Yimin Math Centre. 4 Unit Math Homework for Year 12 (Worked Answers) 4.1 Further Geometric Properties of the Ellipse and Hyperbola...

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1 4 Uni Mah Homework for Year 12 (Worked Answers) Suden Name: Grade: Dae: Score: Table of conens 4 Topic 2 Conics (Par 4) Furher Geomeric Properies of he Ellipse and Hyperbola The Recangular Hyperbola Caresian Equaion of he Recangular Hyperbola Parameric Equaions of he Recangular Hyperbola Equaion of a Chord of he Hyperbola xy = c Equaions of Tangens and Normals o he hyperbola xy = c Equaion of he Chord of Conac of Tangen from an Exernal Poin Furher geomeric properies of he recangular hyperbola Locus Problem and he Recangular Hyperbola This ediion was prined on June 17, Camera ready copy was prepared wih he L A TEX2e ypeseing sysem. Copyrigh ( 4 Uni Mah Homework for Year 12

2 Year 12 Topic 4 Worked Answers Page 1 of 13 4 Topic 2 Conics (Par 4) 4.1 Furher Geomeric Properies of he Ellipse and Hyperbola Definiion: 1. The segmen of he angen o an ellipse or hyperbola beween he poin of he conac and he direcrix subends a righ angle a he corresponding focus. (P S ST ) 2. The angen a a poin P on he ellipse or hyperbola is equally inclines o he focal chords hrough P. ( SP T = S P T ) 3. The normal a a poin P on he ellipse or hyperbola is equally inclined o he focal chords hrough P. Copyrigh (

3 Year 12 Topic 4 Worked Answers Page 2 of 13 Example P (a sec θ, b an θ) lies on he hyperbola x2 y2 = 1. The angen a P mees he a 2 b 2 angens a he ends of he major axis a Q and R. Show ha QR subends a righ angle a eiher focus. Deduce ha if P is he poin 5, 4 x2 on he hyperbola 3 9 y2 = 1, wih foci S and S, hen Q, S, S, S are concyclic and find he equaion of he circle hrough hese poins. Soluion: Le he angen a P mee x = a, x = a in he Q, R respecively. Le QR mee he y-axis in C. Tangen PR has equaion x sec θ = 1. a b Hence Q has coordinaes [ b a, (sec θ 1)] and R has coordinaes [ a, b an θ b(sec θ 1 Gradien QS.gradien RS = b(secθ+1 y an θ a an θ(1 e) a an θ(1+e) = b2 (sec2 θ 1). a 2 (1 e 2 ) an 2 θ Then b 2 = a 2 (e 2 1) gradien QS gradien RS = 1, QS RS Similarly, replacing e by e, QS RS. an θ (sec θ + 1)] Hence QR subends angles of 90 a each of S and S, and Q, S, R, S are noncyclic, wih QR he diameer of he circle hrough he poins. The y-axis is he perpendicular bisecor of he chord SS, Hence he cenre of his circle is he poin C where he diameer QR mees he y-axis. If P (5, 4 x2 lies in he hyperbola y = 1, hen QR has equaion 5x 4y he y-axis in C ( 3 0, 4). Also b 2 = a 2 (e 2 1) gives e = 10, and S ahs coordinaes ( 10, 0). 9 Hence CS 2 = and he circle hrough Q, R, S, S has equaion x 2 + ( y = 1 and mees ) 2 = Copyrigh (

4 Year 12 Topic 4 Worked Answers Page 3 of The Recangular Hyperbola Caresian Equaion of he Recangular Hyperbola Definiion: A hyperbola is called recangular if is asympoes mee a righ angles. The asympoes of x2 y2 = 1 have equaions y = ± b x. a 2 b 2 a Hence asympoes perpendicular ( b ) ( b = 1), a = b. a a Then b 2 = a 2 (e 2 1), e = (1 + b2 ), e = 2. a 2 Hence a recangular hyperbola, wih major axis along he x-axis, has equaion x 2 y 2 = a 2, eccenriciy e = 2 and asympoes y = ±x Parameric Equaions of he Recangular Hyperbola Definiion: The sandard parameric from of he recangular hyperbola xy = c 2 is x = c and y = c, where he value of parameer depends on he posiion of P (c, c ) on he curve as in he figure shown below: Copyrigh (

5 Year 12 Topic 4 Worked Answers Page 4 of 13 Exercise For he recangular hyperbola xy = 4, find he parameric equaion. Soluion: For he hyperbola yxy = 4 we have c 2 = 4, c = 2. Hence he hyperbola equaion xy = 4 has parameric equaions x = c, y = c, x = 2, y = For he recangular hyperbola x = 4, y = 4, find he Caresian equaion. Soluion: The hyperbola x = 4, y = 4 has Caresian equaion xy = 4, xy = Find he parameric equaion of he recangular hyperbola xy = 25. Soluion: For he hyperbola xy = 25 we have c 2 = 25 c = 5 Hence he hyperbola xy = 25 has parameric equaions x = c, y = c, x = 5, y = Find he Caresian equaion of he recangular hyperbola x = 3, y = 3. Soluion: The hyperbola x = 3, y = 3 has Caresian equaion xy = 3 3, xy = 9. Copyrigh (

6 Year 12 Topic 4 Worked Answers Page 5 of Equaion of a Chord of he Hyperbola xy = c 2 Caresian form: Le P (x 1, y 1 ), Q(x 2, y 2 ) lie on he ( hyperbola) xy = c 2. The gradien of PQ is y 2 y 1 x 2 x 1 = 1 c 2 x 2 x 1 c2 = c2 x 2 x 2 1 x 1 x 2 chord PQ has equaion c 2 x + x 1 x 2 y = k, k is consan. k = c 2 (x 1 + x 2 ). The chord from P (x 1, y 1 ) o Q(x 2, y 2 ) on xy = c 2 has equaion c 2 x + x 1 x 2 y = c 2 (x 1 + x 2 ). Parameric form: ( ) ( ) Le P cp, c, Q cq, c lie on he hyperbola xy = c 2. p q The gradien of PQ is c( 1 q 1 p ) = 1 c(q p) pq chord PQ has equaion x + pqy = k, k is consan. c P (cp, ) lies on P Q k = c(p + q). p The chord from P (cp, c) o Q(cq, c ) on xy = c2 p q has equaion x + pqy = c(p + q). Example The poins P (cp, c p ) and Q(cq, c q ) lie on he recangula hyperbola xy = c2. The chord PQ subends a righ angle a he anoher poin R(cr, c ) on he hyperbola. Show he he r normal a R is parallel o PQ. Soluion: The gradien of PR is c( 1 p 1 r ) = 1, c(p r) pr he gradien of QR is c( 1 q 1 r ) c(q r) = 1 qr. Therefore P Q QR, gradien P R gradien QR = 1, 1 pqr 2 = 1 r 2 = 1 pq. The normal a he poin R(cr, c r ) has gradien of r2, he gradien of PQ is c( 1 p 1 q = 1. c(p q) pq Since r 2 = 1, hen gradien of he normal a R equal o gradien of PQ. pq Thus he normal a he poin R is parallel o he chord PR. Copyrigh (

7 Year 12 Topic 4 Worked Answers Page 6 of Equaions of Tangens and Normals o he hyperbola xy = c 2 Caresian from of equaion of angen: By implici differeniaion, xy = c 2, y + x dy dx = 0 dy dx = y x Gradien of angen a P (x 1, y 1 ) on he hyperbola is y 1 x 1 angen a P has equaion y 1 x + x 1 y = k, k is consan. By P lies on angen k = 2x 1 y 1 = 2c 2. The angen o xy = c 2 ap (x 1, y 1 ) has equaion xy 1 + yx 1 = 2c 2 Caresian from of equaion of normal: Also he normal a P (x 1, y 1 ) and equaion y y 1 = x 1 y 1 (x x 1 ). The normal o xy = c 2 a P (x 1, y 1 ) has equaion xx 1 yy 1 = x 2 1 y 2. Parameric form of equaion of angen: x = c, dx = c, and y = c, dy = c d d dy = 1 dx 2 Gradien of angen a P (cp, c ) on he hyperbola is 1, p p 2 angen a P has equaion x + p 2 y = k, k is consan. Bu P lies on angen k = ( 2cp. ) The angen o xy = c 2 a P cp, c has equaion x + p 2 y = 2cp. p Parameric form of equaion ( of normal: ) Also he normal a P cp, c has gradien P 2 and equaion y c = p P p2 (x cp). ( ) ( The normal o xy = c 2 a P cp, c has equaion px 1y = c p 2 ). 1 p p p 2 Example Find he equaion of he angen and he normal o he recangular hyperbola x = 2, y = 2 a he poin where = 4. 2, Soluion: For he hyperbola x = 2, y = 2, where we have c = 2. Hence he angen o he hyperbola x = 2, y = 2 a poin where = 4 has equaion x + 2 y = 2c, x + 16y = 16 and he normal has equaion x 1 y = c ( ) 32x 2y = 255. Copyrigh (

8 Year 12 Topic 4 Worked Answers Page 7 of 13 Exercise Find he equaions of he angen and he normal o he recangular hyperbola xy = 12 a he poin (-3, -4). Soluion: For he hyperbola xy = 12 we have c 2 = 12. Hence he angen o he hyperbola xy = 12 a he poin P ( 3, 4) has equaion xy 1 + yx 1 = 2c 2 4x + 3y = 24 and he normal has equaion xx 1 yy 1 = x 2 1 y1, 2 3x 4y = Find he equaions of he angen and he normal o he recangular hyperbola x = 3, y = 3 a he poin = 1. Soluion: For he hyperbola x = 3, y = 3, we have c = 3. Hence he angen o he hyperbola x = 3, y = 3 a he poin where = 1 has equaion x + 2 y = 2c, x + y = 6, and he normal has equaion x y = c ( ) x = y. 3. Find he equaion of chord of conac of angen from he poin (1, -2) ro he hyperbola xy = 6. Soluion: For he hyperbola xy = 6, c 2 = 6. Hence he chord of anac of angens from he poin T (x 0, y 0 ) = T (1, 2) o he hyperbola xy = 6 has equaion xy 0 + yx 0 = 2c 2 2x y = Find he equaion of he angen and he noram o he recangular hyperbola xy = 8 a he poin (4, 2). Soluion: For he hyperbola xy = 8 we have c 2 = 8. Hence he angen o he hyperbola xy = 8 a he poin P (x 1, y 1 ) = P (4, 2) has equaion xy 1 + yx 1 = 2c 2 x + 2y = 8 and he normal has equaion xx 1 yy 1 = x 2 1 y1 2 2x y = 6. Copyrigh (

9 Year 12 Topic 4 Worked Answers Page 8 of Equaion of he Chord of Conac of Tangen from an Exernal Poin Definiion: T lies on he angen PT and QT, x 0 y 1 +y 0 x 1 = 2x 2 and x 0 y 2 +y 0 x 2 = 2c 2. Hence P (x 1, y 1 ) and Q(x 2, y 2 ) boh saisfy x 0 y + y 0 x = 2c 2. his linear equaion mus be he equaion of PQ. The chord conac of angen from an exernal poin T (x 0, y 0 ) o he hyperbola xy = c has equaion xy 0 + yx 0 = 2c 2. Example Find he equaion of he chord of conac of angen from he poin (2, 1) o xy = 10. Soluion: For he hyperbola xy = 10, we have c 2 = 10. Hence he chord of conac of angens from he poin T (x 0, y 0 ) = T (2, 1) o he hyperbola xy = 10 has equaion xy 0 + yx 0 = 2c 2 x + 2y = 20. Exercise Find he equaion of he chord of conac of angens from he poin ( 1, 3) o he recangular hyperbola xy = 4. Hence find he coordinaed of heir poins of conac and he equaions of hese angens. Soluion: The chord of conac of angen from he poin (-1, -3) o he hyperbola xy = 4 has equaion xy 0 + yx 0 = 2c 2, 3x + y = 8 Le T (x 0, y 0 ) be a poin of conac. Then T lies on he chord 3x 0 + y 0 = 8, T lies on he hyperbola x 0 y 0 = 4. Hence x 0 ( 8 3x 0 ) = 4 3x x = 0 (3x + 2)(x + 2) = 0 x 0 = 2, y 3 0 = 8 3x 0 = 6 or x 0 = 2, y 0 = 8 3x 0 = 2. Equaion of angen a he poin T (x 0, y) 0 ) is xy 0 + yx 0 = 2c 2. he angen form he poin ( 1, 3) o he hyperbola xy = 4 are y = x 4, wih poin of conac P ( 2, 2) and y = 9x 12, wih poin of conac P ( 2, 6). 3 Copyrigh (

10 Year 12 Topic 4 Worked Answers Page 9 of Furher geomeric properies of he recangular hyperbola Definiion: 1. The area of he riangle bounded by a angen and he asympoes is a consan. Le he angen a P (c, c on xy = c2 mee he x-axes and y-axes in R and T respecively. The angen has equaion x + 2 y = 2c. A T, x = 0, y = 2c Area OT R = 1 2 2c 2c = 2c The lengh of he inercep cu off a angen by he asympoes is wice he disance from he poin of he inersecion of he asympoes o he poin of he conac of he angen. T R 2 = ( 2c2 ) + (2c)= 4c 2 ( ), and OP 2 (c) 2 + ( c )2 = c 2 ( ). T R 2 = 4OP 2, T R = 2OP. 3. The produc of he focal disance of a poin P on a recangular hyperbola is equal o he square of he disance from P o he poin of inersecion of he asympoes. Le xy = c 2 have foci S and S, and direcrices m and m. Le M, M be he fee of he perpendiculars from P (c, c o m, m respecively. Then P S P S = ep M ep M = 2P M P M (since e = 2), P S P S = 2 c+ c 2c 2 c+ c + 2c 2 = (c + c )2 ( 2c) 2 P S P S = (c) 2 + ( c )2 = OP 2. Copyrigh (

11 Year 12 Topic 4 Worked Answers Page 10 of 13 Example For he recangular hyperbola xy = 16, find (a) he eccenriciy; (b) he coordinaes if he foci; (c) he equaion s of he direcrices; (d) he equaions of he asympoes. Skech he hyperbola. Soluion: For he hyperbola xy = 16 we have c 2 = 16 c = 4. Hence he hyperbola xy = 16 has eccenriciy e = 2 Foci: S(c 2, c 2), S(4 2, 4 2) and S ( c 2, c 2), S ( 4 2, 4 2), Dircires: x + y = ±c 2, x + y = ±4 2, Asympoes: x = 0 and y = 0. Exercise The poin P (c, c ), where = -1 lies on he recangular hyperbola xy = c2. The angen and normal a P mee he line y = x a T and N respecively. Show ha OT ON = 4c 2. Soluion: The angen o he hyperbola xy = c 2 a he poin P (c, c ) has equaion x + 2 y = 2c. As y = x he poin T has coordinaes ( 2c 1+ 2, 2c 1+ 2 ). The normal o he hyperbola xy = c 2 a he poin P (c, c ) has equaion x y = c(2 1 ). 2 Therefore he poin N has coordinaes ( c(2 +1), c(2 +1) ). OT = 2c 2, ON = c(2 +1) 2, OT ON = 4c Copyrigh (

12 Year 12 Topic 4 Worked Answers Page 11 of Locus Problem and he Recangular Hyperbola Example P (c, c) is a poin on he recangular hyperbola xy = c2. N is he foo of he perpendicular from P o he x-axis and M is he midpoin of P N. Find he Caresian equaion of he locus of M as he posiion of P varies. and describe his locus geomerically. Soluion: M has coordinaes (c, c 2 ). Hence he locus of M has parameric equaions x = c, and y = c 2 and Caresian equaion xy = 1 2 c2. M races a recangular hyperbola as he posiion of P varies. Example P (3p, 3) and Q(3q, 3 ) are poin on differen branches of he hyperbola xy = 9. p q Find he coordinaes of he poin of inersecion T of he angens a P and Q. Find he locus of T if he posiions of P and Q very so ha he chord P Q passes hrough (0, 4). { x + p 2 y = 6p (1) Soluion: T lies on he angen a P and Q. A T, x + q 2 y = 6q (2) (1) (2) (p 2 q 2 )y = 6(p q), p q y = ( ) 6 p+q Then (1) x = 6pq, T has coordinaes 6pq, 6. p+q p+q p+q If (0, 4) lies on chord PQ, wih equaion ( ) x + pqy = 3(p + q), 9 hen 4pq = 3(p + q) and T is T, pq P, Q on differen branches pq < 0, locus of T is x = 9, y < 0 2 Copyrigh (

13 Year 12 Topic 4 Worked Answers Page 12 of 13 Exercise P and Q are variable poins on he recangular hyperbola xy = 9. The angens a P and Q mee a R. If P Q passes hrough he poin (6, 2). find he equaion of he locus of R. Soluion: Le R has coordinaes (x 0, y 0 ). PQ is he chord of conac of angens from R o hyperbola xy = 9. Hence PQ has equaion xx 0 + yy 0 = 18. The poin (6, 2) lies on P Q. Therefore x 0 + 3y 0 = 9. Thus he locus of R has equaion x + 3y = The poin P (c, c ) lies on he recangular hyperbola xy = c2. The angen a P cus he x-axis a X and he y-axis a Y. Show ha he area of Y OX is independen of. Soluion: The angen o he hyperbola xy = c 2 a he poin P ( c, c has equaion x + 2 y = 2c. Hence he poin X has coordinaes (2c, 0) and he poin Y has coordinaes ( 0, 2c The area of Y OX = 1 OX OY = 1 2c 2c = 2c Thus he area of Y OX is independen of. 3. On he recangular hyperbola xy = c 2 here are variable poins P and Q. The angens a P and Q mee a R. Find he equaion of he locus of R if PQ passes hrough he poin (a, 0). Soluion: Le R has coordinaes (x 0, y 0 ). PQ is he chord of conac of angens from R o he hyperbola xy = c 2. Hence PQ has equaion xy 0 + yx 0 = 2c 2. Then (a, 0) lies on PQ. herefore xy 0 + yx 0 = 2c 2, ay)o = ac 2. Thus he locus of R has equaion y = 2c2 a. ) ). 4. The poin P (c, c ) lies on he recangular hyperbola xy = c2. The angen a P cus he x-axis a X and he y-axis a Y. Show ha PX = PY. Soluion: The angen o he hyperbola xy = c 2 a he poin P (c, c ) has equaion x + 2 y = ac. Hence he poin x has coordinaes xy = c 2 and he poin Y has coordinaes (0, 2c ). P X 2 = (c 2c) 2 + ( c )2 = c 2 ( ) and P X 2 = (c) 2 + ( c 2c )2 = c 2 ( ). Therefore P X = P Y. Copyrigh (

14 Year 12 Topic 4 Worked Answers Page 13 of 13 Exercise The poin P (c, c ) lies on he recangular hyperbola xy = c2. Show ha he normal a P cus he hyperbola again a he poin Q wih coordinaes ( c 3, c 3 ). Hence find he coordinaes of he poin R where he normal a Q cus he hyperbola again. Soluion: The normal o he hyperbola xy = c 2 a poin P (c, c ) has equaion x y = c ( ) ( ) The poin Q cq, c lies on he normal. q Hence cq c = q c(2 1 ). ( 2 ) Therefore (q 2 ) = 0. 3 q Since Q P, hen q. Thus q = 1 and Q has coordinaes ( c, c 3). 3 3 Similarly he normal a Q cus he hyperbola again a R(cr, c) wih r = 1 = 9. r q 3 So R has coordinaes ( ) c 9 c, The poin P (c, c ) lies on he recangular hyperbola xy = c2. The normal a P mees he x-axis a A and he angen a P mees he y-axis a B. M is he midpoin of AB. Find he equaion of he locus of M as P moves on he hyperbola. Soluion: The normal o he hyperbola xy = c 2 a he poin P (c, c ) has he equaion x y = c ( ). The normal a P mees he x-axis a A( c (2 1 2 ), 0). The angen o he hyperbola xy = c 2 a he P (c, c ) has equaion x + 2 y = 2c. Hence he angen mees he y-axis a B(0, 2c). If M(x, y) is he midpoin of AB, hen x = c 2 (2 1 ) and y = c. 2 Thus = c and consequenly, x = y ( c2 y2 ). y 2 y 2 c 2 Therefore he locus of N has equaion 2c 2 xy = c 4 y 4. Copyrigh (