3F2 This can be pictured as a string of length P1F1+ P1F2being attached by nails to a board at F1 and F2and, considering the path mapped out by a pencil, extending the string so that it is taut, and moving ‘around’ the two points. Let the coordinates of Pbe ( x,y). PF 1= (x− m)2+ y2and PF 2= (x+ m)2+ y2 and assume PF 1+ PF 2= k Then (x+ m)2+ y2+ (x+ m)2+ y2= k Rearranging and squaring gives (x+ m)2+ y2= k2− 2k (x− m)2+ y2+ (x+ m)2+ y2 ∴ 4mx = k2− 2k (x− m)2+ y2 Rearranging and squaring again gives 4k2(x− m)2+ 4k2y2= k4− 8k2mx + 16 m2x2 Collecting like terms 4(k2− 4m2)x2+ 4k2y2= k2(k2− 4m2) ∴ 4x2 k2 + 4y2 k2− 4m2= 1 Let a= k 2,then x2 a2+ y2 a2− m2= 1

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 36 Essential Specialist Mathematics The points F1and F2are called the foci of the ellipse. The constant k= 2ais called the focal sum . Ifa= 3 and m = 2, the ellipse with equation x2 9 + y2 5 = 1 is obtained. For an ellipse with equation x2 a2+ y2 b2= 1 and a> b, the foci are at ( ± (a2− b2),0) Given an equation of the form Ax 2+ By 2+ Cx + Ey + F= 0, where Aand Bare both positive (or both negative), the corresponding graph is an ellipse or a point. If A= Bthe graph is that of a circle. In some cases, as for the circle, no pairs ( x, y ) will satisfy the equation. Hyperbolas The curve with equation x2 a2− y2 b2= 1 is a hyperbola with centre at the origin. The axis intercepts are ( a, 0) and ( −a, 0). The hyperbola has asymptotes y= b axand y=− b ax. An informal argument for this is as follows. The equation x2 a2− y2 b2= 1 can be rearranged: y2 b2= x2 a2− 1 ∴ y2= b2x2 a2  1− a2 x2  But as x→±∞ ,a2 x2→ 0 ∴ y2→ b2x2 a2 i.e. y→± bx a (a, 0) (_a, 0) y =_b ax y = ax x b y 0 x2 a2− y2 b2= 1 The general equation for a hyperbola is formed by suitable translations. The curve with equation (x− h)2 a2 − (y− k2) b2 = 1 is a hyperbola with centre ( h, k ). The asymptotes are y− k=± b a(x− h) This hyperbola is obtained from the hyperbola with equation x2 a2− y2 b2= 1bythe translation defined by ( x,y)→ (x+ h,y+ k).

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 37 Example 30 For each of the following equations, sketch the graph of the corresponding hyperbola, give the coordinates of the centre and the axes intercepts, and the equations of the asymptotes. a x2 9 − y2 4 = 1 b y2 9 − x2 4 = 1 c (x− 1)2− (y+ 2)2= 1 d (y− 1)2 4 − (x+ 2)2 9 = 1 Solution a x2 9 − y2 4 = 1 ∴ y2= 4x2 9  1− 9 x2  Equation of asymptotes: y = _2 3x y = 2 3x (_3, 0) (3, 0) x y 0 y=± 2 3x When y= 0, x2= 9 and therefore x=± 3. Axes intercepts (3, 0) and ( −3, 0), centre (0, 0). b y2 9 − x2 4 = 1 is the reflection of x2 9 − y2 4 = 1 in the line y= x. ∴asymptotes are x=± 2 3y i.e. y=± 3 2x The y-axis intercepts are (0, 3) and (0, −3). y = _3 2x y = 3 2x (0, _3) (0, 3) x y 0 x y y = _xy = x (1, 0) (_1, 0) 0 c (x− 1)2− (y+ 2)2= 1. The graph of x2− y2= 1 is sketched first. The asymptotes are y= xand y=− x. This hyperbola is called a rectangular hyperbola as its asymptotes are perpendicular. The centre is (0, 0) and the axes intercepts are at (1, 0) and ( −1, 0)

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 38 Essential Specialist Mathematics A translation of ( x,y)→ (x+ 1, y− 2) is applied. The new centre is (1, −2) and the asymptotes have equations y+ 2=± (x− 1), i.e. y= x− 3 and y=− x− 1. x y y = _x _1 _1 _3 _1 y = x _ 3 3 0 (1, _2) (2, _2) (0, _2) √5, 0) (1 _ √5, 0) (1 + When x= 0, y=− 2 and when y= 0 (x− 1)2= 5 x= 1± √5 d (y− 1)2 4 − (x+ 2)2 9 = 1 is obtained by translating the hyperbola y2 4 − x2 9 = 1 through the translation defined by ( x,y)→ (x− 2, y+ 1). x y 2 3_x y = 2 3_x y = _ 0, _2) (0, 2) 0 y2 4 x2 9 _ =1 x y (y _ 1) 2 4 (x + 2) 2 9 – =1 2 3_ 7 3_ x + y = 2 3_ 1 3_ x _ y = _ (_2, _1) (_2, 1) (_2, 3) 0 Note that the asymptotes for y2 4 − x2 9 = 1 are the same as for those of the hyperbola x2 9 − y2 4 = 1. The two hyperbolas are called conjugate hyperbolas . Defining a hyperbola Hyperbolas can be defined in a manner similar to the methods discussed earlier in this section for circles and ellipses. Consider the set of all points, P, such that PF 1− PF 2= kwhere kis a constant and F1and F2are points with coordinates ( m, 0) and ( −m, 0) respectively. Then the equation of the curve defined in this way is x2 a2− y2 m2− a2= 1 k= 2a

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 39 Exercise 1G 1 Sketch the graph of each of the following. Label the axes intercepts. State the coordinates of the centre. a x2 9 + y2 16 = 1 b 25 x2+ 16 y2= 400 c (x− 4)2 9 + (y− 1)2 16 = 1 d x2+ (y− 2)2 9 = 1 e 9x2+ 25 y2− 54 x− 100 y= 44 f 9x2+ 25 y2= 225 g 5x2+ 9y2+ 20 x− 18 y− 16 = 0 h 16 x2+ 25 y2− 32 x+ 100 y− 284 = 0 i (x− 2)2 4 + (y− 3)2 9 = 1 j 2(x− 2)2+ 4(y− 1)2= 16 2 Sketch the graphs of each of the following. Label the axes intercepts and give the equations of the asymptotes. a x2 16 − y2 9 = 1 b y2 16 − x2 9 = 1 c x2− y2= 4 d 2x2− y2= 4 e x2− 4y2− 4x− 8y− 16 = 0 f 9x2− 25 y2− 90 x+ 150 y= 225 g (x− 2)2 4 − (y− 3)2 9 = 1 h 4x2− 8x− y2+ 2y= 0 i 9x2− 16 y2− 18 x+ 32 y− 151 = 0 j 25 x2− 16 y2= 400 3 Find the coordinates of the points of intersection of y= 12xwith: a x2− y2= 1 b x2 4 + y2= 1 4 Show that there is no intersection point of the line y= x+ 5 and the ellipse x2+ y2 4 = 1. 5 Find the coordinates of the points of intersection of the curves x2 4 + y2 9 = 1 and x2 9 + y2 4 = 1. Show that the points of intersection are the vertices of a square. 6 Find the coordinates of the points of intersection of x2 16 + y2 25 = 1 and the line with equation 5 x= 4y. 7 On the one set of axes sketch the graphs of x2+ y2= 9 and x2− y2= 9. 8 Sketch each of the following regions: a x2− y2\b 1 b x2− y2≥ 4 c y2\b x2 4 − 1 d x2 9 + y2 4 < 1

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 40 Essential Specialist Mathematics e x2− y2\b 1 and x2+ y2\b 4 f (x− 3)2 16 + y2 9 \b 1 g x2− y2\b 4 and x2 9 + y2\b 1 h x2− y2> 1 and x2+ y2\b 4 i (x− 2)2 9 + y2\b 4 j x2 4 + y2\b 1 and y\b x 1.8 Parametric equations of circles, ellipses and hyperbolas Circles It is sometimes useful to express the rule of a relation in terms of a third variable, called a parameter . We have already seen in the work on circular functions that the unit circle can be expressed in cartesian form, i.e. {(x,y):x2+ y2= 1}or in the form {(x,y):x= cos t, y= sin t, with t∈[0, 2 ]}. The latter is called the set of parametric equations of the unit circle which give the coordinates ( x,y) of all points on the unit circle. The restriction for the values of tis unnecessary in the representation of the graph as {(x,y):x= cos t,y= sin t, with t∈R}gives the same points with repetitions since cos(2 + t)= cos tand sin(2 + t)= sin t. If the set of values for tis the interval [0, ], only the top half of the circle is obtained. The set notation is often omitted, and in the following this will be done. The next three diagrams illustrate the graphs resulting from the parametric equations x= cos tand y= sin t for three different sets of values of t. x y 0 x y 0 y x 0 t∈[0,2] t∈[0,] t∈ 0, 2 In general, x2+ y2= a2, where a> 0, is the cartesian equation of a circle with centre at the origin and radius a. The parametric equations are x= acos tand y= asin t. The minimal interval of tvalues to yield the entire circle is [0, 2 ]. The domain and range of the corresponding cartesian relation can be determined by the parametric equation determining the xvalue and the yvalue respectively. The range of the function with rule x= acos t,t∈[0, 2 ]is[ −a,a] and hence the domain of the relation x2+ y2= a2is [ −a,a]. The range of the function with rule y= asin t,t∈[0, 2 ]is[ −a,a] and hence the range of the relation x2+ y2= a2is [ −a,a].

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 41 Example 31 A circle is defined by the parametric equations x= 2+ 3 cos and y= 1+ 3 sin for ∈[0, 2 ], Find the corresponding cartesian equation of the circle and state the domain and range of this relation. Solution The range of the function with rule x= 2+ 3 cos is [ −1, 5] and hence the domain of the corresponding cartesian relation is [ −1, 5]. The range of the function with rule y= 1+ 3 sin is [ −2, 4] and hence the range of the corresponding cartesian relation is [ −2, 4]. Rewrite the equations as x− 2 3 = cos and y− 1 3 = sin . Square both sides of each of these equations and add: (x− 2)2 9 + (y− 1)2 9 = cos 2+ sin 2and therefore (x− 2)2 9 + (y− 1)2 9 = 1 i.e. ( x− 2)2+ (y− 1)2= 9 Ellipses It has been shown in the previous section that x2 a2+ y2 b2= 1, where aand bare positive real numbers, is the cartesian equation of an ellipse with centre at the origin, and x-axis intercepts (±a,0)and y-axis intercepts (0, ±b). The parametric equations for such an ellipse are x= acos tand y= bsin t. The minimal interval of tvalues to yield the entire ellipse is [0, 2 ]. The domain and range of the corresponding cartesian relation can be determined by the parametric equation determining the xvalue and the yvalue respectively. The range of the function with rule x= acos tis [ −a,a] and hence the domain of the relation x2 a2+ y2 b2= 1is [−a,a]. The range of the function with rule y= bsin tis [ −b,b] and hence the range of the relation x2 a2+ y2 b2= 1is[ −b,b]. The proof that the two forms of equation yield the same graph uses the Pythagorean identity sin 2t+ cos 2t= 1. Let x= acos t and y= bsin t. Therefore x a= cos t and y b= sin t. Squaring both sides of each of these equations yields x2 a2= cos 2t and y2 b2= sin 2t Now, since sin 2t+ cos 2t= 1 it follows that x2 a2+ y2 b2= 1. Example 32 Find the cartesian equation of the curve with parametric equations x= 3+ 3 sin t, y= 2− 2 cos twith t∈Rand describe the graph.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 42 Essential Specialist Mathematics Solution x= 3+ 3 sin tand y= 2− 2 cos t,therefore x− 3 3 = sin tand 2− y 2 = cos t. Square both sides of each equation and add: (x− 3)2 9 + (2 − y)2 4 = sin 2t+ cos 2t Hence (x− 3)2 9 + (2 − y)2 4 = 1 But (2 – y)2= (y–2) 2so this equation is more neatly written as (x− 3)2 9 + (y− 2)2 4 = 1 Clearly this is an ellipse, with centre at (3, 2), and axes intercepts at (3, 0) and (0, 2). Hyperbolas The general cartesian equation for a hyperbola with ‘centre’ at the origin is x2 a2− y2 b2= 1. The parametric equations are x= asec tand y= btan twhere sec t= 1 cos tand t∈ − 2 , 2  gives the right-hand branch of the hyperbola. For the function with rule x= asec tand domain − 2 , 2  the range is [ a,∞ ). (The sec function is discussed further in Chapter 3.) a 0 x = π 2 x = –π 2 x y The graph of y= asec xis shown for the interval − 2 , 2  . For the function with rule y= btan tand domain − 2 , 2  the range is R. The left branch of the hyperbola can be obtained for t∈  2,3 2  . The proof that the two forms of equation can yield the same graph uses a form of the Pythagorean identity sin 2t+ cos 2t= 1. Divide both sides of this identity by cos t. This yields tan 2t+ 1= sec 2t. Consider x= asec tand y= btan t. Therefore x a= sec tand y b= tan t. Square both sides of each equation to obtain x2 a2= sec 2tand y2 b2= tan 2t. Now, since tan 2t+ 1= sec 2t, it follows that y2 b2+ 1= x2 a2. Therefore x2 a2− y2 b2= 1.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 43 Example 33 Find the cartesian equation of the curve with parametric equations x= 3 sec t,y= 4 tan t, where t∈  2,3 2  and describe the curve. Solution Now x= 3 sec tand y= 4 tan t. Therefore x 3= sec tand y 4= tan t. Square both sides of each equation to obtain x2 9 = sec 2tand y2 16 = tan 2t. Add these two equations to obtain y2 16 + 1= x2 9. So the cartesian form of the curve is x2 9 − y2 16 = 1. The range of the function with rule x= 3 sec tfor t∈  2,3 2  is ( −∞ ,−3]. Hence the domain for the graph is ( −∞ ,−3]. This is the left branch of a hyperbola, with centre at the origin, and xintercept at (−3,0) and with asymptotes with equations y= 4x 3 and y=− 4x 3. Example 34 Give parametric equations for each of the following: a x2+ y2= 9 b x2 16 + y2 4 = 1 c (x− 1)2 9 − (y+ 1)2 4 = 1 Solution a The parametric equations are x= 3 cos tand y= 3 sin tor x= 3 sin tand y= 3 cos t. There are infinitely many pairs of equations which determine the curve given by the cartesian equation x2+ y2= 9. Others are x=− 3 cos(2 t) and y= 3 sin(2 t). For x= 3 cos tand y= 3 sin tit is sufficient for tto be chosen for the interval [0, 2 ] to obtain the whole curve. For x=− 3 cos(2 t) and y= 3 sin(2 t) it is sufficient for tto be chosen in the interval [0, ]. b The obvious solution is x= 4 cos tand y= 2 sin t c The obvious solution is x− 1= 3 sec tand y+ 1= 2 tan t Using a graphics calculator with parametric equations Press MODE and select Par from the fourth row of the menu. Ensure that the calculator is in radian mode.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 44 Essential Specialist Mathematics In the Y = screen enter X 1T = 2 cos 3T and Y1T = 2 sin 3T. The T is obtained by pressing the key XT N. The period of both functions is 2 3 . Press WINDOW and complete the settings as shown. From the ZOOM menu select Zsquare and activate TRACE . The TRACE can be used to see the order of plotting. Using a CAS calculator with parametric equations Press MODE and select PARAMETRIC as shown. Enter the functions in the Y= screen. From WINDOW establish the settings as shown. From the ZOOM menu choose ZoomSquare. Activate TRACE . Exercise 1H 1 Find the cartesian equation of the curve determined by the parametric equations x= 2 cos 3 tand y= 2 sin 3 t, and determine the domain and range of the corresponding relation.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Chapte r1—A toolbox 45 2 Determine the corresponding cartesian equation of the curve determined by each of the following parametric equations and sketch the graph of each of these. a x= sec t,y= tan t,t∈  2,3 2  b x= 3 cos 2 t,y=− 4 sin 2 t c x= 3− 3 cos t,y= 2+ 2 sin t d x= 3 sin t,y= 4 cos t,t∈ − 2, 2 e x= sec t,y= tan t,t∈  − 2, 2  f x= 1− sec (2 t),y= 1+ tan(2 t), t∈  4,3 4  3 Give parametric equations corresponding to each of the following: a x2+ y2= 16 b x2 9 − y2 4 = 1 c (x− 1)2+ (y+ 2)2= 9 d (x− 1)2 9 + (y+ 3)2 4 = 9 4 A circle has centre (1, 3) and radius 2. If parametric equations for this circle are x= a+ bcos(2 t) and y= c+ dsin(2 t), where a,b,cand dare positive constants, state the values of a,b,cand d. 5 An ellipse has x-axis intercepts ( −4, 0) and (4, 0) and y-axis intercepts (0, 3) and (0, −3). State a possible pair of parametric equations for this ellipse. 6 The graph of the circle with parametric equations x= 2 cos 2 tand y= 2 sin 2 tis dilated by a factor 3 from the xaxis. For the image curve, state: a a possible pair of parametric equations b the cartesian equation 7 The graph of the ellipse with parametric equations x= 3− 2 cos t 2  and y= 4+ 3 sin t 2  is translated 3 units in the negative direction of the xaxis and 2 units in the negative direction of the yaxis. For the image curve state: a a possible pair of parametric equations b the cartesian equation 8 Sketch the graph of the curve with parametric equations x= 2+ 3sin(2 t) and y= 4+ 2cos(2 t) for: a t∈0,14  b t∈0,12  c t∈0,32  For each of these, state the domain and range.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Review 46 Essential Specialist Mathematics Summary of circles, ellipses and hyperbolas Circles The circle with centre at the origin and radius ais the graph of the equation x2+ y2= a2. The circle with centre ( h,k) and radius ais the graph of the equation (x− h)2+ (y− k)2= a2. In general, x2+ y2= a2, where a> 0, is the cartesian equation of a circle with centre at the origin and radius a. The parametric equations are x= acos tand y= asin t. The minimal interval of tvalues to yield the entire circle is [0, 2 ]. The circle with centre ( h,k) and radius acan be described through the parametric equations x= h+ acos tand y= k+ asin t. Ellipses The curve with equation x2 a2+ y2 b2= 1 is an ellipse with centre the origin, x-axis intercepts (−a, 0) and ( a,0)and y-axis intercepts (0, −b) and (0, b). For a> bthe ellipse will appear as shown in the diagram on the left. If b> athe ellipse is as shown in the diagram on the right. y x 0 A' _b B' b A _a a B y x 0 A' B' B A _a _b a b The curve with equation (x− h)2 a2 + (y− k)2 b2 = 1 is an ellipse with centre ( h,k). The ellipse with centre at the origin, and x-axis intercepts ( ±a,0)and y-axis intercepts (0, ±b) has parametric equations x= acos tand y= bsin t. The minimal interval of tvalues to yield the entire ellipse is [0, 2 ]. The ellipse with centre ( h,k) formed by translating the ellipse with equation x2 a2+ y2 b2= 1 can be described through the parametric equations x= h+ acos tand y= k+ bsin t.

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Review Chapte r1—A toolbox 47 Hyperbolas The curve with equation x2 a2− y2 b2= 1isa hyperbola with centre at the origin. The axis intercepts are ( a, 0) and ( −a, 0). The hyperbola has asymptotes y= b axand y=− b ax. x x y (_a, 0) (a, 0) y = ba _ x y = ba 0 The curve with equation (x− h)2 a2 − (y− k)2 b2 = 1 is a hyperbola with centre ( h,k). The hyperbola has asymptotes y− k= b a(x− h) and y− k=− b a(x− h). The parametric equations for the hyperbola shown above are x= asec tand y= btan t where sec t= 1 cos t. The hyperbola with centre ( h,k) formed by translating the hyperbola with equation x2 a2− y2 b2= 1 can be described through the parametric equations x= h+ asec tand y= k+ btan t. Multiple-choice questions 1 The 3rd term of a geometric sequence is 4. If the 8th term is 128, then the 1st term is: A 2 B 1 C 32 D 5 E none of these 2 If the numbers 5, xand yare in arithmetic sequence then: A y= x+ 5 B y= x− 5 C y= 2x+ 5 D y= 2x− 5 E none of these 3 If 2 cos x◦− √2= 0, the value of the acute angle x◦is: A 30 ◦ B 60 ◦ C 45 ◦ D 25 ◦ E 27 .5◦ 4 The equation of the graph shown is: 2π π x 1 0 –1 y A y= sin 2  x−  4  B y= cos  x+  4  C y= sin(2 x) D y=− 2 sin( x) E y= sin  x+  4  5 The exact value of the expression sin 2 3  × cos  4  × tan  6  is: A 1√2 B 1√3 C √2 4 D √3 2 E none of these

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Review 48 Essential Specialist Mathematics 6 In the diagram, A,B,Cand D are points on the circle. ∠ABD = 35 ◦and ∠AX B = 100 ◦. The magnitude of ∠XDC is: B A D C X 35° 100 ° A 35 ◦ B 40 ◦ C 45 ◦ D 50 ◦ E 55 ◦ 7 In a geometric sequence t2= 24 and t4= 54. The sum of the first 5 terms, if the common ratio is positive, is: A 130 B 211 C 238 D 316.5 E 810 8 In a triangle ABC ,a= 30 ,b= 21 and cos C = 5153. The value of cto the nearest whole number is: A 9 B 10 C 11 D 81 E 129 9 The coordinates of the centre of the circle with equation x2− 8x+ y2− 2y= 8 are: A (−8,−2) B (8, 2) C (−4,−1) D (4, 1) E (1, 4) 10 The equation of the graph shown is: 2 4 11 0 –7 –4 y x A (x+ 2)2 27 − y2 108 = 1 B (x− 2)2 9 − y2 34 = 1 C (x+ 2)2 81 − y2 324 = 1 D (x− 2)2 81 − y2 324 = 1 E (x+ 2)2 9 − y2 36 = 1 Short-answer questions (technology-free) 1 For the difference equation fn= 5fn−1,f0= 1, find fnin terms of n. 2 AP and BP are tangents to the circle with centre O.If AP = 10 cm, find OP . A B P O 2α°

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Review Chapte r1—A toolbox 49 3 Write down the equation of the ellipse shown. (0, 3) (_2, 7) x y O 4 Find sin ◦. 7 8 θ° 5 Find x. 9 cm x cm 30 ° 6 A circle has a chord of length 10 cm situated 3 cm from its centre. Find: a the radius length b the angle subtended by the chord at the centre 7 a Find the exact value of cos 315 ◦. b Given that tan x◦= 34and 180 < x< 270, find an exact value of cos x◦. c Find an angle A(A = 330) such that sin A= sin 330 ◦. 8 In the diagram, AD is a tangent to the circle with centre O,AC intersects the circle at B, and BD = AB . a Find ∠BCD in terms of x. b IfAD = ycm, AB = acm and BC = bcm, express yin terms of aand b. x° A B D C O 9 ABC is a horizontal right-angled triangle with the right angle at B.Pis a point 3 cm directly above B. The length of AB is 1 cm and the length of BC is 1 cm. Find, the angle which the triangle AC P makes with the horizontal. A B C P 10 a Solve 2 cos(2 x+ )− 1= 0, −\b x\b . b Sketch the graph of y= 2 cos(2 x+ )− 1, −\b x\b , clearly labelling axes intercepts. c Solve 2 cos(2 x+ )< 1, −\b x\b .

P1: GHM 0521609992c01.xml CUAT007-EVANS July 19, 2006 19:34 Review 50 Essential Specialist Mathematics 11 The triangular base ABC of a tetrahedron has side lengths AB = 15 cm, BC = 12 cm and AC = 9 cm. If the apex D is 9 cm vertically above C, then find: a the angle Cof the triangular base b the angles that the sloping edges make with the horizontal 12 Two ships sail from port at the same time. One sails 24 nautical miles due east in three hours, and the other sails 33 nautical miles on a bearing of 030 ◦in the same time. a How far apart are the ships three hours after leaving port? b How far apart would they be in five hours if they maintained the same bearings and constant speed? 13 Find x. 30 ° 45 ° 18 cm x cm 14 An airport Ais 480 km due east of airport B. A pilot flies on a bearing of 225 ◦from Ato C and then on a bearing of 315 ◦from Cto B. a Make a sketch of the situation. b Determine how far the pilot flies from Ato C. c Determine the total distance the pilot flies. 15 Find the equations of the asymptotes for the hyperbola with rule x2− (y− 2)2 9 = 15. 16 A curve is defined by the parametric equations x= 3 cos(2 t)+ 4 and y= sin(2 t)− 6. Give the cartesian equation of the curve. 17 a Find the value of x. 2x° x° b Find a,b,cand d, given that PR is a tangent to the circle with centre O. P R a° b° 60 ° d° c° O 18 A curve is defined by the parametric equations x= 2 cos( t) and y= 2 sin( t)+ 2. Give the cartesian equation of the curve.

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