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x2 y2 Standard Equation of hyperbola is a 2 – b2 = 1 (i) Definition hyperbola : 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 (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity. The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called conjugate hyperbola. Note : (i) If e1 and e2 are the eccentricities of the (ii) Vertices : The point A and A where the curve meets the line joining the foci S and S hyperbola and its conjugate then 1 + e 2 e 1 = 1 2 are called vertices of hyperbola. (iii) Transverse and Conjugate axes : The straight line joining the vertices A and A is called transverse axes of the hyperbola. Straight line perpendicular to the transverse axes and passes through its centre called conjugate axes. (iv) Latus Rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its transverse axes is called 2b2 latus rectum. Length of latus rectum = a . (ii) The focus of hyperbola and its1 conju2gate are concyclic. Standard Equation and Difinitions Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and eccentricity 3 . Sol. Let P (x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix. Then by definition SP = e PM (v) Eccentricity : For the hyperbola x2 y2 a 2 – b2 = 1, (SP)2 = e2(PM)2 2x y 12 b2 = a2 (e2 – 1) (x–1)2 + (y–2)2 = 3 Conjugate axes 2 5(x2 + y2 – 2x – 4y + 5} = e = = 1 Transverse axes 3(4x2 + y2 + 1+ 4xy – 2y – 4x) 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0 (vi) Focal distance : The distance of any point on the hyperbola from the focus is called the focal distance of the point. Note : The difference of the focal distance of a point on the hyperbola is constant and is equal to the length of the transverse axes. |SP – SP| = 2a (const.) which is the required hyperbola. Ex.2 Find the lengths of transverse axis and conjugate axis, eccentricity and the co- ordinates of foci and vertices; lengths of the latus rectum, equations of the directrices of the hyperbola 16x2 – 9y2 = –144 Sol. The equation 16x2 – 9y2 = – 144 can be Sol. y= m1(x –a),y= m2(x + a) where m1m2 = k, given x 2 written as 9 x2 y 2 – 16 = – 1. This is of the form y2 In order to find the locus of their point of intersection we have to eliminate the unknown m1 and m2. Multiplying, we get y2 = m1m2 (x2 – a2) or y2 = k(x2–a2) a 2 – b2 = – 1 a2 = 9, b2 = 16 a = 3, b = 4 or x – y 1 k = a2 which represents a hyperbola. Length of transverse axis : The length of transverse axis = 2b = 8 Length of conjugate axis : The length of conjugate axis = 2a = 6 5 Ex.5 T

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x2 y2 Standard Equation of hyperbola is a 2 – b2 = 1 (i) Definition hyperbola : 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 (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity. The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called conjugate hyperbola. Note : (i) If e1 and e2 are the eccentricities of the (ii) Vertices : The point A and A where the curve meets the line joining the foci S and S hyperbola and its conjugate then 1 + e 2 e 1 = 1 2 are called vertices of hyperbola. (iii) Transverse and Conjugate axes : The straight line joining the vertices A and A is called transverse axes of the hyperbola. Straight line perpendicular to the transverse axes and passes through its centre called conjugate axes. (iv) Latus Rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its transverse axes is called 2b2 latus rectum. Length of latus rectum = a . (ii) The focus of hyperbola and its1 conju2gate are concyclic. Standard Equation and Difinitions Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and eccentricity 3 . Sol. Let P (x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix. Then by definition SP = e PM (v) Eccentricity : For the hyperbola x2 y2 a 2 – b2 = 1, (SP)2 = e2(PM)2 2x y 12 b2 = a2 (e2 – 1) (x–1)2 + (y–2)2 = 3 Conjugate axes 2 5(x2 + y2 – 2x – 4y + 5} = e = = 1 Transverse axes 3(4x2 + y2 + 1+ 4xy – 2y – 4x) 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0 (vi) Focal distance : The distance of any point on the hyperbola from the focus is called the focal distance of the point. Note : The difference of the focal distance of a point on the hyperbola is constant and is equal to the length of the transverse axes. |SP – SP| = 2a (const.) which is the required hyperbola. Ex.2 Find the lengths of transverse axis and conjugate axis, eccentricity and the co- ordinates of foci and vertices; lengths of the latus rectum, equations of the directrices of the hyperbola 16x2 – 9y2 = –144 Sol. The equation 16x2 – 9y2 = – 144 can be Sol. y= m1(x –a),y= m2(x + a) where m1m2 = k, given x 2 written as 9 x2 y 2 – 16 = – 1. This is of the form y2 In order to find the locus of their point of intersection we have to eliminate the unknown m1 and m2. Multiplying, we get y2 = m1m2 (x2 – a2) or y2 = k(x2–a2) a 2 – b2 = – 1 a2 = 9, b2 = 16 a = 3, b = 4 or x – y 1 k = a2 which represents a hyperbola. Length of transverse axis : The length of transverse axis = 2b = 8 Length of conjugate axis : The length of conjugate axis = 2a = 6 5 Ex.5 T

- 1. Total no.of questions in Hyperbola are- Inchapter examples..................................................10 Solved Examples............................................... ......14 Total no. of questions.............................................. 24 HYPERBOLA
- 2. (i) Definition hyperbola : 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 (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity. (ii) Vertices : The point A and A where the curve meets the line joining the foci S and S are called vertices of hyperbola. (iii) Transverse and Conjugate axes : The straight line joining the vertices A and A is called transverse axes of the hyperbola. Straight line perpendicular to the transverse axes and passes through its centre called conjugate axes. (iv) Latus Rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its transverse axes is called latus rectum. Length of latus rectum = a b 2 2 . (v) Eccentricity : For the hyperbola 2 2 a x – 2 2 b y = 1, b2 = a2 (e2 – 1) e = 2 a 2 b 2 1 = 2 axes Transverse axes Conjugate 1 (vi) Focal distance : The distance of any point on the hyperbola from the focus is called the focal distance of the point. Note : The difference of the focal distance of a point on the hyperbola is constant and is equal to the length of the transverse axes. |SP – SP| = 2a (const.) z' y z P(x, y) S (ae, 0) x A L' L M M' K' K A'(–a, 0) T ' –x T (a, 0) (0, 0) O (–ae, 0) S' 2. CONJUGATE HYPERBOLA The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called conjugate hyperbola. Note : (i) If e1 and e2 are the eccentricities of the hyperbola and its conjugate then 2 1 e 1 + 2 2 e 1 = 1 (ii) The focus of hyperbola and its conjugate are concyclic. Examples based on Standard Equation and Difinitions Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and eccentricity 3 . Sol. Let P (x,y) be any point on the hyperbola. Draw PM perpendicular from P on the directrix. Then by definition SP = e PM (SP)2 = e2(PM)2 (x–1)2 + (y–2)2 = 3 2 1 4 1 y x 2 5(x2 + y2 – 2x – 4y+ 5} = 3(4x2 + y2 + 1+ 4xy – 2y – 4x) 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0 which is the required hyperbola. Ex.2 Find the lengths of transverse axis and conjugate axis, eccentricity and the co- ordinates of foci and vertices; lengths of the latus rectum, equations of the directrices of the hyperbola 16x2 – 9y2 = –144 1. STANDARD EQUATION AND DEFINITIONS Standard Equation of hyperbola is 2 2 a x – 2 2 b y = 1
- 3. Sol. The equation 16x2 – 9y2 = – 144 can be written as 9 x2 – 16 y2 = – 1. This is of the form 2 2 a x – 2 2 b y = – 1 a2 = 9, b2 = 16 a = 3, b = 4 Length of transverse axis : The length of transverse axis = 2b = 8 Length of conjugate axis : The length of conjugate axis = 2a = 6 Eccentricity : e = 2 2 b a 1 = 16 9 1 = 4 5 Foci : the co- ordinates of the foci are (0,± be), i.e., (0, ± 4) Length of Latus rectum : The length of latus rectum = b a 2 2 = 4 ) 3 ( 2 2 = 2 9 Equation of directrices : The equation of directrices are y = ± e b y = ± 4 / 5 4 = ± 5 16 Ex.3 Find the coordinates of foci, the eccentricity and latus- rectum, equations of directrices for the hyperbola 9x2–16y2–72x+96y–144=0 Sol. Equation can be rewritten as 2 2 4 ) 4 x ( – 2 2 3 ) 3 y ( = 1 so a = 4, b = 3 b2 = a2 (e2 – 1) gives e = 5/4 Foci : X = ± ae, Y = 0 gives the foci as (9,3), (– 1,3) Centre : X = 0, Y = 0 i.e. (4,3) Directrices : X = ± a/e i.e. x – 4 = ± 16/5 directrices are 5x – 36 = 0; 5x – 4 = 0 Latus- rectum = 2b2/ a = 2 x 9/4= 9/2 Ex.4 Two straight lines pass through the fixed points (± a,0) and have gradients whose product is k. Show that the locus of the points of inter-section of the lines is a hyperbola. Sol. y= m1(x –a),y= m2(x+a) wherem1m2 =k, given In order to find the locus of their point of intersection we have to eliminate the unknown m1 and m2. Multiplying, we get y2 = m1m2 (x2 – a2) or y2 = k(x2–a2) or 1 x2 – k y2 = a2 which represents a hyperbola. Ex.5 The eccentricity of the conic represented by x2 – y2 – 4x + 4y + 16 = 0 is (A) 1 (B) 2 (C) 2 (D)1/2 Sol.[B] We have x2 – y2 – 4x +4y+16 = 0 or, (x2 – 4x)– (y2 – 4y) = 16 or, (x2 – 4x + 4) – (y2 – 4y+ 4) = – 16 or, (x–2)2 – (y – 2)2 = – 16 or, 2 2 4 ) 2 x ( – 2 2 4 ) 2 y ( = 1 Shifting the origin at (2,2) , we obtain 2 2 4 X – 2 2 4 Y = – 1, where x = X + 2, y = Y + 2 This is a rectangular hyperbola, whose eccentricity is always 2 . Ex.6 The equation 16x2 = 3y2 – 32x + 12y – 44= 0 represents a hyperbola (A) the length of whose transverse axis is 3 4 (B) the length of whose conjugate axis is 4 (C) whose centre is (– 1,2) (D) whose eccentricity is 3 19 Sol.[D] We have, 16(x2 – 2x) – 3 (y2 – 4y) = 44 16 (x –1)2 – 3 (y– 2)2 = 48 3 ) 1 x ( 2 – 16 ) 2 y ( 2 = 1 This equation represents a hyperbola with eccentricity given e = 2 axis Transverse axis Conjugate 1 = 2 3 4 1 = 3 19 3. Parametric equation of the Hyperbola Let the equation of ellipse in standard form will be given by 2 2 a x – 2 2 b y = 1
- 4. Then the equation of ellipse in the parametric form will be given by x= a sec , y = b tan where is the eccentric angle whose value vary from 0 < 2. Therefore coordinate of any point P on the ellipse will be given by (a sec, b tan ). 4. POSITION OFAPOINT P(x1, y1) WITH RESPECT TO HYPERBOLA The quantity 2 2 1 a x – 2 2 1 b y = 1 is positive, zero or negative according as the point (x1, y1) lies inside on or outside the hyperbola 2 2 a x – 2 2 b y = 1. 5. LINE & HYPERBOLA ‘’The straight line y = mx + c is a sacant, a tangent or passes outside the hyperbola 2 2 a x – 2 2 b y = 1 according as c2 > = < a2m2 – b2 Examples based on Line and Hyperbola Ex.7 Find the position of the point (5, – 4) relative to the hyperbola 9x2 – y2 = 1. Sol. Since 9(5)2 –(–4)2 –1=225–16 –1 =208 > 0 so the point ( 5,–4) lies inside the hyperbola 9x2– y2 = 1 S.No. Particulars Hyperbola Conjugate Hyperbola 2 2 a x – 2 2 b y = 1 – 2 2 a x + 2 2 b y = 1 1. Co-ordinate of the centre (0, 0) (0, 0) 2. Co-ordinate of the vertices (a, 0) & (–a, 0) (0, b) & (0, –b) 3. Co-ordinate of foci (+ ae, 0) (0, + be) 4. Length of the transverse axes 2a 2b 5. Length of the conjugate axes 2b 2a 6. Equation of directrix x = + a/e y = + b/e 7. Eccentricity e = 2 2 a b 1 e = 2 2 b a 1 8. Length of latus rectum a b 2 2 b a 2 2 9. Equation of transverse axes y = 0 x = 0 10. Equation of conjugate axes x = 0 y = 0 Ex.8 Show that the line x cos + y sin = p touches the hyperbola 2 2 a x – 2 2 b y = 1 if a2 cos2 – b2 sin2 = p2. Sol. The given line is x cos + y sin = p y sin = – x cos + p y = – x cot + p cosec Comparing this line with y = mx + c m = – cot , c = p cosec Since the given line touches the hyperbola 2 2 a x – 2 2 b y = 1 then c2 = a2m2 – b2 p2 cosec2 = a2 cot2 – b2 or p2 = a2 cos2 – b2 sin2 Ex.9 The line 5x + 12y = 9 touches the hyperbola x2 – 9y2 = 9 at the point (A) (– 5,4/3) (B) (5, – 4/3) (C) (3, – 1/2) (D) None of these Sol.[B] We have : m = Slope of the tangent = – 12 5 If a line of slope m is tangent to the hyperbola 2 2 a x – 2 2 b y = 1, then the coordinates of the point of contact are
- 5. l 2 x x a + l 2 y y b = a2 + b2 = a2 e2. (ii) The equation of normal at (a sec , b tan ) to the ellipse 2 2 a x – 2 2 b y = 1 is ax cos + by cot = a2 + b2. (iii) The equation of the normal to the hyperbola 2 2 a x – 2 2 b y = 1 in terms of the slope m of the normal is given by y = mx – 2 2 2 2 2 m b a ) b a ( m Note : In general four normals can be drawn from a point (x1, y1) to the hyperbola 2 2 a x – 2 2 b y = 1 8. EQUATION OF PAIR OF TANGENTS If P (x1, y1) be any point out side the hyperbola 2 2 a x – 2 2 b y = 1 and a pair of tangents PA, PB can be drawn to it from P. then the equation of pair of tangents of PA and PB is SS1 = T2 where S 2 2 a x – 2 2 b y – 1 = 0 S1 = 2 2 1 a x – 2 2 1 b y – 1 = 0 T = 2 1 a x x – 2 1 b y y – 1 = 0 2 2 2 2 2 2 2 2 b m a b , b m a m a Here, a2 = 9, b2 = 1 and m = – 5/12 So, points of contact are 3 4 , 5 i.e. 3 4 , 5 and 3 4 , 5 . Out of these two points 3 4 , 5 lies on the line 5x + 12y = 9. Hence, 3 4 , 5 is the required point. 6. EQUATION OFTANGENT (i) The equation of tangents of slope m to the hyperbola 2 2 a x – 2 2 b y = 1 are y = mx + 2 2 2 b m a and the co-ordinates of the point of contacts are 2 2 2 2 2 2 2 2 b m a b , b m a m a (ii) Equation of tangent to the hyperbola 2 2 a x – 2 2 b y = 1 at the point (x1, y1) is 2 1 a xx – 2 1 b yy = 1 (iii) Equation of tangent to the hyperbola 2 2 a x – 2 2 b y = 1 at the point (a sec, b tan) is a x sec – b y tan = 1 Note : In general two tangents can be drawn from an external point (x1, y1) to the hyperbola and they are y – y1 = m1 (x – x1) and y – y1 = m2 (x – x1), where m1 and m2 are roots of (x1 2 – a2) m2 – 2x1y1 + y1 2 + b2 = 0 7. EQUATION OF NORMAL (i) The equation of normal to the hyperbola 2 2 a x – 2 2 b y = 1 at (x1, y1) is
- 6. 9. CHORD OF CONTACT If PA and PB be the tangents through point P (x1, y1) to the hyperbola 2 2 a x – 2 2 b y = 1, then the equation of the chord of contact AB is 2 1 a x x – 2 1 b y y = 1 or T = 0 (at x1, y1) 10. EQUATION OFACHORDWHOSE MIDDLE POINT IS GIVEN Equation of the chord of the hyperbola 2 2 a x – 2 2 b y = 1, bisected at the given point (x1, y1) is T = S1. 2 1 a xx – 2 1 b yy – 1 = 2 2 1 a x – 2 2 1 b y – 1 11. DIRECTOR CIRCLE The locus of the intersection of tangents which are at right angles is known as director circle of the hyperbola. The equation of the director circle is x2 + y2 = a2 – b2. Note: The circle described on transverse axis of the hyperbola as diameter is called auxiliary circle and so its equation is x2 + y2 = a2 Examples based on Director Circle Ex.10 Find the locus of the mid points of the chords of the circle x2 + y2 = 16, which are tangent to the hyperbola 9x2 – 16y2 = 144. Sol. Any tangent to hyperbola 16 x2 – 9 y2 = 1 is y = mx + (16m2 – 9) ...(1) Let (x1,y1) be the mid- point of the chord of the circle x2 +y2 = 16, then equation of the chord is (T= S1) xx1 + yy1 – (x1 2 + y1 2) = 0 ...(2) Since (1) and (2) are same, comparing, we get 1 x m = – 1 y 1 = ) y x ( 9 m 16 2 1 2 1 2 m = – 1 1 y x , ( 2 1 x + 2 1 y )2 = 2 1 y (15m2 – 9) Eliminating m and generalizing (x1, y1), required locus is (x2 + y2)2 = 16x2 – 9y2
- 7. Ex.1 If e and e be the eccentricities of a hyperbola and its conjugate then the value of 2 e 1 + 2 ' e 1 = (A) 0 (B) 1 (C) 2 (D) 4 Sol.[B] Hyperbola 2 2 a x – 2 2 b y = 1 b2 = a2 (e2 – 1) or e2 = 1 + 2 2 a b = 2 2 2 a b a Conjugate hyperbola 2 2 a x – 2 2 b y = –1 i.e.Transverse axis is along y-axis and conjugate along x-axis. a2 = b2 (e'2 – 1) e'2 = 2 2 2 b b a 2 e 1 + 2 ' e 1 = 2 2 2 2 b a b a = 1 Ex. 2. If the foci of the ellipse 16 x2 + 2 2 b y = 1 and hyperbola 144 x 2 – 81 y2 = 25 1 coincide, then the value of b2 is- (A) 1 (B) 5 (C) 7 (D) 9 Sol.[C] For hyperbola 144 x 2 – 81 y2 = 25 1 e2 = 1 + 2 2 a b = 1 + 144 81 = 144 225 e = 12 15 = 4 5 i.e., e > 1 Hence the foci are (±ae, 0) 0 , 4 5 . 5 12 = (±3, 0) Now the foci coincide therefore for ellipse ae = 3 or a2 e2 = 9 or a2 2 2 a b 1 = 9 a2 – b2 = 9 or 16 – b2 = 9 b2 = 7 Ex. 3. The equation of the common tangents to the parabola y2 = 8x and the hyperbola 3x2 – y2 = 3 is- (A) 2x± y+1 = 0 (B) x ± y + 1 = 0 (C) x ± 2y+1 = 0 (D) x ± y + 2 = 0 Sol.[A] Parabola y2 = 8x 4a = 8 a = 2 Any tangent to the parabola is y = mx + m 2 ...(i) If it is also tangent to the hyperbola 1 x2 – 3 y2 = 1 i.e. a2 = 1, b2 = 3 then c2 = a2 m2 – b2 2 m 2 = 1.m2 – 3 or m4 – 3m2 – 4 = 0 (m2 –4) (m2 + 1) = 0 m = ±2 puting for m in (i), we get the tangents as 2x ± y + 1 = 0 Ex. 4. The locus of the point of intersection of the lines 3 x–y–4 3 k=0 and 3 kx +ky – 4 3 = 0 for different values of k is- (A) Ellipse (B) Parabola (C) Circle (D) Hyperbola Sol.[D] 3 x – y = 4 3 k ...(i) K( 3 x + y) = 4 3 ...(ii) To find the locus of their point of intersection eliminate the variable K between the equations from (i) K = 3 4 y x 3 and putting in (ii), we get ( 3 x – y) ( 3 x + y) = 3 (4)2 3x2 – y2 = 48 or 16 x2 – 48 y2 = 1 Hence the locus is hyperbola Ex. 5. The area of a triangle formed by the lines x – y = 0, x + y = 0 and any tangent to the hyperbola x2 – y2 = a2 is- (A) a2 (B) 2a2 (C) 3a2 (D) 4a2 Sol.[A] Any tangent to the hyperbola at P(a sec , a tan ) is x sec – y tan = a ...(i) Also x – y = 0 ...(ii) x + y = 0 ...(iii) Solving the above three lines in pairs, we get the point A, B, C as tan sec a , tan sec a , SOLVED EXAMPLES
- 8. tan sec a , tan sec a and (0, 0) Since the one vertex is the origin therefore the area of the triangle ABC is 2 1 (x1 y2 – x2 y1 ) = 2 a 2 2 2 2 2 tan sec 1 tan sec 1 = 2 a 2 (–2) = –a2 = a2 Ex.6. The locus of the mid point ofthe chords of the circle x2 + y2 = 16, which are tangent to the hyperbola 9x2 – 16y2 = 144 is- (A) x2 + y2 = a2 – b2 (B) (x2 + y2 )2 = a2 – b2 (C) (x2 + y2 )2 = a2 x2 – b2 y2 (D) (x2 + y2 )2 = a2 + b2 Sol.[C] Let (h, k) be the mid point of the chord of the circle x2 + y2 = a2 , so that its equation by T = S1 is hx + ky = h2 + k2 or y = – k h x + k k h 2 2 i.e. the form y = mx + c It will touch the hyperbola if c2 = a2 m2 – b2 2 2 2 k k h = a2 2 k h – b2 (h2 + k2 )2 = a2 h2 – b2 k2 Generalising, the locus of the mid-point (h, k) is (x2 + y2 )2 = a2 x2 – b2 y2 Ex.7 The eccentricity of the conic represented by x2 – y2 – 4x + 4y + 16 = 0 is- (A) 1 (B) 2 (C) 2 (D) 2 1 Sol.[B] We have x2 – y2 – 4x + 4y + 16 = 0 or (x2 – 4x) – (y2 – 4y) = – 16 or (x2 – 4x + 4) – (y2 – 4y + 4) = – 16 or (x – 2)2 – (y – 2)2 = – 16 or 2 2 4 ) 2 x ( – 2 2 4 ) 2 y ( = – 1 Shifting the origin at (2, 2), we obtain 2 2 4 x – 2 2 4 y = –1, where x = X + 2, y = Y + 2 This is rectangular hyperbola, whose eccentricity is always 2 . i.e. e2 = 1+ 2 2 a b = 1 + 2 2 4 4 ; e = 2 Ex.8 The equation 9x2 – 16y2 – 18x + 32y – 151 = 0 represent a hyperbola - (A) The length of the transverse axes is 4 (B) Length of latus rectum is 9 (C) Equation of directrix is x= 5 21 and x = – 5 11 (D) None of these Sol.[C] We have 9x2 – 16y2 – 18x + 32y – 151 = 0 9(x2 – 2x) – 16(y2 – 2y) = 151 9(x2 –2x+1) – 16(y2 – 2y + 1) = 144 9(x – 1)2 – 16(y – 1)2 = 144 16 ) 1 x ( 2 – 9 ) 1 y ( 2 = 1 Shifting the origin at (1, 1) without rotating the axes 16 x2 – 9 y2 = 1 where x = X + 1 and y = Y + 1 This is of the form 2 2 a x – 2 2 b y = 1 where a2 = 16 and b2 = 9 so The length of the transverse axes = 2a = 8 The length of the letus rectum = a b 2 2 = 2 a The equaiton of the directrix x = ± a e x – 1 = ± 5 16 x = ± 5 16 + 1 x = 5 21 ; x = – 5 11 Ex.9 The equations to the common tangents to the two hyperbola 2 2 a x – 2 2 b y = 1 & 2 2 a y – 2 2 b x = 1 are (A) y = ± x ± 2 2 a b (B) y = ± x ± 2 2 b a (C) y = ± x ± (a2 – b2 ) (D) y = ± x ± 2 2 b a Sol.[B] Any tangent to the hyperbola 2 2 a x – 2 2 b y = 1 is y = mx ± 2 2 2 b m a or y = mx + c where c = ± 2 2 2 b m a This will touch the hyperbola 2 2 a y – 2 2 b x = 1 if the equation 2 2 a ) c mx ( – 2 2 b x = 1 has equal roots or x2 (b2 m2 – a2 ) + 2b2 mcx + (c2 – a2 )b2 = 0 is an quadratic equation have equal roots
- 9. 4b4 m2 c2 = 4(b2 m2 – a2 ) (c2 – a2 )b2 c2 = a2 – b2 m2 a2 m2 – b2 = a2 – b2 m2 m2 (a2 + b2 ) = a2 + b2 m = ±1 Hence, the equations of the common tangents are y = ±x ± 2 2 b a Ex.10 For what value of does the line y = 2x + touches the hyperbola 16x2 – 9y2 = 144? Sol. Equation of hyperbola is 16x2 – 9y2 = 144 or 9 x2 – 16 y2 = 1 comparing this with 2 2 a x – 2 2 b y = 1, we get a2 = 9, b2 = 16 and comparing this line y = 2x + with m = mx + c ; m = 2 & c = If the line y = 2x + touches the hyperbola 16x2 – 9y2 = 144 then c2 = a2 m2 – b2 = 9(2)2 – 16 = 36 – 16 = 20; = ±25 Ex.11 Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the line x – y + 4 = 0. Sol. Let m be the slope of the tangent since the tangent is perpendicular to the line x – y + 4 = 0. m x 1 = –1 m = –1 since x2 – 4y2 = 36 or 36 x2 – 9 y2 = 1 Comparing this with 2 2 a x + 2 2 b y = 1; a2 = 36 & b2 = 9 so the equation of tangents are y = (– 1) x ± 9 ) 1 ( x 36 2 y = –x ±27 or x + y ± 33 = 0 EX.12 Find the locus of the point of intersection of tangents to the hyperbola 4x2 – 9y2 = 36 which meet at a constant angle /4. Sol. Let the point of intersection of tangents be P (x1 ,y1 ). Then the equation of pair of tangents from P(x1 ,y1 ) to the given hyperbola is (4x2 – 9y2 – 36) (4x1 2 – 9y1 2 – 36) = [4x1 x – 9y1 y – 36]2 . ...(1) [SS1 = T2 ] or x2 (y1 2 +4) +2x1 y1 xy+y2 (x1 2 – 9)+..= 0 ...(2) Since angle between the tangent is /4. tan 4 = 9 x 4 y 9 x 4 y y x 2 2 1 2 1 2 1 2 1 2 1 2 1 Hence locus of P (x1 , y1 ) is (x2 + y2 – 5)2 = 4(9y2 – 4x2 + 36) Ex.13 Prove that the locus of the middle points of chords of hyperbola 2 2 a x – 2 2 b y =1, passing through the fixed point (h,k) is a hyperbola whose centre is the point ( 2 1 h, 2 1 k) Sol. Using T= S1 , equation of chord having (x1 ,y1 ) as mid point is 2 1 a x x – 2 1 b y y = 2 2 1 a x – 2 2 1 b y It passes through (h,k) so 2 1 a h x – 2 1 b k y = 2 2 1 a x – 2 2 1 b y Locus of (x, y) is 2 2 a h 2 1 x – 2 2 b k 2 1 y = 4 1 2 2 2 2 b k a h which is a hyperbola whose centre is k 2 1 , h 2 1 . Ex.14 From the points on the circle x2 + y2 = a2 , tangents are drawn to the hyperbola x2 – y2 = a2 ; prove that the locus of the middle points of the chords of contact is the curve (x2 – y2 )2 = a2 (x2 + y2 ). Sol. Since any point on the circle x2 + y2 = a2 is (a cos, a sin) chord of contact of this point w.r.t hyperbola x2 – y2 = a2 is x (a cos) – y(a sin) = a2 or x cos – y sin = a ...(1) If its mid point be ( h,k) , then it is same as T = S1 i.e. hx – ky– a2 = h2 –k2 – a2 or hx – ky = h2 – k2 ...(2) Comparing (1) & (2) , we get h cos = k sin = ) k h ( a 2 2 or (h2 – k2 ) cos = ah ...(3) and (h2 – k2 ) sin= ak ...(4) Squaring and adding (3) and (4), we get (h2 – k2 )2 = a2 h2 + a2 k2 (h2 – k2 )2 = a2 (h2 + k2 ) Hence the required locus is (x2 – y2 )2 = a2 (x2 + y2 )