Special Mathematics Notes
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1 Special Mathematics Notes Tetbook: Classroom Mathematics Stds 9 & 10 CHAPTER 6 Trigonometr Trigonometr is a stud of measurements of sides of triangles as related to the angles, and the application of this theor. Let ABC be right-angled so that angles A and B are acute (less than 90 o ). A c b C θ a B Let ABC = θ. We define (i) the sine of θ as sinθ = b = AC = opposite c AB hpotenuse (ii) the cosine of θ as cos θ = a = CB = adjacent c AB hpotenuse and (iii) the tangent of θ as tan θ = b = AC = opposite. a CB adjacent The three functions defined above are called trigonometric functions. We can deduce from (iii) that tan θ = sinθ. In fact all other trigonometric functions cosθ are defined in terms of sine and cosine onl. The other three trig functions are defined as follows: (iv) the cosecant of θ is csc θ = 1, sinθ (v) the secant of θ is sec θ = 1, cosθ (vi) the cotangent of θ is sec θ = cosθ. sinθ Now sin 2 θ + cos 2 θ = b2 +a 2 = c2 = 1; thus sin 2 θ = 1 cos 2 θ and cos 2 θ = c 2 c 2 1 sin 2 θ. Also sin(90 o θ) = a/c = cos θ and cos(90 o θ) = b/c = sinθ. If and are acute angles such that + = 90 o then the are called complimentar angles. In such a case sin = cos and cos = sin. Eample : If cos = 1/2 and and are complimentar angles, find sin. Soln. sin = cos = 1/2. Some special (acute) angles (in degrees) viz. 0; 30; 45; (i) sin0 o = 0, cos 0 o = 1. Since 0 and 90 o are complimentar angles, we con- 1
2 clude that sin90 o = cos0 o = 1 and cos 90 o = sin0 o = 0. (ii) Consider a right-angled isosceles triangle as shown: A 45 o c a C 45 a o B Then sin45 o = a/c = cos45 o. Also a 2 + a 2 = c 2 implies c = 2a. Thus sin45 o = 1/ 2 = cos 45 o. Hence tan 45 o = 1. (iii) Consider the right-angled triangle with acute angles of 30 o and 60 o and a = c/2 as shown: A 30 o c b C 60 a o B Then b 2 = c 2 (c/2) 2 = (3/4)c 2, impling b = 4c/2. Thus sin60 o = 3 = cos30 o since = 90. Also b = 4c/2. Thus 2 cos 60 o = 1 = sin 2 30o since = 90. Trig functions for angles > 90 o The -plane is divided into 4 regions called quadrants: the 1st quadrant is formed b the +ve -ais and the +ve -ais; the 2nd quadrant is formed b the +ve -ais and the -ve -ais; the 3rd quadrant is formed b the -ve -ais and the -ve -ais; the 4th quadrant is formed b the -ve -ais and the +ve -ais; see diagram below: 2
3 2nd quad. 1st quad. 3rd quad. 4th quad. With this in mind, an acute angle is usuall referred to as an angle in the first quad; an obtuse angle (between 90 o and 180 o ) is usuall referred to as an angle in the 2nd quad; an angle between 180 o and 270 o is usuall referred to as an angle in the 3rd quad; and an angle between 270 o and 360 o is usuall referred to as an angle in the 4th quad. Let P(, ) be a point such that OP makes an angle θ with the +ve -ais. Denote the length of OP b r. Then we define (i) sinθ = ; (ii) cos θ = ; (iii) tan θ =. The other trig functions r r are defined as before. P(, ) P(, ) r r θ r θ r P(, P(, ) ) Observe that r 2 = in each quadrant. We sa an angle is positive if it is measured counterclockwise. It is negative when measured clockwise. Now consider the above diagram: (1) sinθ = /r 3
4 (ii) Since 180 o θ is in the 2nd quad, sin(180 o θ) = /r = sinθ b using the coordinates of P in the 2nd quad.; (iii) Since 180 o +θ is in the 3rd quad, sin(180 o +θ) = /r = sinθ b using the coordinates of P in the 3rd quad.; (iv) Since 360 o θ is in the 4th quad, sin(360 o θ) = /r = sinθ b using the coordinates of P in the 4th quad.; Similarl (v) cos θ = /r (vi) Since 180 o θ is in the 2nd quad, cos(180 o θ) = /r = cosθ b using the coordinates of P in the 2nd quad.; (vii) Since 180 o +θ is in the 3rd quad, cos(180 o +θ) = /r = cos θ b using the coordinates of P in the 3rd quad.; (viii) Since 360 o θ is in the 4th quad, cos(360 o θ) = /r = cos θ b using the coordinates of P in the 4th quad.; NOTE: From above we see that (a) both sine and cosine are +ve in the first quad (b) In the 2nd quad sine is +ve while cosine is -ve (c) In the 3rd quad both sine and cosine are -ve (d) In the 4th quad sine is -ve while cosine is +ve sine +ve & cos -ve. sine +ve & cos +ve. sine -ve & cos -ve. sine -ve & cos +ve. Eamples: 1. Find sin120 o without a calculator. Soln. sin120 o = sin(180 o 60 o ) = sin60 o = 3 b (ii) above 2 2. Find cos225 o without a calculator. Soln. cos 225 o = cos(180 o + 45 o ) = cos45 o = 1 2 b (vii) above 4
5 3. Find cos300 o without a calculator. Soln. cos 300 o = cos(360 o 60 o ) = cos 60 o = 1 b (viii) above 2 4. Find sin315 o without a calculator. Soln. sin315 o = sin(360 o 45 o ) = sin45 o = 1 2 b (iv) above Note: sin(180 o ) = sin(180 o 0 o ) = sin0 o = 0 cos(180 o ) = cos(180 o 0 o ) = cos0 o = 1 sin(270 o ) = sin(180 o + 90 o ) = sin90 o = 1 cos(270 o ) = cos(180 o + 90 o ) = cos 90 o = 0 sin(360 o ) = sin(360 o 0 o ) = sin0 o = 0 cos(360 o ) = cos(360 o 0 o ) = cos0 o = 1 Periodicit of sine and cosine Sine and cosine are periodic with period 360 o since after 360 O the start to repeat themselves, i.e. sin(+360 o ) = sin and cos(+360 o ) = cos. Hence after an multiple of 360 o, sine and cosine repeat themselves, i.e. sin(+360 o n) = sin and cos( o n) = cos where n is an integer, e.g. 3 or -3. Eamples: 1. Find sin765 o without using a calculator. Soln. sin765 o = sin(2 360 o + 45 o ) = sin45 o = Given sec θ = 5/2, find cos θ, sin θ and cot θ without a calculator. Soln. sec θ = 5/2 implies cosθ = 2/5. Therefore r = 5 and = 2. Therefore 2 = r 2 2 = 25 4 = 21 b the Theorem of Pthagoras. Thus = 21. Since this cosine is +ve, the angle θ is either in the 1st or 4th quad. See diagram below: 5
6 P(2, 21) 5 θ 5 P(2, 21) Case θ is in the first quad: sinθ = 21 and cotθ = Case θ is in the 4th quad: sinθ = 21 and cot θ = Negative Angles An angle is negative when measured clockwise. P(, ) r θ θ r P(, ) sin( θ) = /r = sinθ cos( θ) = /r = cosθ Eamples: 1. cos( 45 o ) = cos(45 o ) = sin( 690 o ) = sin(690 o ) = sin(720 o 30 o ) [ sin30 o ] = sin30 o = 1 2 6
7 Trigonometric Identities Recall: 1. sin 2 + cos 2 = 1 (1), for an angle. This is an eample of a trig identit. 2. Divide eqn (1) b cos to get tan = sec 2 (2a), and divide eqn (1) b sin to obtain 1 + cot 2 = csc 2 (2b) We give the following identities without proof: 3. sin( + ) = sincos + cos sin (3) 4. cos( + ) = cos cos sinsin (4) We derive the following: 5. In (3), letting = we get sin2 = 2sin cos (5) 6. In (4), letting = we get cos2 = cos 2 sin 2 (6) Eample: (a) Find sin 15 o cos15 o Soln. B (5) above sin 15 o cos15 o = sin2(15o ) = sin(30o ) = 1/4 2 2 (b) Find cos o sin o Soln. B (6) above cos o sin o = cos2(105 o ) = cos 210 o = cos(180 o + 30 o ) = cos30 o = 3 2 (7) Now (6) implies cos 2 = cos 2 (1 cos 2 ) = 2cos 2 1, impling cos 2 1+cos 2 = (7), (double angle formula); 2 8) Also (6) implies cos2 = 1 sin 2 sin 2 = 1 2sin 2, impling sin 2 1 cos 2 = (8), (double angle formula); 2 Eample: (a) Find sin 15 o without using a calculator. Soln. sin 2 15 o = 1 cos 2(15o ) = 1 32 = 2 3, thus sin 15 o 2 3 = (b) Find cos(22, 5) o without using a calculator. Soln. cos 2 (22, 5) o 1+cos 45o = = 1+(1/ 2) = , thus cos(22, 2 5)o = Add the following eqns: sin( + ) = sincos + cossin (3a) and sin( ) = sincos cos sin (3b) to get 2sin cos = sin( + ) + sin( ) (9) 10. Add the following eqns: cos( + ) = coscos sinsin (4a) and cos( ) = cos cos + sinsin (4b) to get 2cos cos = cos( + ) + cos( ) (10) 11. Subtracting in the above eqns (4b - 4a) we have 2sinsin = cos( ) cos( + ) (11) Eample: sin75 o cos 15 o = sin90o +sin 60 o = 2+ 3 b eqn (9) 2 4 More Eamples 7
8 Eample 1. Let tan θ = 4/3, 270 o < θ < 360 o. Find sin 2θ; cos 2θ and tan 2θ. Soln. θ r = 5 P(3, 4) From Pthagoras, the hpothenuse is r 2 = = 25, hence r = 5. Therefore sin2θ = 2sinθ cosθ = 2( 4/5)(3/5) = 24/25 cos 2θ = cos 2 θ sin 2 θ = (3/5) 2 ( 4/5) 2 = (9 16)/25 = 7/25 Proving given Identities In proving identities, take one side viz the right hand side (rhs) or the left hand side (lhs) and show that it equals the other side. It is advisable to take the more complicated side and simplif it until it gives the other side. Before giving eamples, we derive an identit involving tan θ: We want to show that tan(θ + α) = tanθ+tan α Soln. tan(θ + α) = sin(θ+α) cos(θ+α) = 1 tan θ sinθ cosα+cosθ sinα). cos θ cosα sinθ sinα cos θ cos α, then tan(θ + α) = tanθ+tan α 1 tan θ In the above eqn let θ = α, then tan 2θ = 2tanθ 1 tan 2 θ. Eample: Find tan 75 o without using a calculator. Soln. tan 75 o = tan(30 o + 45 o ) = tan30o +tan 45 o More Eamples: sin 2 1. Prove that = tan 1+cos2 Soln. lhs = sin2 = 2sin cos = 1+cos 2 1+cos 2 sin 2 tan = rhs 1 sin 2 2. Prove that = sincos sin cos = tan30 o tan45 o sin cos cos 2 +sin 2 +cos 2 sin 2 Now divide the rhs b = 2sin cos 2cos cos = 8
9 Soln. lhs = sin2 +cos 2 2 sin cos sin cos 2 cot 3. Prove that = sin2 1+cot 2 Soln. lhs = 2cos / sin 1+cos 2 /sin 2 4. Prove that 1 1+cos + 1 = 2sin cos sin 2 +cos 2 Soln. lhs = 1 cos+1+cos 1 cos 2 5. Prove that 1 cos 2 = tan2 sin 2 / cos 2 = (sin cos)(sin cos) sin cos = sin2/1 = rhs = 1 cos 2csc2 = 2/sin 2 = 2csc 2 = rhs 1+tan 2 Soln. rhs = = sin 2 /1 = 1 cos 2 = lhs (cos 2 +sin 2 )/cos 2 csc 6. Prove that = sec 1+cot 1+tan 1/sin = 1 (sin+cos )/sin sin+cos Soln. lhs = lhs. 7. Prove that (sec tan ) 2 = 1 sin 1+sin and rhs = 1/ = sin cos = rhs cos (sin+cos)/cos = 1 sin cos = (1 sin) Soln. rhs = 2 = 1 2sin+sin2 = 1/cos 2 2sin /cos 2 + (1+sin)(1 sin) 1 sin 2 sin 2 /cos 2 = sec 2 2tan sec + tan 2 = (sec tan) 2 = lhs. 8. Prove that cot = sin2 1 cos 2 2 sin cos Soln. rhs = = cot = rhs. Trig Eqns = 2 sincos 1 cos 2 +sin 2 2 sinsin Eamples : 1. Solve 2cos = 1, [0 o, 360 o ] without the use of a calculator. Soln. cos = 1/2 is negative in 2nd and 3rd quadrants. Thus the soln is = 180 o 60 o = 120 o or = 180 o + 60 o = 240 o. 2. Solve 2cos = 1, [ 180 o, 360 o ] without the use of a calculator. Soln. cos = 1/2 is negative in 2nd and 3rd quadrants. Thus the soln is = 180 o 60 o = 120 o or = 180 o + 60 o = 240 o or = 120 o. 3. Find a general solution of the eqn 2cos = 1, without the use of a calculator. Soln. cos = 1/2 is negative in 2nd and 3rd quadrants. Thus the general soln is = 120 o o n or = 240 o o n where n is an integer. (NOTE: = 120 o o n is the same as = 240 o o n and = 240 o o n is the same as = 120 o o n) A general soln caters for all possible values of angle. 4. Find a general soln of 2sin 2 sin 1 = 0, without the use of a calculator. Soln. (2sin + 1)(sin 1) = 0 implies sin = 1/2 or sin = 1. sin is 9
10 negative in 3rd and 4th quadrants. Thus = 210 o +360 o n or = 330 o +360 o n or = 90 o o n where n is an integer. 5. Solve sec = 2, [ 180 o, 180 o ] without the use of a calculator. Soln. cos = 1/ 2 is +ve in 1st and 4th quadrants. Thus the soln is = 45 o or = 45 o. 6. Solve cos = sin20 o, [0 o, 360 o ], without the use of a calculator. Soln. cos = cos 70 o > 0 since = 90. Thus the soln is = 70 o or = 360 o 70 o = 290 o. Using a calculator If cos = 0.35 then to find a solution from a calculator, enter the number 0.35 and then press cos 1. Then adjust b 180 o or 360 o to find a correct answer. Eamples: 1. Find a general soln of the eqn 3cos 2 + 5sin = 1, using a calculator. Soln. 3cos 2 +5sin = 3(1 sin 2 )+5sin = 1 implies 3sin 2 5sin 2 = 0. Thus (3sin + 1)(sin 2) = 0. Thus sin = 1/3 since the last bracket is impossible. Therefore = 19, 5 o o o n = 199, 5 o o n or = 360 o 19, 5 o o n = 340, 5 o o n 2. Solve 4cot + 3csc = 0, [0 o, 360 o ] using a calculator. Soln. 4 cos + 3/sin = 0 implies cos = 3/4 is negative in 2nd and sin 3rd quadrants. Using a calculator, we have = 180 o 41, 4 o = 138, 6 o or = 180 o + 41, 4 o = 221, 4 o. Graphs of Trigonometric Functions Here we look at graphs of = sin ; = cos and = tan. Note that sin and cos are periodic with period 360 o while tan has period 180 o. Thus the graph of each trig function repeats itself after its period. 1. Graph of = sin: Consider the table of points on the graph: Thus the graph of = sin is as follows: 10
11 The maimum of sin = 1 and is called the amplitude of the graph or of the function = sin. In general the graph of = k sin has amplitude = k. Secondl the graph of = sin(n) has period = 360 o /n. Similarl for the cosine function; i.e. = k cos n has amplitude = k and its graph has period = 360 o /n. = k sin n has amplitude = k and period = 360 o n. Similarl for the cosine function. 2. Graph of = 3sin 2: Its period = 360/2 = 180 o, so the graph repeats itself after intervals of 180 o. Its amplitude = 3. Consider the table of points on the graph: Thus the graph of = sin is as follows: 11
12 Graph of = 2sin(/2): Its period = = 720 o, so the graph repeats itself after intervals of 720 o. Its amplitude = 2. Consider the table of points on the graph: Thus the graph of = sin is as follows: 12
13 Graph of = cos: Its period = 360 o, so the graph repeats itself after intervals of 360 o. Its amplitude = 1. Consider the table of points on the graph: Thus the graph of = cos is as follows: 13
14 Graph of = tan = tan n has period = 180/n and NO amplitude is defined for it. 5. Graph = tan. Soln. Period = 180 o. This graph has asmptotes at points such that cos = 0. Thus the asmptotes are the lines = ±90 o ; = ±270 o and further multiples of 90 o. Consider the table of points on the graph: Thus the graph of = tan is as follows: 14
15 Graph = tan 2. Soln. Period = 180 o /2 = 90 o. This graph has asmptotes at points such that cos 2 = 0. Thus the asmptotes are the lines = ±45 o ; = ±135 o and further multiples of 45 o. Consider the table of points on the graph: Thus the graph of = tan 2 is as follows: 15
16
17 CHAPTER 7 Analtic Geometr Distance between two points: Let (a, b) and (, ) be an two points in a plane. The distance between them is d = (a ) 2 + (b ) 2. See the accompaning diagram: B(a, b) b A(, ) a C(a, ) B the Theorem of Pthagors, we have AB 2 = AC 2 +CB 2 = (a ) 2 +(b ) 2, hence AB = (a ) 2 + (b ) 2. Eamples : 1. Find the distance between A( 3, 1) and B(2, 4). Soln. The distance is AB = ( 3 2) 2 + (1 4) 2 = Calculate the value of for which M( 2, 1) and N(, 7) are equidistant from P(1, 4). Soln. 17
18 M( 2, 1) N(, 7) N(, 7) P(1, 4) MP = PN implies MP 2 = PN 2. Using the distance formula we have ( 3) = ( 1) 2 +( 3) 2, impling ( 1) 2 = 25, hence 1 = ±5. Thus = 4 or = A(2, 1), B( 3, 4) and C(4, 5) are vertices of ABC. Find the perimeter of the triangle. Sa whether the triangle is equilateral, isosceles, scalene or right-angled. 18
19 B( 3, 4) C(4, 5) A(2, 1) Soln. (i) Perimeter is P = AB+BC+CA = ( 5) (7) = ( 2) 2 + ( 6) 2 = (ii) Triangle is isosceles since AB = BC = 5 2 (iii) Triangle is not right-angled since the sides do not satisf the theorem of Pthagoras. 4. Show that P( 1, 1), Q(0, 4), R(3, 5) and S(2, 2) are vertices of a rhombus. (A rhombus is a 4-sided figure having that all its sides are of equal length). 19
20 Q(0, 4) R(3, 5) P( 1, 1) S(2, 2) Soln. QR = = 10; PS = = 10; PQ = = 10; RS = = 10. Thus the given 4-sided figure is a rhombus since all 4 sides are equal. 5. Show that M(2, 1) is the circumcentre of P QR having vertices P(6, 2), Q( 1, 3) and R( 1, 5). (Show that PM = QM = RM). Soln. 20
21 Q( 1, 3) P(6, 2) M(2, 1) R( 1, 5) PM = = 5 = QM = RM, as required. Mid-point of a line segment Let A( 1, 1 ) and B( 2, 2 ) be points in a plane. M is the mid-point of the line segment if AM = BM and M lies on the same line as A and B. We want to determine the coordinates of M in terms of those of A and B. Suppose we use (a, b) for the coordinates of M. See diagram below: 21
22 B( 2, 2 ) M(a, b) A( 1, 1 ) D( 2, 1 ) Using congruence of triangles, we ma show that (a, b) = ( 1+ 2, 1+ 2 ). 2 2 Eamples : 1. Find the midpoint of the line segment joining A( 3, 1) to (4, 2). Soln. The midpoint is M = ( 3+4, 1+2) = 2 2 (1, 3) Find and if ( 3, 2) is the mid-point of the line segment joining ( 1, 5) and (, ). 22
23 A( 1, 5) M( 3, 2) B(, ) Soln. Now 3 = ( 1)/2 implies = 5, and 2 = (5 + )/2 implies = 1. Recall: If L 1 : = m 1 + c 1 and L 2 : = m 2 + c 2, such that (i) L 1 L 2, then m 1 = m 2, (ii) L 1 L 2, then m 1 m 2 = 1. Eamples : 1. Show that A( 2, 3), (0, 2) and C(1, 3/2) are collinear. Soln. Note that collinear means ling on the same line. The slope of AB is M AB = 2 3 = 1/2 and slope of BC is M 0 ( 2) BC = 3/2 2 = 1/2 = M 1 AB, thus AB BC. Since the point B belongs to both AB and BC, we conclude that A, B and C are collinear. 2. A( 3, 1), B( 2, 2), C(1, 3) and D(0, 0) are vertices of a quadrilateral (four-sided figure). (a) Show that the quadrilateral is a parallelogram; (b) The diagonals of the quad. bisect each other at right angles; (c) what kind of a quad. is this? 23
24 B( 2, 2) C(1, 3) A( 3, 1) D(0, 0) Soln. (a) Slope of AB is M AB = 2 ( 1) = 3 and the slope of DC is M 2 ( 3) DC = = M AB, impling AB DC. Slope of BC is M BC = 3 2 = 1/3 and the 1 ( 2) 1 0 = slope of AD is M AD = 0 ( 1) 0 ( 3) = 1/3 = M BC, impling BC AD. Therefore ABCD is a parallelogram. (Opposite sides are parallel). (b) M AC = = 1 and M BD = 2/2 = 1, impling M AC M BD = 1. Thus AC BD, so the diagonals intersect at right angles since we alread know that the diagonals of a parallelogram bisect each other. (c) The parallelogram is a rhombus. (Check that all the sides are of equal length 10). The Circle A circle is a set of points that are equidistant from a fied point called the center of the circle. The distance from an of the points on the circle to the center is the radius of the circle. The distance from one point to the opposite 24
25 one through the center is the diameter of the circle. Hence the diameter is twice the radius. See diagram below: r ( o, o ) Circumference of a circle is its perimeter or length. It is given b the formula C = 2πr where r is the radius. The area of the circle is A = πr 2. The equation of a circle of radius r centered at ( 0, 0 ) is ( 0 ) 2 +( 0 ) 2 = r 2. Eamples : 1. (3, a) lies on the circle = 18. Find a. Soln. Substitute the coordinates of the point (3, a) into the eqn of the circle since it lies on the circle. Thus we have 9 + a 2 = 18 impling a = ±3. 2. A chord of the circle = 50 has mid-point (6, 3). Determine the line of which the chord is a segment. (A chord is line segment joining two points on a circle). Soln. 25
26 A O M(6, 3) B The circle has radius 50 which is approimatel 7 units. OMB OMA (S,S,S). Thus angle OMA = angle OMB = 90 o hence OM AB. Slope of OM is = 1/2, impling slope of AB is = 2. Therefore the eqn of the 6 0 1/2 required line is 3 = 2( 6) = , impling = Determine the eqn of the circle that passes through (3, 2) and (5, 4) with its centre on the line = 1. Soln. The general eqn is of form ( o ) 2 + ( o ) 2 = r 2. Since ( o, o ) is the centre, it lies on the given line, impling o = o 1 ( ). Substitute into the circle eqn to get ( o ) 2 + ( o + 1) 2 = r 2. Replacing (, ) with (3, 2) and (5, 4) respectivel, we get (3 o ) 2 + (3 o ) 2 = r 2 impling 2(3 o ) 2 = r 2 (1) and (5 o ) 2 + (3 + o ) 2 = r 2 = 2(3 o ) 2 b (1). This gives o = 2 and o = 3 upon using ( ). Substitute this into eqn (1) to get r 2 = 50. Therefore the eqn of the circle is (+2) 2 +(+3) 2 =
27 Tangents and Normals to the Circle l : tangent P L : secant O A line L passing through two points on the circle is called a secant. A line l touching the circle at onl one point P is a tangent to the circle at P. A line segment such as OP from the centre to the point P of tangenc is alwas perpendicular to the tangent l. Thus an angle between OP and l is 90 o. Eamples: 1. Find an equation of the tangent to the circle = 10 at ( 3, 1). Soln. 27
28 P l O 10 Let l be tangent to the circle at ( 3, 1). OP l. Slope of OP is M OP = 1 1/3 implies slope of l is = 3. Therefore eqn of l is 1 = 3( + 3) = 1/ , impling = Determine the eqn of the tangent to the circle = 5 at ( 2, 1). Soln. We complete the squares to epress the eqn in standard form: = 5, impling ( 1) 2 +(+2) 2 = 10. This is a circle of radius 10 centered at (1, 2). Let l be a line tangent to the circle at P( 2, 1). 28
29 O 10 P l 1+2 OP l. Slope of OP is M OP = 1/3. Thus the slope of l is 3. Therefore 2 1 the eqn of l is + 1 = 3( + 2) = 3 + 6, impling = L : = + 2 cuts the circle = 20 at A and B. (a) Find A and B; (b) the length of the chord AB; (c) find M, the mid-point of AB; (d) show that OM AB, where O is the origin ; (e) find the tangents at A and B; (f) find the point C of intersection of the tangents in (e). 29
30 L B(2, 4) l 2 l 1 M( 1, 1) O 10 A( 4, 2) Soln. (a) To find points of intersection of L and the circle, solve their eqns simultaneousl: So we have 2 + ( + 2) 2 = 20, impling = 0 = ( + 4)( 2), thus = 4 or = 2, and = 2 or = 4. Therefore A = ( 4, 2) and B = (2, 4). (b) AB = = 6 2; (c) M = ( 2 4 ) = ( 1, 1);, (d) Slope of OM is M OM = 1 and slope of AB is M AB = 4+2 = 1. This 2+4 implies M OM M AM = 1, thus OM AB. (e) Let l 1 and l 2 be tangent at A and B respectivel. So OB l 1 and OB l 2. Slope of OA is M OA = 2 4 = 1/2, impling slope of l 1 is 2. Also slope of OB is M OB = 4 2 = 2, impling slope of l 2 is 1/2. Therefore l 1 is given b + 2 = 2( + 4) = 2 8, thus = 2 10 is the eqn of l 1. Also l 2 is given b 4 1/2( 2) = ( 1/2) + 1, thus the eqn of l 2 is = ( 1/2) + 5. (f) We solve = 2 10 and = ( 1/2) + 5 simultaneousl. Thus 30
31 2 10 = ( 1/2)+5, impling = 10. Hence = ( 1/2)( 10)+5 = 10. Thus C = ( 10, 10). 31
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