Chapter 24.3 PDF - Conic Sections, Hyperbola
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This document provides a comprehensive introduction to hyperbolas, including their definitions, equations, and properties. It details various aspects of hyperbolas, such as standard equations, conjugate hyperbolas, special forms, auxiliary circles, and applications in ballistics. Examples and formulas are used to illustrate different concepts.
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Complex Numbers 223 Chapter E3 60 5.3 CONTENTS 5.3.2 Standard equation of the hyperbola 5.3.3 Conjugate hyperbola 5.3.4 Special form of hyperbola Y ID L1 X Auxiliary circle of hyperbola 5.3.6 Position of a point with respect to a hyperbola U 5.3.5 Intersection of a line and a hyperbola 5.3.8 Equati...
Complex Numbers 223 Chapter E3 60 5.3 CONTENTS 5.3.2 Standard equation of the hyperbola 5.3.3 Conjugate hyperbola 5.3.4 Special form of hyperbola Y ID L1 X Auxiliary circle of hyperbola 5.3.6 Position of a point with respect to a hyperbola U 5.3.5 Intersection of a line and a hyperbola 5.3.8 Equations of tangent in different forms 5.3.9 Equation of pair of tangents 5.3.10 Equations of normal in different forms 5.3.11 Equation of chord of contact of tangents drawn from a point to a hyperbola 5.3.12 Equation of the chord of the hyperbola whose mid point (x1, y1) is given 5.3.13 Equations of the chord joining two points on the hyperbola U Pole and Polar ST 5.3.14 D YG 5.3.7 5.3.15 Diameter of the hyperbola 5.3.16 Subtangent and Subnormal of the hyperbola 5.3.17 Reflection property of the hyperbola 5.3.18 Asymptotes of hyperbola 5.3.19 Rectangular or equilateral hyperbola 5.3.20 Intersection of a circle and rectangular hyperbola Assignment (Basic and Advance Level) Answer Sheet of Assignment (–ae,0)S L1 M M L P Q (x,y) N axis (–a,0) A Z C x=–a/e Y CZ A (a,0) Directrix Definition Directrix 5.3.1 S (ae,0) X L Q x=a/e Apollonius writes Conics in which he introduces the terms "parabola", " ellipse" and "hyperbola". De Beaune writes Notes brieves which contains the many results on "Cartesian geometry", in particular giving the now familiar equations for hyperbolas, parabolas and ellipses. The hyperbola is also useful for describing the path of an alpha particle in the electric field of the nucleus of an atom. Hyperbola has its application in the field of Ballistics. Suppose a gun is fired. If the sound reaches two listening posts, situated at two foci of the hyperbola at different times, from the time difference, the distance between the two listening posts (two foci) can be calculated. 224 Conic Section : Hyperbola 60 5.3.1 Definition. A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point in the same plane to its distance from a fixed line is always constant Z which is always greater than unity. P M A hyperbola is the particular case of the conic ax 2 2hxy by 2 2 gx 2 fy c 0 Directrix E3 Fixed point is called focus, fixed straight line is called directrix and the constant ratio is called eccentricity of the hyperbola. Eccentricity is denoted by e and e > 1. S(Focus) Z ID When , abc 2 fgh af 2 bg 2 ch 2 0 i.e., 0 and h 2 ab. Let S (h, k ) is the focus, directrix is the line ax by c 0 and the eccentricity is e. Let P( x 1 , y1 ) be a point which moves such that SP e.PM (x 1 h)2 (y 1 k )2 e. ax1 by 1 c U a2 b 2 D YG (a 2 b 2 )[(x 1 h)2 (y1 k )2 ] e 2 (ax1 by1 c)2 Hence, locus of (x 1 , y 1 ) is given by (a 2 b 2 )[(x h)2 (y k )2 ] e 2 (ax by c)2 Which is a second degree equation to represent a hyperbola (e > 1). The equation of the conic with focus at (1, – 1), directrix along x y 1 0 and with eccentricity Example: 1 (a) x y 1 2 (b) xy 1 [EAMCET 1994; DCE 1998] (c) 2 xy 4 x 4 y 1 0 (d) 2 xy 4 x 4 y 1 0 U 2 2 is Solution: (c) Here, focus (S) = (1, –1), eccentricity (e)= 2 ST From definition , SP e PM (x 1) 2 (y 1) 2 2.( x y 1) 12 12 (x 1)2 (y 1)2 = (x y 1)2 2 xy 4 x 4 y 1 0 , which is the required equation of conic (Rectangular hyperbola) Example: 2 The centre of the hyperbola 9 x 2 36 x 16 y 2 96 y 252 0 is (a) (2, 3) Solution: (a) (b) (– 2, – 3) (c) (–2, 3) [Karnataka CET 1993] (d) (2, – 3) Here a 9, b 16 , h 0 , g 18 , f 48 , c 252 (0) (48 ) (16 ) (18 ) (18 ) (0) (9) (48 ) hf bg gh af = (2, 3) , , Centre of hyperbola = = 2 2 (9) (16 ) 0 (9)(16 ) 0 ab h ab h 5.3.2 Standard equation of the Hyperbola. Let S be the focus, ZM be the directrix and e be the eccentricity of the hyperbola, then by definition, Conic Section : Hyperbola 225 SP e (SP )2 e 2 (PM)2 PM x2 y2 1 a 2 a 2 (e 2 1) P L Q N (x,y) X (–ae,0)S x2 y2 1 , where b 2 a 2 (e 2 1) a2 b 2 L1 (–a,0) A Z C Z A (a,0) C x=–a/e Y axis S (ae,0) L X Q x=a/e 60 2 M M Directrix a (x a.e ) 2 (y 0) 2 e 2 x e Y L1 Directrix This is the standard equation of the hyperbola. x2 y2 1 a2 b 2 (1) Centre : All chords passing through C are bisected at C. Here C (0,0 ) E3 Some terms related to hyperbola : Let the equation of hyperbola is x2 y2 1 a2 b 2 D YG (4) Eccentricity : For the hyperbola U ID (2) Vertex: The point A and A where the curve meets the line joining the foci S and S are called vertices of hyperbola. The co-ordinates of A and A are (a, 0) and (– a, 0) respectively. (3) Transverse and conjugate axes : The straight line joining the vertices A and A is called transverse axis of the hyperbola. The straight line perpendicular to the transverse axis and passing through the centre is called conjugate axis. Here, transverse axis = A A 2a Conjugate axis = B B 2b 2 2 2b Conjugate axis We have b a (e 1) , e 1 1 2a Transverse axis (5) Double ordinates : If Q be a point on the hyperbola, QN perpendicular to the axis of the hyperbola and produced to meet the curve again at Q . Then Q Q is called a double ordinate at Q. 2 2 2 U b b If abscissa of Q is h, then co-ordinates of Q and Q are h, h 2 a 2 and h, h2 a2 a a (6) Latus-rectum : The chord of the hyperbola which passes through the focus and is transverse axis is called latus-rectum. 2b 2 b2 2a(e 2 1) and end points of latus-rectum L ae, a a ST Length of latus-rectum L L L1 L1 respectively. perpendicular to its b2 ; L ae, a ; b2 b2 respectively. ; L1 ae, L1 ae , a a (7) Foci and directrices: The points S (ae , 0 ) and S (ae , 0 ) are the foci of the hyperbola and ZM and Z M are a a two directrices of the hyperbola and their equations are x and x respectively. e e Distance between foci S S 2ae and distance between directrices ZZ 2a / e. (8) Focal chord : A chord of the hyperbola passing through its focus is called a focal chord. (9) Focal distance : The difference of any point on the hyperbola from the focus is called the focal distance of the point. a a From the figure, SP ePM e x 1 ex 1 a , S P eP M e x 1 = ex 1 a e e 226 Conic Section : Hyperbola The difference of the focal distance of a point on the hyperbola is constant and is equal to the length of transverse | S P SP | 2a A A Transverse axis Example: 3 The eccentricity of the hyperbola which passes through (3, 0) and (3 2 , 2) is (a) Solution: (b) 13 3 (b) (13) [UPSEAT 2000] 13 4 (c) Let equation of hyperbola is x 2 / a 2 y 2 / b 2 1. Point (3, 0) lies on hyperbola (d) None of these 60 axis. 3 ( 2 ) 2 (2)2 9 (3)2 0 2 1 2 1 or 2 1 or a 2 9 and point (3 2 , 2) also lies on hyperbola. So, 2 a2 b a a b 4 4 18 4 4 2 2 1 or 2 2 1 or 2 1 2 or 2 1 or b 4 Put a 9 we get, 9 b2 b b b We know that b 2 a 2 (e 2 1). Putting values of a 2 and b 2 4 9(e 2 1) or e 2 1 13 4 4 or e 2 1 or e (1 4 / 9) or e (13 ) / 9 . 9 9 3 The foci of the hyperbola 9 x 2 16 y 2 144 are (a) ( 4, 0) (b) (0, 4 ) 2 Solution: (c) The equation of hyperbola is 2 x y 1. 16 9 If the foci of the ellipse (d) (0, 5) 5 5 . Hence foci are ( ae, 0) = 4. , 0 i.e., ( 5, 0) 4 4 x2 x2 y2 y2 1 coincide, then the value of b 2 is 2 = 1 and the hyperbola 144 81 25 16 b (b) 5 D YG (a) 1 Solution: (c) ( 5, 0) U Now, b 2 a 2 (e 2 1) 9 16 (e 2 1) e Example: 5 (c) (c) 7 [MNR 1992; UPSEAT 2001; AIEEE 2003] (d) 9 x2 y2 1 For hyperbola, 144 81 25 81 B2 81 225 5 , e1 1 2 1 25 144 144 4 A 12 5 . , 0 ( 3, 0 ). Therefore foci of ellipse i.e., (4 e, 0) (3, 0) Therefore foci = (ae 1 , 0 ) 5 4 A 144 ,B 25 x2 y2 1 such that CPQ is an equilateral triangle, C being the centre of the a2 b 2 hyperbola. Then the eccentricity e of the hyperbola satisfies [EAMCET 1999] If PQ is a double ordinate of hyperbola (a) 1 e 2 / 3 (b) e 2 / 3 (c) e 3 / 2 (d) e 2 / 3 Let P(a sec , b tan ) ; Q(a sec , b tan ) be end points of double ordinates and C(0, 0) is the centre of the hyperbola ST Solution: (d) (For ellipse a 4 ) 9 3 , Hence b 2 16 1 7. 16 4 U e Example: 6 [MP PET 2001] ID Example: 4 E3 So, Now PQ 2b tan ; CQ CP a 2 sec 2 b 2 tan 2 P (a sec , b tan ) Since CQ CP PQ , 4 b 2 tan 2 a 2 sec 2 b 2 tan 2 3b 2 tan 2 a 2 sec 2 3b 2 sin 2 a 2 3a 2 (e 2 1) sin 2 a 2 3 (e 2 1) sin 2 1 1 ( sin 2 1) sin 2 1 3(e 1) 2 1 4 1 2 e2 e 3 e2 1 3 3 e 1 3 C(0, 0) 2 5.3.3 Conjugate Hyperbola. Q(a sec ,–b tan) Conic Section : Hyperbola 227 The hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of a given hyperbola is called conjugate hyperbola of the given hyperbola. Y x y 1 a2 b 2 (0, 0) 2a 2b ( ae, 0) x2 y2 x y 2 2 1 or 2 2 1 a b a b x a / e y b / e Fundamentals Centre Length of transverse axis Length of conjugate axis Foci Equation of directrices Eccentricity 2 a2 b 2 e 2 a Length of latus rectum 2 2 (0, 0) 2b 2a (0, be ) Z B(0,b) y= b/e X a2 b 2 e b2 Focal radii SP ex 1 a & S P ex 1 a SP ey1 b & S P ey 1 b 2a 2b U x a, x a y b, y b y 0 x 0 D YG Equation of the transverse axis Note : If e and e S (0,–b) Y y 0 x 0 Equation of the conjugate axis X y= –b/e B (0,–b) ID Parametric co-ordinates 2a 2 b (b sec , a tan ), 0 2 Tangents at the vertices C Z 2b 2 a (a sec , b tan ) , 0 2 Difference of focal radii (S P SP ) S (0,–b) 60 2 E3 Hyperbola are the eccentricities of a hyperbola and its conjugate, then 1 1 1. e 2 e 2 The foci of a hyperbola and its conjugate are concyclic. The eccentricity of the conjugate hyperbola of the hyperbola x 2 3 y 2 1 , is Example: 7 (b) U (a) 2 Solution: (a) The given hyperbola is (c) 4 (d) 3 4 3 x2 y2 1 1. Here a 2 1 and b 2 3 1 1/3 Since b 2 a 2 (e 2 1) ST 2 [MP PET 1999] 2 1 4 1 (e 2 1) e 2 e 3 3 3 If e is the eccentricity of the conjugate hyperbola, then 1 e2 1 e 2 =1 1 e 2 1 1 e2 1 3 1 e 2. 4 4 5.3.4 Special form of Hyperbola. If the centre of hyperbola is (h, k) and axes are parallel to the co-ordinate axes, then its equation is (x h)2 (y k )2 1. By shifting the origin at (h, k) without rotating the co-ordinate axes, the above equation reduces a2 b2 to X2 Y 2 1 , where x X h, y Y k. a2 b 2 Example: 8 The equation of the hyperbola whose foci are (6, 4) and (– 4, 4) and eccentricity 2 is given by (a) 12 x 4 y 24 x 32 y 127 0 2 2 [MP PET 1993] (b) 12 x 4 y 24 x 32 y 127 0 2 2 228 Conic Section : Hyperbola (c) Solution: (a) (d) 12 x 2 4 y 2 24 x 32 y 127 0 12 x 2 4 y 2 24 x 32 y 127 0 Foci are (6, 4) and (– 4, 4) and e 2. 6 4 4 4 , Centre is (1, 4 ) 2 2 So, ae 1 6 ae 5 a 3 (x 1) 2 (y 4 ) 2 1 or 12 x 2 4 y 2 24 x 32 y 127 0 25 / 4 (75 / 4 ) 60 Hence, the required equation is 5 5 and b 2 2 The equations of the directrices of the conic x 2 2 x y 2 5 0 are (a) Solution: (c) x 1 (b) (x 1) 2 y 2 1 5 0 Equation of directrices of y2 b 2 y 2 (c) (x 1) y 1 4 4 2 x2 a 2 2 b e 1 are y 2 (d) x 3 y 2. ID Here b 2 , e 1 1 2. Hence, y y 2 E3 Example: 9 2 5.3.5 Auxiliary circle of Hyperbola. 2 2 Y Q 90o X (– a,0)A (0,0)C D YG x y auxiliary circle of the hyperbola 2 1 a b U x2 y2 1 be the hyperbola with centre C and transverse axis A A. a2 b 2 Therefore circle drawn with centre C and segment A A as a diameter is called Let Equation of the auxiliary circle is x 2 y 2 a 2 (x,y) P N A(a,0) X Y Let QCN Here P and Q are the corresponding points on the hyperbola and the auxiliary circle (0 2 ) (1) Parametric equations of hyperbola : The equations x a sec and y b tan are known as the x2 y2 1. This ( a sec , b tan ) lies on the hyperbola for all values of . a2 b 2 U parametric equations of the hyperbola ST Position of points Q on auxiliary circle and the corresponding point P which describes the hyperbola and 0 2 varies from 0 to 2 to 2 3 to 2 3 to 2 2 Q(a cos φ, a sin φ) P(a sec φ, b tan φ) I I II III III II IV IV Conic Section : Hyperbola 229 Note : The equations x a cosh and y b sin h are also known as the parametric equations of the hyperbola and the co-ordinates of any point on the hyperbola e e 2 and sin h e e . 2 The distance between the directrices of the hyperbola x 8 sec , y 8 tan is (a) 16 2 (b) (c) 2 Equation of hyperbola is x 8 sec , y 8 tan sec 2 tan 2 1 x y sec , tan 8 8 x2 y2 1 82 82 Here a 8, b 8. Now e 1 b2 82 1 2 2 2 a 8 Distance between directrices = (d) 8 2 28 2a 8 2. = e 2 4, 2 ID Solution: (c) [Karnataka CET 2003] E3 Example: 10 60 (a cos h , b sin h ), where cos h x2 y2 1 are expressible as a2 b 2 Let the hyperbola be x2 y2 1. a2 b 2 U 5.3.6 Position of a point with respect to a Hyperbola. according as x 12 2 a D YG Then P(x 1 , y 1 ) will lie inside, on or outside the hyperbola y12 2 b X A C A Y 2 x y 1 through (4, 1) is 4 3 (a) 1 (b) 2 (c) 0 (d) 3 16 1 1 0 ; Number of tangents through (4, 1) is 0. Since the point (4, 1) lies inside the hyperbola 4 3 The number of tangents to the hyperbola U Solution: (c) x 2 y2 1 a2 b 2 P (outside) P (on) P(inside) X 1 is positive, zero or negative. 2 Example: 11 Y [AMU 1998] 5.3.7 Intersection of a Line and a Hyperbola. ST The straight line y mx c will cut the hyperbola x 2 y2 1 in two points may be real, coincident or imaginary a2 b 2 according as c 2 , , a 2 m 2 b 2. Condition of tangency : If straight line y mx c touches the hyperbola x 2 y2 1 , then c 2 a 2 m 2 b 2. a2 b 2 5.3.8 Equations of Tangent in Different forms. (1) Point form : The equation of the tangent to the hyperbola xx yy x2 y2 2 1 at (x 1 , y 1 ) is 21 21 1. 2 a b a b (2) Parametric form : The equation of tangent to the hyperbola x y sec tan 1 a b x2 y2 1 at (a sec , b tan ) is a2 b 2 230 Conic Section : Hyperbola (3) Slope form : The equations of tangents of slope m to the hyperbola a 2m b2 and the co-ordinates of points of contacts are , a 2m 2 b 2 a 2m 2 b 2 touches the hyperbola . x2 y2 1 , then a 2 l 2 b 2 m 2 n 2. a2 b 2 If the straight line x cos y sin p touches the hyperbola x2 y2 1, a2 b 2 Important Tips 2 For hyperbola y x x y 1 , the equation of common tangent is y x a 2 b 2 , points of contacts are 1 and a2 b 2 a2 b 2 and length of common tangent is 2 2 a b a2 b 2 ; b2 If the line y mx a 2m 2 b 2 touches the hyperbola Example: 12 (a 2 b 2 ) a2 b 2. x 2 y2 b 1 at the point (a sec , b tan ) , then sin 1 . a2 b 2 am The value of m for which y mx 6 is a tangent to the hyperbola (a) Solution: (a) 2. ID a2 U 2 17 20 (b) 20 17 D YG 2 E3 then a 2 cos 2 b 2 sin 2 p 2 Two tangents can be drawn from an outside point to a hyperbola. 2 60 Note : If the straight line lx my n 0 x2 y2 1 are y mx a 2 m 2 b 2 a2 b 2 (c) x2 y2 1 , is 100 49 3 20 [Karnataka CET 1993] (d) 20 3 For condition of tangency, c 2 a 2 m 2 b 2. Here c 6 , a 10, b 7 Then, (6)2 (10 )2. m 2 (7)2 36 100 m 2 49 100 m 2 85 m 2 17 20 x 2 y2 1 which pass through the point (6, 2), then 25 16 11 20 48 24 (a) m1 m 2 (b) m 1m 2 (c) m1 m 2 (d) m 1m 2 11 20 11 11 The line through (6, 2) is y 2 m (x 6) y mx 2 6 m If m1 and m 2 are the slopes of the tangents to the hyperbola U Example: 13 17 m 20 Solution: (a, b) Now, from condition of tangency (2 6m)2 25 m 2 16 ST 36 m 2 4 24 m 25 m 2 16 0 11m 2 24 m 20 0 20 24 Obviously, its roots are m1 and m 2 , therefore m1 m 2 and m1m 2 11 11 Example: 14 Solution: (a) The points of contact of the line y x 1 with 3 x 2 4 y 2 12 is (a) (4, 3) (b) (3, 4) (c) (4, – 3) The equation of line and hyperbola are y x 1.....(i) and 3 x 2 4 y 2 12.....(ii) From (i) and (ii), we get 3 x 2 4(x 1)2 12 3 x 2 4(x 2 2 x 1) 12 or x 2 8 x 16 0 x 4 From (i), y 3 so points of contact is (4, 3) a 2m b2 , Trick : Points of contact are a 2m 2 b 2 a 2m 2 b 2 . Here a 2 4 , b 2 3 and m 1. So the required points of contact is (4, 3). [BIT Ranchi 1996] (d) None of these Conic Section : Hyperbola 231 x2 y2 1, N is the foot of the perpendicular from P on the transverse axis. The tangent to the a2 b 2 hyperbola at P meets the transverse axis at T. If O is the centre of the hyperbola, then OT.ON is equal to P is a point on the hyperbola e2 (a) a2 This meets x-axis at T , 0 ; x1 Solution: (c) b2 (d) b2 a2 Let P(x 1 , y1 ) be a point on the hyperbola. Then the co-ordinates of N are (x 1 , 0). The equation of the tangent at (x 1 , y1 ) is Example: 16 (c) xx 1 a2 yy 1 b2 OT.ON P (x1,y1 ) Y 1 60 Solution: (b) (b) a 2 X 2 a x1 a 2 x1 O T N (x1,0) Y x2 y2 If the tangent at the point (2 sec , 3 tan ) on the hyperbola 1 is parallel to 3 x y 4 0 , then the value of is 4 9 (a) 45 o (b) 60 o Here x 2 sec and y 3 tan (c) Gradient of tangent 30 o (d) 75 o ID Differentiating w.r.t. dx dy 2 sec tan and 3 sec 2 d d dy dy / d 3 sec 2 ; dx dx / d 2 sec tan dy 3 cosec dx 2.....(i) But tangent is parallel to 3 x y 4 0 ; Gradient m 3 y2 x2 x2 y2 1 and 1 are 9 16 9 16 (c) 1, 2 D YG Solution: (b) cosec 2 , 30 o The slopes of the common tangents to the hyperbola (a) – 2, 2......(ii) U 3 From (i) and (ii), cosec 3 2 Example: 17 X E3 Example: 15 (b) – 1, 1 y2 x2.....(i) and 1 9 16 x2 y2 Given hyperbola are 1 9 16 Any tangent to (i) having slope m is y mx 9 m 2 16 [Roorkee 1997] (d) 2, 1.....(ii).....(iii) Putting in (ii), we get, 16[mx 9m 2 16 ]2 9 x 2 144 (16 m 2 9)x 2 32 m ( 9 m 2 16 ) x 144 m 2 256 144 0 (16 m 2 9)x 2 32 m ( 9 m 2 16 ) x (144 m 2 400 ) 0 U.....(iv) If (iii) is a tangent to (ii), then the roots of (iv) are real and equal. ST Discriminant = 0; 32 32 m 2 (9m 2 16 ) = 4 (16 m 2 9)(144 m 2 400 ) = 64 (16 m 2 9)(9 m 2 25 ) 16 m 2 (9m 2 16 ) (16 m 2 9)(9m 2 25 ) 144 m 4 256 m 2 144 m 4 481 m 2 225 225 m 2 225 m 2 1 m 1 5.3.9 Equation of Pair of Tangents. If P(x 1 , y 1 ) be any point outside the hyperbola x2 y2 1 then a pair of tangents PQ, PR can be drawn to it from P. a2 b 2 Y The equation of pair of tangents PQ and PR is SS 1 T 2 where, S 2 2 x 12 2 y12 2 xx yy x y 2 1 , S1 1 , T 21 21 1 2 a b a b a b T (h,k) X Q X A C P A (x1,y1) R Y 232 Conic Section : Hyperbola Director circle : The director circle is the locus of points from which perpendicular tangents are drawn to the given x2 y2 2 1 is x 2 y 2 a 2 b 2 2 a b hyperbola. The equation of the director circle of the hyperbola Y P (h, k) 90o A A C X Y Solution: (a) The locus of the point of intersection of tangents to the hyperbola 4 x 2 9 y 2 36 which meet at a constant angle / 4 , is (a) (x 2 y 2 5)2 4(9 y 2 4 x 2 36 ) (b) (x 2 y 2 5) 4(9 y 2 4 x 2 36 ) (c) 4 (x 2 y 2 5)2 (9 y 2 4 x 2 36 ) (d) None of these E3 Example: 18 60 X Let the point of intersection of tangents be P(x 1 , y1 ). Then the equation of pair of tangents from P(x 1 , y1 ) to the given hyperbola is (4 x 2 9 y 2 36 ) (4 x 12 9 y12 36 ) [4 x 1 x 9 y1 y 36 ]2 ID......(i) From SS 1 T 2 or x 2 (y12 4 ) 2 x 1 y1 xy y 2 (x 12 9) ..... 0.....(ii) Since angle between the tangents is / 4. 2 [x 12 y12 (y12 4 ) (x 12 9)] y12 4 x 12 9. Hence locus of P(x 1 , y1 ) is (x 2 y 2 5)2 4(9 y 2 4 x 2 36 ). U tan( / 4 ) D YG 5.3.10 Equations of Normal in Different forms. (1) Point form : The equation of normal to the hyperbola a2 x b 2y x2 y2 a2 b 2. ( x , y ) at is 1 1 1 x1 y1 a2 b 2 (2) Parametric form: The equation of normal at (a sec , b tan ) to the hyperbola x2 y2 1 is a2 b 2 ax cos by cot = a 2 b 2 Y P(x1,y1) (3) Slope form: The equation of the normal to the hyperbola ST U m (a 2 b 2 ) x2 y2 in terms of the slope m of the normal is 1 y mx a2 b 2 a 2 b 2m 2 m(a 2 b 2 ) a 2 m 2b 2 or c 2 C A Y (4) Condition for normality : If y mx c is the normal of then c X x2 y2 1 a2 b 2 m 2 (a 2 b 2 )2 , which is condition of normality. (a 2 m 2 b 2 ) a2 (5) Points of contact : Co-ordinates of points of contact are , a 2 b 2m 2 2 2 2 a b m mb 2 A Tangent Normal X Conic Section : Hyperbola 233 Note : If the line lx my n 0 will be normal to the hyperbola x2 y2 1 ,then a2 b 2 a2 b 2 (a 2 b 2 )2. l2 m 2 n2 Important Tip In general, four normals can be drawn to a hyperbola from any point and if , , , be the eccentric angles of these four co-normal points, then is an odd multiple of . If , , are the eccentric angles of three points on the hyperbola. If the normal at P meets the transverse axis in G, then SG e. SP. Also the tangent and normal bisect the angle between the focal distances of P. x2 y2 The feet of the normals to 2 2 1 from (h, k ) lie on a 2 y(x h) b 2 x (y k ) 0. a b 60 x2 y2 1 , the normals at which are concurrent, then, a2 b 2 Example: 19 The equation of the normal to the hyperbola 3 x 2 y 25 (a) 2 Solution: (d) From (b) x2 y2 1 at the point (8, 3 3 ) is 16 9 x y 25 2 a x b y a2 b 2 x1 y1 y 2 x 25 (d) x2 y2 1 meets transverse axis at G, then AG. AG a2 b 2 (Where A and A are the vertices of the hyperbola) (a) a 2 (e 4 sec 2 1) (b) (a 2 e 4 sec 2 1) (c) a 2 (1 e 4 sec 2 ) (d) None of these Example: 20 If the normal at ' ' on the hyperbola Solution: (a) The equation of normal at (a sec , b tan ) to the given hyperbola is ax cos by cot (a 2 b 2 ) a2 b 2 This meets the transverse axis i.e., x-axis at G. So the co-ordinates of G are a vertices A and A are A(a, 0) and A(a, 0) respectively. a2 b 2 sec = a sec , 0 and the co-ordinates of the 2 sec 2 a 2 = (ae 2 )2 sec 2 a 2 = a 2 (e 4 sec 2 1) The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axis at L and M respectively, then the locus of the middle point of LM is a hyperbola whose eccentricity is e e e (a) (b) (c) (d) None of these 2 2 4 2 a e 1 e 1 e 1 a2 b 2 AG. AG a a U 2 2 sec a a b a ST Example: 21 2 x 3 y 25 16 x 9y 16 9 i.e., 2 x 3 y 25. 8 3 3 D YG (c) Solution: (a) [MP PET 1996] U Here a 2 16 , b 2 9 and (x1 , y1 ) (8, 3 3 ) ID E3 sin( ) sin( ) sin( ) 0 The equation of the normal at P(a sec , b tan ) to the hyperbola is ax cos by cot a 2 b 2 a 2 e 2 a 2 e 2 tan It meets the transverse and conjugate axes at L and M, then L(ae 2 sec , 0) ; M 0, b Let the middle point of LM is (, ) ; then and 2b a 2 e 2 tan tan 2 2 2b a e ae 2 sec 2 sec 2 2 ae.....(i)......(ii) 234 Conic Section : Hyperbola 1 sec 2 tan 2 ; 1 4 2 4 b 2 2 x2 4 4 , Locus of (, ) is 2 4 a2e 4 a e a e 4 It is a hyperbola, let its eccentricity e 1 a2e 4 a4 e 4 4 4b 2 a2e 4 4 1 a2 b 2 y2 a4 e 4 4b2 a2 b 2 b 2 1 a2e 2 a (e 2 1) 2 ; e e1 e 1. 2 60 5.3.11 Equation of Chord of Contact of Tangents drawn from a Point to a Hyperbola. Let PQ and PR be tangents to the hyperbola x2 y2 1 drawn from any external point P ( x 1 , y 1 ). a2 b 2 E3 Then equation of chord of contact QR is xx 1 yy 1 or 2 1 a2 b or T 0 (At x 1 , y 1 ) X Y Q ID A C P (x1,y1) X A R Y 5.3.12 Equation of the Chord of the Hyperbola whose Mid point (x1, y1) is given h i.e., T S 1 Note a2 yy 1 b2 1 = x 12 a2 a2 y 12 y2 b2 1 , bisected at the given b2 Y Q (x2,y2) 1 X D YG point (x 1 , y 1 ) is xx 1 x2 U Equation of the chord of the hyperbola P X A (x1,y1) R (x3,y3) C Y : The length of chord cut off by hyperbola x2 y2 1 from the line y mx c a2 b 2 is 2 ab [c 2 (a 2 m 2 b 2 )](1 m 2 ) (b 2 a 2 m 2 ) ST U 5.3.13 Equation of the Chord joining Two points on the Hyperbola. The equation of the chord joining the points P(a sec 1 , b tan 1 ) and Q(a sec 2 , b tan 2 ) is b tan 2 b tan 1 y b tan 1 (x a sec 1 ) a sec 2 a sec 1 x 2 y 1 2 2 cos 1 sin cos 1 a 2 b 2 2 Note : If the chord joining two points (a sec 1 , b tan 1 ) and (a sec 2 , b tan 2 ) passes through the focus of the hyperbola Example: 22 The equation of the chord of contact of tangents drawn from a point (2, –1) to the hyperbola 16 x 2 9 y 2 144 is (a) Solution: (a) 1e x2 y2. 2 1 , then tan 1 tan 2 2 2 2 1e a b 32 x 9 y 144 From T 0 i.e., (b) 32 x 9 y 55 (c) 32 x 9 y 144 0 2 2 x y xx 1 yy 1 2 1. Here, 16 x 2 9 y 2 144 i.e., 1 9 16 a2 b (d) 32 x 9 y 55 0 Conic Section : Hyperbola 235 So, the equation of chord of contact of tangents drawn from a point (2, –1) to the hyperbola is 2 x (1) y 1 9 16 i.e., 32 x 9 y 144 Example: 23 x2 y2 1 at the points where it is intersected by the line a2 b 2 The point of intersection of tangents drawn to the hyperbola lx my n 0 is xx 1 yy 1 2 1 a2 b......(i) The given line is lx my n 0.....(ii) Equation (i) and (ii) represent the same line x1 a 2l y1 b 2m a 2l b 2m 1 x1 ; Hence the required point is , y1 h n n a 2l b 2m n , n . ID What will be equation of that chord of hyperbola 25 x 2 16 y 2 400 , whose mid point is (5, 3) [UPSEAT 1999] (b) 125 x 48 y 481 (a) 115 x 117 y 17 Solution: (b) a 2n b 2n , (d) m l a 2n b 2n , l m Let (x 1 , y1 ) be the required point. Then the equation of the chord of contact of tangents drawn from (x 1 , y1 ) to the given hyperbola is Example: 24 (c) E3 Solution: (a) a 2l b 2m , (b) n n a 2l b 2m n , n 60 (a) (c) (d) 15 x 121 y 105 127 x 33 y 341 According to question, S 25 x 16 y 400 0 2 2 Equation of required chord is S 1 T U......(i) Here S 1 25 (5) 16 (3) 400 = 625 144 400 81 and T 25 xx 1 16 yy 1 400 , where x 1 5 , y 1 3 2 2 25 x (5) 16 y(3) 400 125 x 48 y 400 Example: 25 D YG So, from (i) required chord is 125 x 48 y 400 81 125 x 48 y 481. The locus of the mid-points of the chords of the circle x 2 y 2 16 which are tangent to the hyperbola 9 x 2 16 y 2 144 is (a) (c) (x 2 y 2 ) 2 16 x 2 9 y 2 (x 2 y 2 ) 2 16 x 2 9 y 2 2 Solution: (a) The given hyperbola is (b) (x 2 y 2 ) 2 9 x 2 16 y 2 (d) None of these 2 x y 1 16 9 ……(i) Any tangent to (i) is y mx 16 m 2 9 ……(ii) U Let (x 1 , y 1 ) be the mid point of the chord of the circle x 2 y 2 16 Then equation of the chord is T S 1 i.e., xx 1 yy 1 (x 12 y 12 ) 0 ……(iii) Since (ii) and (iii) represent the same line. m 1 16 m 2 9 x1 y1 (x 12 y 12 ) m Locus of (x 1 , y 1 ) is (x 2 y 2 ) 2 16 x 2 9 y 2. ST x1 x2 and (x 12 y 12 ) 2 y 12 (16 m 2 9) (x 12 y 12 ) 2 16. 12 y 12 9 y 12 = 16 x 12 9 y 12 y1 y1 5.3.14 Pole and Polar g Let P be any point inside or outside the hyperbola. If any straight line drawn through P interesects the hyperbola at A and B. Then the locus of the point of intersection of the tangents to the hyperbola at A and B is called the polar of the given point P with respect to the hyperbola and the point P is called the pole of the polar. The equation of the required polar with (x 1 , y 1 ) as its pole is A (h,k) Q Polar A X Q Pole P(x1,y1) B B X 236 Conic Section : Hyperbola xx 1 yy1 2 1 a2 b Note : Polar of the focus is the directrix. Any tangent is the polar of its point of contact. x2 y2 1 is Q a 2(h, k) bB2 60 (1) Pole of a given line : The pole of a given line lx my n 0 with respect to the hyperbola A a 2l b 2m (x 1 , y 1 ) , n n X Polar X E3 Pole P (x1,y1) A Q B U ID (2) Properties of pole and polar (i) If the polar of P(x 1 , y 1 ) passes through Q(x 2 , y 2 ) , then the polar of Q(x 2 , y 2 ) goes through P(x 1 , y 1 ) and such points are said to be conjugate points. (ii) If the pole of a line lx my n 0 lies on the another line l x m y n 0 then the pole of the second line will lie on the first and such lines are said to be conjugate lines. (iii) Pole of a given line is same as point of intersection of tangents as its extremities. Important Tips x2 If the polars of (x1 , y1 ) and (x 2 , y 2 ) with respect to the hyperbola Example: 26 a2 D YG x2 If the polar of a point w.r.t. a 2 y2 b y2 b2 x2 y2 U ST This touches a2 x2 a2 y2 b2 xx 1 1 is b2 Locus of (x 1 , y 1 ) is a2 b2 1 if y1 x2 a2 y2 b2 yy 1 y2 b2 1 , then the locus of the point is [Pb. CET 1999] (b) Ellipse (d) None of these b2 x2 a 4 y2 b 4 1 a 2 1 2 a2. b 2 x1 a2y 1 b Let (x 1 , y 1 ) be the pole w.r.t. Then equation of polar is hx a 2 x2 a xx b2x b2 1 21 2 1 x y1 a a y1 2 2 2 2 2 4 2 4 2 b 2 b a b x1 b 2 b b x 1 1 x 1 y 1 1 y12 a4 y12 a2 b 2 y 12 a 2 y 12 1. Which is the same hyperbola. x2 (b) 2 b2 y1 1 i.e., y The locus of the poles of the chords of the hyperbola (a) Solution: (a) a 2 x1 x 2 a4 4 0 y1 y 2 b Let (x 1 , y 1 ) be the given point. Its polar w.r.t. Example: 27 x2 1 touches the hyperbola 2 (a) Given hyperbola (c) Circle Solution: (a) 1 are at right angles, then 2 ky b2 a 2 y2 b2 1 1 y2 b 2 1 a 2 x2 a 2 1 b 2 y2 b2 1 , which subtend a right angle at the centre is (c) x2 a......(i).....(ii) 4 y2 b 4 1 a 2 1 b 2 (d) x2 a 4 y2 b4 1 a2 1 b2 Conic Section : Hyperbola 237 The equation of lines joining the origin to the points of intersection of (i) and (ii) is obtained by making homogeneous (i) with the 2 x 2 y 2 hx 1 1 ky h2 k 2 2hk help of (ii), then 2 2 2 2 x 2 2 4 y 2 2 4 2 2 xy 0 b a b a b a b a b a Since the lines are perpendicular, then coefficient of x 2 coefficient of y 2 0 a 2 h2 a 4 1 b 2 k2 b 4 0 or h2 a 4 k2 b 4 1 a 2 1 b 2. Hence required locus is x2 a 4 y2 b 4 1 a 2 1 b2 60 1 5.3.15 Diameter of the Hyperbola. The locus of the middle points of a system of parallel chords of a hyperbola is called a diameter and the point where the diameter intersects the hyperbola is called the vertex of the diameter. (x1,y1) Y P E3 x2 y2 Let y mx c a system of parallel chords to 2 2 1 for different chords a b X X ID C b2x then the equation of diameter of the hyperbola is y 2 , which is passing through Q R(h,k) a m (x2,y2) Y (0, 0) Conjugate diameter : Two diameters are said to be conjugate when each bisects all chords parallel to the others. Note : b2. a2 U If y m 1 x , y m 2 x be conjugate diameters, then m 1 m 2 If a pair of diameters be conjugate with respect to a hyperbola, they are conjugate with respect to D YG its conjugate hyperbola also. In a pair of conjugate diameters of a hyperbola. Only one meets the curve in real points. 2 The condition for the lines AX 2 HXY BY 2 0 to be conjugate diameters of x2 y2 1 is a2 b 2 a2 A b 2 B. Important Tips If CD is the conjugate diameter of a diameter CP of the hyperbola x2 a 2 y2 b2 1 , where P is (a sec , b tan ) then coordinates of D is U (a tan , b sec ) , where C is (0, 0). If a pair of conjugate diameters meet the hyperbola and its conjugate in P and D respectively, then CP 2 CD 2 a2 b 2 ST Example: 28 (a) Solution: (b) (b) a 2 b 2 (c) a2 b2 (d) None of these Coordinates of P and D are (a sec , b tan ) and (a tan , b sec ) respectively. Then (CP ) 2 (CD) 2 = a 2 sec 2 b 2 tan 2 a 2 tan 2 b 2 sec 2 Example: 29 If the line lx my n 0 passes through the extremities of a pair of conjugate diameters of the hyperbola (a) Solution: (a) = a 2 (sec 2 tan 2 ) b 2 (sec 2 tan 2 ) = a 2 (1) b 2 (1) = a 2 b 2. a2l 2 b 2m 2 0 (b) a 2 l 2 b 2 m 2 0 The extremities of a pair of conjugate diameters of x2 a 2 (c) y2 b2 a2l 2 b 2m 2 n 2 x2 a2 y2 b2 1 then (d) None of these 1 are (a sec , b tan ) and (a tan , b sec ) respectively. According to the question, since extremities of a pair of conjugate diameters lie on lx my n 0 238 Conic Section : Hyperbola l(a sec ) m (b tan ) n 0 l (a tan ) m(b sec ) n 0 ……(i) Then from (i), al sec bm tan n or a 2 l 2 sec 2 b 2 m 2 tan 2 2ablm sec tan n 2 ……(ii) And from (ii), al tan bm sec n or a 2 l 2 tan 2 b 2 m 2 sec 2 2ablm sec tan n 2 ……(iii) Then subtracting (ii) from (iii) a 2 l 2 (sec 2 tan 2 ) b 2 m 2 (tan 2 sec 2 ) 0 or a 2 l 2 b 2 m 2 0. 60 5.3.16 Subtangent and Subnormal of the Hyperbola a Let the tangent and normal at P( x 1 , y 1 ) meet the x-axis at A and B respectively. Y (x1, y1) P a Length of subtangent AN CN CA x 1 x1 Length of subnormal BN CB CN (a 2 b 2 ) a2 E3 2 X 2 C b x 1 x 1 = 2 x 1 (e 2 1)x 1 a A B N X Y 5.3.17 Reflection property of the Hyperbola a ID If an incoming light ray passing through one focus (S) strike convex side of the hyperbola then it will get reflected towards other focus (S ) M Light ray Y TPS LPM Reflected ray (–ae,0)S A T A C D YG X P U L S (ae,0) Normal N X Tangent Y Example: 30 A ray emanating from the point (5, 0) is incident on the hyperbola 9 x 2 16 y 2 144 at the point P with abscissa 8; then the equation of reflected ray after first reflection is (Point P lies in first quadrant) (a) Solution: (a) 3 3 x 13 y 15 3 0 (b) 3 x 13 y 15 0 (c) 3 3 x 13 y 15 3 0 (d) None of these Given hyperbola is 9 x 2 26 y 2 144. This equation can be rewritten as x2 y2 1 16 9.....(i) U Since x coordinate of P is 8. Let y-coordinate of P is (8, ) lies on (i) 64 2 1; 16 9 Y Y M light ray L Reflected ray 27 ( P lies in first quadrant) ST Normal Tangent X S(–ae,0) A C T A S(ae,0) N X 3 3 Hence coordinate of point P is (8, 3 3 ) Y Equation of reflected ray passing through P(8, 3 3 ) and S (5, 0 ); Its equation is y 3 3 or 13 y 39 3 3 3 x 24 3 or 3 3 x 13 y 15 3 0 03 3 (x 8 ) 5 8 5.3.18 Asymptotes of a Hyperbola a An asymptote to a curve is a straight line, at a finite distance from the origin, to which the tangent to a curve tends as the point of contact goes to infinity. The equations of two asymptotes of the hyperbola x2 y2 b x y 2 1 are y x or 0. 2 a b a a b Conic Section : Hyperbola 239 Note : The combined equation of the asymptotes of the hyperbola x2 y2 x2 y2 0. 1 is a2 b 2 a2 b 2 When b a i.e. the asymptotes of rectangular hyperbola x 2 y 2 a 2 are y x , which are at right angles. A hyperbola and its conjugate hyperbola have the same asymptotes. The equation of the pair of asymptotes differ the hyperbola and the conjugate hyperbola by the same constant only i.e. Hyperbola – Asymptotes = Asymptotes – Conjugated hyperbola or, 60 Y x 2 y2 x 2 y2 x 2 y2 x 2 y2 2 2 1 2 2 2 2 2 2 1. a a b b a b a b Asymptotes B X A A C x2 E3 B The asymptotes pass through the centre of the hyperbola. Y The bisectors of the angles between the asymptotes are the coordinate axes. a x2 a 2 y2 2 y2 b2 1 1 b2 x2 y2 b 2 1 is 2 tan 1 or 2 sec 1 e. 2 a b a ID The angle between the asymptotes of the hyperbola S 0 i.e., X C Asymptotes are equally inclined to the axes of the hyperbola. U Important Tips The parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes and is of constant area. Y D YG Area of parallelogram QR Q R 4(Area of parallelogram QDCP) = 4 ab Constant RM The product of length of perpendiculars drawn from any point on the hyperbola asymptotes is a2 b 2 y b 2 1 to the Q D P X C P C X R Q D Y x2 y2 x2 y2 2 1 tangents are drawn to the hyperbola 2 2 2. The area cut-off by the 2 a a b b chord of contact on the asymptotes is equal to From any point on the hyperbola, ab 2 ST (a) Solution: (d) a 2 2. U Example: 31 a2b 2 x 2 (b) ab Let P(x 1 , y1 ) be a point on the hyperbola (c) And x y =0 a b x y =0 a b (d) 4 ab x2 y2 x2 y2 2 1 , then 12 12 1 2 a b a b The chord of contact of tangent from P to the hyperbola The equation of asymptotes are 2ab xx 1 yy 1 x2 y2 2 2 2 is a2 b a2 b 2.....(i)......(ii)......(iii) 2a 2b 2a 2b , (0, 0) ; , , The point of intersection of the asymptotes and chord are x 1 / a y1 / b x 1 / a y1 / b x 1 / a y 1 / b x 1 / a y 1 / b 240 Conic Section : Hyperbola Solution: (d) 1 1 | ( x 1 y 2 x 2 y1 ) | = 2 2 8a b x 2 / a2 y 2 / b 2 1 1 4 ab. The combined equation of the asymptotes of the hyperbola 2 x 2 5 xy 2y 2 4 x 5 y 0 [Karnataka CET 2002] (a) 2 x 2 5 xy 2y 2 0 (b) 2 x 2 5 xy 2y 2 4 x 5 y 2 0 0 (c) 2 x 2 5 xy 2y 2 4 x 5 y 2 0 (d) 2 x 2 5 xy 2y 2 4 x 5 y 2 0 60 Example: 32 Area of triangle = Given, equation of hyperbola 2 x 2 5 xy 2y 2 4 x 5 y 0 and equation of asymptotes 2 x 2 5 xy 2y 2 4 x 5 y 0......(i) which is the equation of a pair of straight lines. We know that the standard equation of a pair of straight lines is ax 2 2hxy by 2 2 gx 2 fy c 0 5 5 and c . , g 2, f 2 2 E3 Comparing equation (i) with standard equation, we get a 2, b 2 , h We also know that the condition for a pair of straight lines is abc 2 fgh af 2 bg 2 ch 2 0. 9 9 25 25 0 or 2 0 or 8 4 2 2 4 ID Therefore, 4 25 Substituting value of in equation (i), we get 2 x 2 5 xy 2y 2 4 x 5 y 2 0. 5.3.19 Rectangular or Equilateral Hyperbola a U (1) Definition : A hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The eccentricity of rectangular hyperbola is always 2. The general equation of second degree represents a rectangular hyperbola if 0, h 2 ab D YG and coefficient of x 2 + coefficient of y 2 = 0 The equation of the asymptotes of the hyperbola x2 y2 b 2 1 are given by y x. 2 b a a ST U b b 2b / a 2ab a a 2 The angle between these two asymptotes is given by tan . 2 2 b b 1 b /a a b2 1 a a 2ab If the asymptotes are at right angles, then / 2 tan tan 2 tan a 2 b 2 0 2 2 2 a b a b 2a 2b. Thus the transverse and conjugate axis of a rectangular hyperbola are equal and the equation is x 2 y 2 a 2. The equations of the asymptotes of the rectangular hyperbola are y x i.e., y x and y x. Clearly, each of these two asymptotes is inclined at 45 to the transverse axis. (2) Equation of the rectangular hyperbola referred to its asymptotes as the axes of coordinates : Referred to the transverse and conjugate axis as the axes of coordinates, the equation of the rectangular hyperbola is x 2 y 2 a2 …..(i) The asymptotes of (i) are y = x and y = – x. Each of these two asymptotes is inclined at an angle of 45 with the transverse axis, So, if we rotate the coordinate axes through an angle of / 4 keeping the origin fixed, then the axes X Y x X cos( / 4 ) Y sin( / 4 ) coincide with the asymptotes of the hyperbola and and 2 Y X Y y X sin / 4 Y cos( / 4 ) . 2 XY=c2 Substituting the values of x and y in (i), X O Y X Conic Section : Hyperbola 241 2 2 a2 X Y Y X a 2 XY XY c 2 We obtain the 2 2 2 a2. 2 This is transformed equation of the rectangular hyperbola (i). (3) Parametric co-ordinates of a point on the hyperbola XY = c2 : If t is non–zero variable, the coordinates of any point on the rectangular hyperbola xy c 2 can be written as (ct, c / t). The point (ct, c / t) on the hyperbola 60 where c 2 xy c 2 is generally referred as the point ‘t’. For rectangular hyperbola the coordinates of foci are (a 2 , 0) and directrices are x a 2. (5) Equation of tangent in different forms ID E3 For rectangular hyperbola xy c 2 , the coordinates of foci are (c 2 , c 2 ) and directrices are x y c 2. (4) Equation of the chord joining points t1 and t2 : The equation of the chord joining two points c c c t 2 t1 c c 2 ct1 , and ct2 , on the hyperbola xy c is y (x ct1 ) x y t1 t 2 c (t1 t 2 ). t1 ct2 ct1 t2 t1 x y 2 x1 y 1 U (i) Point form : The equation of tangent at (x 1 , y1 ) to the hyperbola xy c 2 is xy1 yx 1 2c 2 or D YG x c (ii) Parametric form : The equation of the tangent at ct, to the hyperbola xy c 2 is yt 2c.On t t x c replacing x 1 by ct and y 1 by on the equation of the tangent at (x 1 , y1 ) i.e. xy1 yx 1 2c 2 we get yt 2c. t t Note : Point of intersection of tangents at ' t1 ' and ' t2 ' is 2ct1 t 2 , 2c t1 t 2 t1 t 2 (6) Equation of the normal in different forms : (i) Point form : The equation of the normal at (x 1 , y1 ) to the y dy 1 hyperbola xy c 2 is xx 1 yy1 x12 y12. As discussed in the equation of the tangent, we have x1 dx ( x 1 , y 1) ST U So, the equation of the normal at (x 1 , y1 ) is y y1 yy1 y12 xx 1 x 12 1 x ( x x 1 ) y y1 1 ( x x 1 ) y1 dy dx ( x 1 , y 1 ) xx 1 yy1 x 12 y12 This is the required equation of the normal at (x 1 , y1 ). c (ii) Parametric form: The equation of the normal at ct, to the hyperbola xy c 2 is xt 3 yt ct4 c 0. t On replacing x 1 by ct and y 1 by c / t in the equation. We obtain xx 1 yy1 x12 y12 , Note : xct yc c2 c 2 t 2 2 xt 3 yt ct 4 c 0 t t c The equation of the normal at ct, is a fourth degree in t. So, in general, four normals can be t drawn from a point to the hyperbola xy c 2 242 Conic Section : Hyperbola 1 c If the normal at ct, on the curve xy c 2 meets the curve again in ' t ' then; t 3. t t c {t t (t 2 t t t 2 ) 1} c {t13 t 23 (t12 t1 t 2 t 22 )} Point of intersection of normals at ' t 1 ' and ' t 2 ' is 1 2 1 1 2 2 , t1 t 2 (t1 t 2 ) t1 t 2 (t1 t 2 ) Important Tips A triangle has its vertices on a rectangular hyperbola; then the orthocentre of the triangle also lies on the same hyperbola. All conics passing through the intersection of two rectangular hyperbolas are themselves rectangular hyperbolas. An infinite number of triangles can be inscribed in the rectangular hyperbola xy c 2 whose all sides touch the parabola y 2 4 ax. If 5 x 2 y 2 20 represents a rectangular hyperbola, then equals (a) 5 Solution: (c) (c) – 5 (b) 4 (d) None of these E3 Example: 33 60 Since the general equation of second degree represents a rectagular hyperbola if 0, h 2 ab and coefficient of x 2 coefficient of y 2 0. Therefore the given equation represents a rectangular hyperbola if 5 0 i.e., 5 If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the mid-point of PN is (a) Circle (b) Parabola (c) Ellipse (d) Hyperbola Solution: (d) Let xy c 2 be the rectangular hyperbola, and let P(x 1 , y1 ) be a point on it. Let Q (h, k ) be the mid-point of PN. Then the y coordinates of Q are x 1 , 1 . 2 y1 k x 1 h and y 1 2k 2 U x 1 h and But (x 1 , y 1 ) lies on xy = c2. h. (2k ) c 2 hk c 2 / 2 D YG ID Example: 34 Y xy=c2 Q (h, k) P(x1, y1) O X N X Hence, the locus of (h, k ) is xy c 2 / 2 , which is a hyperbola. Example: 35 c If the normal at ct, on the curve xy c 2 meets the curve again in t, then t (a) 1 t 3 (b) t 1 t (c) t 1 t (d) 2 t2 1 t2 c The equation of the tangent at ct, is ty t 3 x c t 4 c t U Solution: (a) t Y c If it passes through c t , then t tc 1 t 3 ct ct 4 c t t 3 t 2 t 4 t t t t t 3 t (t t) t 3 t t ST Example: 36 If the tangent and normal to a rectangular hyperbola cut off intercepts a1 and a 2 on one axis and b1 and b 2 on the other axis, then (a) Solution: (c) a1 b1 a 2 b 2 0 (b) a1 b 2 b 2 a1 0 (c) a1 a2 b1 b 2 0 Let the hyperbola be xy c 2. Tangent at any point t is x yt 2 2ct 0 Putting y 0 and then x 0 intercepts on the axes are a1 2ct and b 1 Normal is xt 3 yt ct 4 c 0. Intercepts as above are a 2 c (t 4 1) t 3 , b2 c(t 4 1) t 2c t (d) None of these Conic Section : Hyperbola 243 c (t 4 1) t 3 2c 2 2c 2 2c c(t 4 1) = 2 (t 4 1) 2 (t 4 1) 0 ; t t t t (b) 16 x 2 10 xy y 2 2 (c) (d) None of these 1 k 4 (x h) x 4 x 2 (4 h k ) x 1 0 1 xk 4 x 2 4 hx This has two roots say x 1 , x 2 ; x 1 x 2 Also, A(x 1 , y 1 ) and B(x 2 , y 2 ) 4h k 4......(iv) and x 1 x 2 2x1 x 2 h 3 1......(v) 4 [ P divides AB in the ratio 1 : 2] i.e., 2 x 1 x 2 3h......(vi) (vi) – (iv) gives, x 1 3 h 4h k 8h k 4 4 and x 2 3 h 2. 8h k 2h k 4 2 8 h k 2h k 1 4 2 4 (8h k)(2h k) 2 U Putting in (v), we get 16 h 2 10 hk k 2 2 Required locus of P(h, k ) is 16 x 2 10 xy y 2 2. D YG ......(iii) E3 .....(i) 60......(ii) in Eliminating y between (i) and (ii), we get Solution: (b) 16 x 2 10 xy y 2 4 Let P(h, k ) be any point on the locus. Equation of the line through P and having slope 4 is y k 4 (x h) Suppose this meets xy 1 Example: 38 a1 a 2 b1 b 2 0. A variable straight line of slope 4 intersects the hyperbola xy 1 at two points. The locus of the point which divides the line segment between these two points in the ratio 1 : 2 is [IIT 1997] (a) 16 x 2 10 xy y 2 2 Solution: (a) ID Example: 37 a1 a 2 b1 b 2 2ct PQ and RS are two perpendicular chords of the rectangular hyperbola xy c 2. If C is the centre of the rectangular hyperbola, then the product of the slopes of CP, CQ, CR and CS is equal to (a) – 1 (b) 1 (c) 0 (d) None of these c Let t1 , t 2 , t3 , t4 be the parameters of the points P, Q, R and S respectively. Then, the coordinates of P, Q, R and S are ct1 , , t 1 c , ct 3 , t 3 c and ct 4 , t 4 respectively. c c c c t 2 t1 t4 t3 1 1 1 1 t1t2 t3 t4 1.....(i) ct 2 ct1 ct 4 ct 3 t1 t2 t3 t4 U c ct 2 , t 2 Now, PQRS Product of the slopes of CP, CQ, CR and CS ST 1 t 12 1 t 22 1 t 32 1 t 42 1 t 12 t 22 t 32 t 42 1 [Using (i)] 5.3.20 Intersection of a Circle and a Rectangular Hyperbola a If a circle x 2 y 2 2 gx 2 fy k 0 cuts a rectangular hyperbola xy c 2 in A, B, C and D and the parameters of these four points be t1 , t 2 , t 3 and t 4 respectively; then (1) (i) t1 2g c (ii) t1 t 2 (iii) t1 t 2 t 3 2f c (iv) t1 t 2 t 3 t 4 1 k c2 (v) 1 2f t1 c 244 Conic Section : Hyperbola c c but D is ct 4 , (2) Orthocentre of ABC is H ct4 , t4 t4 Hence H and D are the extremities of a diagonal of rectangular hyperbola. c c 1 f g (3) Centre of mean position of four points is t1 , i.e., , 4 t1 4 2 2 E3 60 Centres of the circles and rectangular hyperbola are (– g, – f) and (0, 0); mid point of centres of circle and f g hyperbola is , . Hence the centre of the mean position of the four points bisects the distance between the centres 2 2 of the two curves (circle and rectangular hyperbola) (4) If the circle passing through ABC meet the hyperbola in fourth points D; then centre of circle is (–g, –f) Example: 39 c1 1 1 ; t1 t 2 t 3 2 t1 t 2 t 3 If a circle cuts a rectangular hyperbola xy c 2 in A, B, C, D and the parameters of these four points be t1 , t2 , t3 and t4 respectively. Then (a) t1t2 t3 t4 (b) t1t2 t3 t4 1 Let the equation of circle be x 2 y 2 a 2 (c) D YG Parametric equation of rectangular hyperbola is x c t, y Put the values of x and y in equation (i) we get c 2 t 2 Hence product of roots t1 t2 t3 t4 Example: 40 t1 t2 U Solution: (b) ID 1 c i.e., t1 t 2 t 3 t1 t 2 t 3 2 [Kurukshetra CEE 1998] (d) t3 t 4......(i) c t c2 a2 c 2t 4 a2t 2 c 2 0 t2 c2 1 c2 If the circle x 2 y 2 a 2 intersects the hyperbola xy c 2 in four points P (x1, y1 ), Q(x 2 , y 2 ), R(x 3 , y3 ) , S (x 4 , y4 ) then x1 x 2 x 3 x 4 0 U (a) Solution: (a,b,c,d) Given, circle is x 2 y 2 a 2 ST from (ii) y (b) y1 y2 y3 y4 0 [IIT 1998] (c) x1 x 2 x 3 x 4 c 4.......(i) and hyperbola be xy c 2 c4 c2. Putting in (i), we get x 2 2 a 2 x 4 a 2 x 2 c 4 0 x x x1 x 2 x 3 x 4 0 , x 1 x 2 x 3 x 4 c 4 Since both the curves are symmetric in x and y, y1 y2 y3 y4 0 ; y1 y 2 y 3 y 4 c 4. *** (d) y1 y 2 y 3 y 4 c.....(ii) 4
