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HYPERBOLA, , HYPERBOLA, 1. DEFINITION, A hyperbola is the locus of a point which moves in a plane so that the ratio of its distances from a fixed point, (called focus) and a fixed line (called directrix) is a constant which is greater than one. This ratio is called, eccentricity and is denoted by e. For a hyperbola e > 1., Let S be the focus, QN be the directrix and P be any point on the hyperbola. Then, by definition, , PS, or PS = e PN, e > 1,, PN, where PN is the length of the perpendicular from P on the directrix QN., An Alternate Definition, A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two, fixed points (called foci) is constant., , 2. EQUATION OF HYPERBOLA IN STANDARD FORM, The general form of standard hyperbola is, x2, a2, , , , y2, b2, , 1, , where a and b are constants., B, M', , P(x, y) L, , M, , N', , Rectum, C, , Z, , x = a/e, , x=-a/e, , X, S(ae, 0), Latus, , (-a, 0), , Z', , A(a, 0), , A', , S', (-ae, 0), , Directrix, , X', , Directrix, Conjugate, , Axis, , N, , L', , B', Y', , 3. TERMS RELATED TO A HYPERBOLA, A sketch of the locus of a moving point satisfying the equation, , x2, a2, , , , y2, b2, ,  1 , has been shown in the figure given, , above., Symmetry Since only even powers of x and y occur in the above equation, so the curve is symmetrical about, both the axes., Foci If S and S' are the two foci of the hyperbola and their coordinatesd are (ae, 0) and (–ae, 0) respectively,, then distance between foci is given by SS' = 2ae., Directries ZM and Z' M' are the two directrices of the hyperbola and their equations are x =, , a, and, e, , a, 2a, respectively, then the distance directrices is given by zz' =, ., e, e, Axes The lines AA' and BB' are called the transverse axis and conjugate axis respectively of the hyperbola., , x = –, , The length of transverse axis = AA' = 2a, The length of conjugate axis = BB' = 2b, 144

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HYPERBOLA, , Centre The point of intersection C of the axes of hyperbola is called the centre of the hyperbola. All chords,, passing through C, are bisected at C., Vertices The points A  (a, 0) and A'  (–a, 0) where the curve meets the line joining the foci S and S', are, called the vertices of the hyperbola., Focal Chord A chord of the hyperbola passing through its focus is called a focal chord., Focal Distances of a Point The difference of the focal distances of any point on the hyperbola is constant, and equal to the length of the transverse axis of the hyperbola. If P is any point on the hyperbola, then, S'P – SP = 2a = Transverse axis., Latus Rectum If LL' and NN' are the latus rectum of the hyperbola then these lines are perpendicular to the, transverse axis AA', passing through the foci S and S' respectively., , , b 2 , L   ae,, , a  ,, , , ,  b 2 , L'   ae,, , a  ,, , , , b 2 , N    ae,, , a  ,, , , ,  b 2 , N'    ae,, , a  ., , , 2b 2, = NN'., a, Eccentricity of the Hyperbola We know that, Length of latus rectum = LL' =, , SP = e PM, or, , SP2 = e2 PM2, , or, , a, , (x – ae) + (y – 0) = e N'   x  , e, , 2, 2, 2, (x – ae) + y = (ex – a), 2, , 2, , 2, , 2, , x2 + a2e2 – 2aex + y2 = e2x 2 – 2aex + a2, x2 (e2 – 1) – y2 = a2 (e2 – 1), , x2, a2, On comparing with, , , , x2, a2, , y2, a 2 (e 2  1), , , y2, b2, ,  1., ,  1 , we get, , b2 = a2 (e2 – 1), , or e =, , 1, , b2, a2, , 4. PARAMETRIC EQUATIONS OF THE HYPERBOLA, Since coordinates x = a sec  and y = b tan  satisfy the equation, x2, a2, , , , y2, b2, , 1, , x2 y2, for all real values of  therefore, x = a sec , y = b tan  are the parametric equations of the hyperbola 2  2  1 ,, a, b, where the parameter 0   < 2., 2, 2, x, y, Hence, the coordinates of any point on the hyperbola 2  2  1 may be taken as (a sec , b tan ). This, a, b, point is also called the point ''., , The angle  is called the eccentric angle of the point (a sec, b tan) on the hyperbola., Equation of Chord The equation of the chord joining the points, P  (a sec1, b tan1) and Q  (a sec 2, b tan 2) is, x,    2  y,    2 ,    2 , cos  1,   sin  1,   cos  1, , a,  2  b,  2 ,  2 , , x, or, , y, , 1, , a sec 1, , b tan 1 1  0, , a sec  2, , b tan  2 1, 145

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HYPERBOLA, Y, S(0, be), , 5. CONJUGATE HYPERBOLA, The hyperbola whose transverse and conjugate axes are respectively the, conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola, , B(0, b) y = b/e, , of the given hyperbola., , Z, X', , The conjugate hyperbola of the hyperbola., , a, , is, , , , 2, , , , x2, a, , 2, , y2, b, , , , B(0, -b) y = -b/e, , 1, , 2, , S'(0,-be), , x2, , X, , C, , , , x2 y2,  1  i.e., 2  2  1, , , b, a, b, , , y2, 2, , Y', , 6. PROPERTIES OF HYPERBOLA AND ITS CONJUGATE, , Hyperbola, , Standard equation, Centre, , x2, a2, , , , y2, , 1, , b2, , Conjugate Hyperbola,  x2, a2, , , , y2, b2, ,  1 or, , x2, a2, , , , y2, b2, ,  1, , (0, 0), , (0, 0), , Equation of transverse axis y = 0, , x = 0, , Equation of conjugate axis x = 0, , y = 0, , Length of transverse axis, , 2a, , 2b, , Length of Conjugate axis, , 2b, , 2a, , Foci, , (± ae, 0), , (0, ± be), , Equation of directrices, , x = ± a/e, , y = ± b/e, , Vertices, , (± a, 0), , (0, ± b), , Eccentricity, , e, , Length of latus rectum, , 2b 2, a, , 2a2, b, , Parameter Coordinates, , (a sec , b tan ), , (b sec , a tan ), , Focal radii, , SP = ex1 – a and S'P = ex1 + a, , SP = ey1 – b and S'P = ey1 + b, , Difference of focal, , 2a, , 2b, , x = ± a, , y = ± b, , a2  b2, a2, , e, , a2  b2, b2, , radii (S'P – SP), Tangent at the vertices, , 146

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HYPERBOLA, , Ex.1, , Find the eccentricity of the conic represented by x2 – y2 – 4x + 4y + 16 = 0., , Sol., , We have x2 – y2 – 4x + 4y + 16 = 0,  (x2 – 4x) – (y2 – 4y) = 16, , , ,  (x – 2)2 – (y – 2)2 = –16, , , , Shifting the origin at (2, 2), we obtain, , (x2 – 4x + 4) – (y2 – 4y + 4) = – 16, , ( x  2) 2, 42, , , , ( y  2)2, 42, , 1, , X2 Y 2, ,  1 , where x = X + 2, y + Y + 2, 42 4 2, , This is rectangular hyperbola, whose eccentricity is always, , 2., , Ex.2, , Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity, , Sol., , Let P(x, y) be any point on the hyperbola and PM is perpendicular from P on the directrix., , 3., , Then by definition, SP = e PM, , , (SP)2 = e2 (PM)2, , , ,  2 x  y  1, , (x – 1) + (y – 2) = 3 ,  4 1 , , , , 5(x2 + y2 – 2x – 4y + 5) = 3(4x2 + y2 + 1 + 4xy – 2y – 4x), , , , 7x2 – 2y2 + 12 xy – 2x + 9y – 22 = 0, , 2, , 2, , 2, , which is the required hyperbola., Ex.3, , Find the eccentricity of the hyperbola 16x2 – 32x – 3y2 + 12y = 44., , Sol., , We have, 16(x2 – 2x) – 3 (y2 – 4y) = 44,  16(x – 1)2 – 3 (y – 2)2 = 48, , , , ( x  1)2 ( y  2)2, , 1, 3, 16, , This equation represents a hyperbola with eccentricty given, 2, ,  4 ,  Conjugate axis , 19,  ,   1  , e  1  , , 3,  Transverse axis ,  3, , 7. POSITION OF A POINT WITH RESPECT TO A HYPERBOLA, The point P(x1, y1) lies outside, on or inside the hyperbola, < 0., , x2, a2, , , , y2, b2, ,  1 according as, , x 12, a2, , , , y 12, b2, ,  1  0. = 0 or, , 8. CONDITION FOR TRANGENCY AND POINTS OF CONTACT, The condition for the line y = mx + c to be a tangent to the hyperbola, and the coordinates of the points of contact are, , a 2m, b2, , ,, , a 2m 2  b 2, a 2m 2  b 2, , , x2, a2, , , , y2, b2, ,  1 is that c2 = a2m2 – b2, , , , , , 147

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HYPERBOLA, , 10. EQUATION OF NORMAL IN DIFFERENT FORMS, Point Form The equation of the normal to the hyperbola, , x2, a2, , , , y2, b2, ,  1 at the point (x , y ) is, 1, 1, , a 2 x b2 y, ,  a2  b2, x1, y1, Parametric Form The equation of the normal to the hyperbola, , x2, a, , 2, , , , y2, b2, ,  1 at the point (a sec, b tan) is, , ax, by, ,  a2  b2, sec  tan , , Slope Form The equation of normal to the hyperbola, , x2, a2, , , , y2, b2, ,  1 in terms of slope 'm' is, , m(a 2  b 2 ), , y = mx ±, , a 2  b 2m 2, , Notes :,  The coordinates of the points of contact are, , , a2, mb 2, , , ,, , 2, 2 2, 2, 2 2 , a, , b, m, , a, b, m, , ,  Number of Normals In general, four normals can be drawn to a hyperbola from a point in its plane i.e.,, , there are four points on the hyperbola, the normals at which will pass through a given point. These four points, are called the co-normal points.,  Tangent drawn at any point bisects the angle between the lines joining the point to the foci, whereas normal, bisects the supplementary angle between the lines., , 11. EQUATION OF THE PAIR OF TANGENTS, The equation of the pair of tangents drawn from a point, x2 y2, P(x1, y1) to the hyperbola 2  2  1 is, a, b, SS1 = T2, S, , where, , x2, , , , y2, , S1 , ,  1,, , a2 b2, xx, yy, T  21  21  1, a, b, , and, , x 12, a2, , , , y12, b2, , 1, , 12. CHORD WITH A GIVEN MID POINT, The equation of chord of the hyperbola, , T, , 13., , xx1, a2, , , , yy1, b2, ,  1 and S1 , , x 12, a2, , , , y12, b2, , x2, a2, , , , y2, b2, ,  1 with P(x , y ) as its middle point is given by T = S where, 1, 1, 1, , 1, , CHORD OF CONTACT, The equation of chord of contact of tangents drawn from a point P(x1, y1) to the hyperbola, T = 0, where T , , xx1, a, , 2, , , , yy1, b2, , x2, a2, , , , y2, b2, ,  1 is, , 1., , 149

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HYPERBOLA, , Ex.5, , 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 tanget to hyperbola, , x2 y2, ,  1 is y = mx +, 16 9, , (16m 2  9), , ....(i), , Let (x1, y1) be the mid-point of the chord of the circel x2 + y2 = 16, then equation of the chord is (T = S1), xx1 + yy1 – (x12 + y12) = 0, , ....(ii), , Since (i) and (ii) are same, comparing, we get, , m, 1, 16m2  9, , , x1, y1  ( x12  y12 ), , x1,  m = – y , (x12 + y12)2 = y12 (15m2 – 9), 1, , Eliminating m and generalizing (x1, y1) required locus is (x2 + y2)2 = 16x2 – 9y2., , 14. POLE AND POLAR, The polar of a point P(x1, y1) w.r.t. the hyperbola, , x2, a, , 2, , , , y2, b, , 2, ,  1 is T = 0, where T , , xx1, a, , 2, , , , yy1, b2, , 1, , Notes :,  Pole of a given line lx + my + n = 0 w.r.t. the hyperbola, x2, , , , y2, ,  1 is, , a2 b2,  Polar of the focus is the directrix., ,   a 2l  b 2 m , , , ,,  n, n , , ,  Any tangent is the polar of its point of contact.,  If the polar of P(x1, y1) passes through Q(x2, y2) then the polar of Q will pass through P and such points, are said to the conjugate points.,  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., , 15. EQUATION OF A DIAMETER OF A HYPERBOLA, The equation of the diameter bisecting chords of slope m of the hyperbola, , x2, a2, , , , y2, b2, ,  1 is y , , b2, a 2m, , ., , 16. CONJUGATE DIAMETERS, Two diameters of a hyperbola are said to be conjugate diameters if each bisects the chord parallel to the other., If m1 and m2 be the slopes of the conjugate diameters of a hyperbola, , x2, a2, , , , y2, b2, ,  1 , then m m =, 1 2, , b2, a2, , 17. ASYMPTOTES OF HYPERBOLA, x2, , y2, , bx, are called the asymptotes of the hyperbola., a, a, b, The curve comes close to these lines as x   or x  – but never meets them. In other words, asymtote, , The lines, , 2, , , , 2, ,  0 i.e., y = ±, , to a curve touches the curve at infinity., Note :, , , , b,  1 is 2tan–1   ., a, a, b, Asymptotes are the diagonals of the rectangle passing through A, B, A', B' with sides parallel to axes., , The angle between the asymptotes of, , x2, 2, , , , y2, 2, , 150

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HYPERBOLA, , , , A hyperbola and its conjugate hyperbola have the same asymptotes., , , , The asymptotes pass through the centre of the hyperbola., , , , The bisector of the angle between the asymptotes are the coordinates axes., , , , The product of the perpendicular from any point on the hyperbola, , x2, a2, , , , y2, b2, ,  1 to its asymptotes is a, , a 2b 2, , constant equal to, , , , ., a2  b2, Any line drawn parallel to the asymptote of the hyperbola would meet the curve only at one point., , , , A hyperbola and its conjugate hyperbola have the same asymptotes., , 18. RECTANGULAR HYPERBOLA, If asymptotes of the standard hyperbola are perpendicular to each other, then it is known as Rectangular Hyperbola., Then, b , ,  b = a or x2 – y2 = a2, a 2, is general form of the equation of the rectangular hyperbola., , 2 tan–1, , If we take the coordinate axes along the asymptotes of a rectangle hyperbola, then equation of rectangular, hyperbola becomes : xy = c2, where c is any constat., In parametric form, the equation of rectangular hyperbola, x = ct, y = c/t, where t is the parameter., The point (ct, c/t) on the hyperbola xy = c2 is generally referred as the point 't'., Properties of Rectangular Hyperbola, x2 – y2 = t2,  The equations of asymptotes of the rectangular hyperbola are y = ± x.,  The transverse and conjugate axes of a rectangular hyperbola are equal in length., 1, , b2, ,  2., a2, Properties of Rectangular hyperbola xy = c2, ,  Eccentricity, e =, ,  Equation of the chord joining 't1' and 't2' is, x + yt1t2 – c (t1 + t2) = 0,  Equation of tangnet at (x1, y1) is xy1 + x1y = 2c2 or,  Equation of tangent at 't' is :, , x, y, , 2, x1 y1, , x, + yt = 2c., t, ,  Point of intersection of tangents at 't1' and 't2' is,  Equation of normal at (x1, y1) is xx1 – yy1 = x12, ,  2ct1t 2 2c , , , ,,  t1  t 2 t 1t 2 , – y12., ,  Equation of normal at 't' is: xt3 – yt – ct4 + c = 0,  The equation of the chord of the hyperbola xy = c 2 whose middle point is (x 1, y1) is T = S1 i.e.,, xy 1 + x 1 y = 2x 1 y 1 .,  The slope of the tangent at the point (ct, c/t) is – 1/t2, which is always negative. Hence tangents drawn at, any point to xy = c2 would always make an obtuse angle with the x-axis.,  The slope of the normal at the point (ct, c/t) is t2 which is always positive. Hence normal drawn to, xy = c2 at any point would always make an acute angle with the x-axis., 151

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HYPERBOLA, , SOLVED EXAMPLES, Ex.1, , If eccentricities of a hyperbola and its conjugate hyperbola are e and e’ respectively, then, [1] 1, , Sol., , [2] 2, , [3] 0, x2, , Let equation of the hyperbola be, , a2, , , , y2, b2, , 1, e2, , 1, , , , e' 2, , is equal to, , [4] 3, ,  1., , The equation of its conjugate hyperbola will be , , x2 y2, , 1, a2 b 2, , Since their eccentricities are e and e’, so, b2, , e2 = 1 +, , a2, , a2, , e’2 = 1 +, , (1), (2) , Ex.2, , e, , 2, , , , 1, , a2  b2, , , , a2  b2, , Sol., , ....(2), , b2, , 1, , e' 2, , Ans.(1), , [2] 2 / 3, , [3] 4, 2, , [4] 4/3, 2, , Equation of the conjugate hyperbola to the hyperbola x – 3y = 1 is, , , , , , –x2 + 3y2 =1, , Ex.3, , ....(1), , a2, , The eccentricity of the conjugate hyperbola to the hyperbola x2 – 3y2 = 1 is[1] 2, , Sol., , 1, , b2, , , , Here, , a2 = 1, b2 = 1/3, , , , eccentricity e =, , x2 y2, , 1, 1 1/ 3, , Ans.(1), , 1  a2 / b2  1  3  2, , The equation of the common tangent to the parabola y2 = 8x and the hyperbola 3x2 – y2 = 3 is[1] 2x ± y + 1 = 0, [2] x ± y + 1 = 0, [3] x ± 2y + 1 = 0, [4] x ± y + 2 = 0, Parabola y2 = 8x,  4a = 8  a = 2, Any tangent to the parabola is, , y = mx +, , 2, m, , ...(i), , If it is also tangent to the hyperbola, , x2 y2, ,  1 i.e., a2 = 1, b2 = 3 then, 1, 3, 2, , 2, 2, c = a m – b     1.m  3, m, 2, , or, , , 2, , 2, , 2, , m4 – 3m2 – 4 = 0  (m2 – 4) (m2 + 1) = 0, m = ± 2 putting for m in (i), we get the tangents as 2x ± y + 1 = 0, , Ans.(1), , 152

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HYPERBOLA, , Ex.4, , The locus of the point of intersection of the lines, values of k is[1] Ellipse, , [2] Parabola, , 3 x  y  4 3 k  0 and, [3] Circle, , 3 kx  ky  4 3  0 for different, [4] Hyperbola, , 3 xy 4 3 k, , ....(i), , K( 3 x  y )  4 3, , ....(ii), , Sol., , To find the locus of their point of intersection eliminate the variable K between the equations from, (i), , K=, , 3x  y, 4 3, , and putting in (ii), we get, , ( 3 x  y)( 3 x  y)  3(4)2, 3x2 – y2 = 48, or, , Ex.5, Sol., , x2 y2, , 1, 16 48, , hence the locus is hyperbola., Ans.(4), 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[1] a2, [2] 2a2, [3] 3a2, [4] 4a2, 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, a, a, , , ,, , , sec, tan, , , , sec, , , tan, , , , , a, a, , ,  sec   tan  , sec   tan   and (0, 0), , , Since the one vertex is the origin therefore the area of the triangle ABC is, , , , Ex.6, , Sol., , 1, (x y – x2y1), 2 1 2, , 1, a2 , 1,  a2, , ( 2)  a 2  a 2, , , 2  sec 2   tan 2  sec 2   tan 2   2, , Ans.(1), , The locus of the mid point of the chords of the circle x2 + y2 = 16, which are tangent to the hyperbola, 9x2 – 16y2 =144 is, [1] x2 + y2 = a2 – b2, [2] (x2 + y2)2 = a2 – b2, [3] (x2 + y2)2 = a2x2 – b2y2[4] (x2 + y2)2 = a2 + b2, Let (h, k) be the mid point of the chord of the circle x2 + y2 = a2, so that its equation by T = S1 is, , h, h2  k 2, x, i.e., the form y = mx + c, k, k, It will touch the hyperbola if c2 = a2m2 – b2, , hx + ky = h2 + k2 or y = –, , 2, , , , 2,  h2  k 2 , ,   a 2   h   b 2,  k, ,  k, , , , (h2 + k2)2 = a2h2 – b2k2, Generalising, the locus of the mid point (h, k) is (x2 + y2)2 = a2x2 – b2y2, , Ans.(3), 153

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HYPERBOLA, , Ex.7, , Sol., , The length of LR of the hyperbola, 9x2 – 16y2 – 18x – 32y – 151 = 0 is, [1] 9/4, [2] 9, [3] 3/2, 2, 2, 9x – 16y – 18x – 32y – 151 = 0, , 9 (x – 1)2 – 16 (y + 1)2 = 144, , , ( x  1)2 ( y  1)2, , 1, 16, 9, , Here, , a2 = 16, b2 = 9, ,  length of LR, , =, , [4] 9/2, , 2b 2 2(9 ) 9, , , a, 4, 2, , Ans.(4), , Ex.8, , The number of normals which can be drawn from an external point on the hyperbola, , Sol., , [1] 2, [2] 4, Equation of any normal to the hyperbola is, , [3] 6, , x2, a2, , , , y2, b2, ,  1 is-, , [4] 8, , m(a 2  b 2 ), , y = mx –, , Ex.9, , , (a2 – b2m2) (y – mx)2 = m2(a2 + b2)2, If it passes through the point (x1, y1) then, (a2 – b2m2) (y1 – mx1)2 = m2(a2 + b2)2, It is 4 degree equation in m, so it gives 4 values of m. Corresponding to there 4 values, for normals can be drawn, from the point (x1, y1)., Ans.(2), 2, 2, x, y, ,  1 . If the eccentricity of the hyperbola be 2, then its equation, If foci of a hyperbola are foci of the ellipse, 25 9, is[1], , Sol., , x2 y2, , 1, 4 12, , For ellipse e =, , [2], , x2 y2, , 1, 12 4, , [3], , x2 y2, , 1, 4 12, , Ans.(1), , Line x cos  + y sin  = p is a normal to the hyperbola, , [1] a sec  – b cosec  =, 2, , [4] none of these, , ae 4,   2, b = 2 4  1  2 3, e, 2, , Hence equation of the hyperbola is, , 2, , x2 y2, , 1, 12 4, , 4, , so foci = (±4, 0), 5, , For hyperbola e = 2, so a =, , Ex.10, , a 2  b 2m 2, , 2, , 2, , [3] a2 cos2  – b2 sin2  =, , (a 2  b 2 ) 2, p2, , (a 2  b 2 ) 2, p2, , x2, a2, , , , y2, b2, ,  1 , if-, , [2] a sec  + b cosec  =, 2, , 2, , 2, , 2, , [4] a2 cos2  + b2 sin2  =, , (a 2  b 2 ) 2, p2, , (a 2  b 2 ) 2, p2, 154

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HYPERBOLA, , Ex.13, , Distance between directrices of the hyperbola x = 8 sec , y = 8 tan , [1] 16 2, , Sol., , [2] 8 2, 2, , [3] 6 2, , [4] 4 2, , 2, , Hyperbola is x – y = 64 which is rectangular hyperbola., So its e  2, , 2a 2(8), 8 2, Here a = 8. Hence distance between directrices  e , 2, Ex.14, , Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the line x – y + 4 = 0., [2] x + y  3 3, , [1] x – y + 3 = 0, Sol., , x2 – 4y2 = 36, , Comparing this with, y = (–1) x ±, , [4] x – y – 3 3, , x2, a2, , , , y2, b2, ,  1;, , or, ,  a2 = 36 & b2 = 9 so the equation of tangents are, ,  y = –x ±, , 36 x( 1)2  9, , x2 y2, , 1, 36 9, , 27, , or x + y ± 3, , 3 =0, , Ans.(2), , The product of the length of perpendiculars drawn from any point on the hyperbola x2 – 2y2 – 2 = 0 to its, asymptotes is, [1], , Sol., , [3] 2x + y  3 3, , Let m be the slope of the tangent since the tangent is perpendicular to the line x – y + 4 = 0., , m × 1 = –1,  m = –1, since, , Ex.15, , Ans.(2), , 1, 2, , 2, 3, , [2], , [3], , 3, 2, , [4] 2, , Any point on the given hyperbola is, , P( 2 sec , tan ), Asymptotes are x –, , 2y  0 ,, , x+, , 2y  0, , Product of perpendiculars from P on these asymptotes, , , , ( 2 sec   2 tan )( 2 sec   2 tan ), 1 2, , , , 2 sec 2   2 tan 2  2, , 3, 3, , Ans.(2), , 156

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HYPERBOLA, , EXERCISE # 1, Q.1, , Foci of the hyperbola, [1] (5, 2) (– 5, 2), , Q.2, , [2] (5, 2) (5, – 2), , [3] (5, 2) (– 5, – 2), , [4] none, , [3] [±, , [4] [0, ±, , The foci of the hyperbola 4x2 – 9y2 – 36 = 0 are, [1] [±, , Q.3, , x 2 (y  2)2, ,  1 are 16, 9, , 11 , 0], , [2] [±, , 12 , 0], , 13 , 0], , 12 ], , If the length of the transverse and conjugate axes of a hyperbola be 8 and 6 respectively, then the difference of, focal distances of any point of the hyperbola will be [1] 8, , Q.4, , Q.5, , [4] 2, , [1] 7x2 + 12xy – 2y2 – 2x + 4y – 7 = 0, , [2] 11x2 + 12xy + 2y2 – 10x – 4y + 1 = 0, , [3] 11x2 + 12xy + 2y2 – 14x – 14y + 1 = 0, , [4] none of these, , The equation of hyperbola whose foci are the foci of the ellipse, , x2 y2, , 1, 4 12, , [2], , x2 y2, , 1, 4 12, , [3], , 3 is -, , x2 y2, ,  1 and the eccentricity is 2, is 25 9, , x2 y2, , 1, 12 4, , [4], , x2 y2, , 1, 12 4, , The eccentricity of an ellipse is 2/3, latus rectum is 5 and centre is (0, 0). The equation of the ellipse is, , [1], , Q.7, , [3] 14, , The equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 1) and eccentricity =, , [1], , Q.6, , [2] 6, , x2 y2, , 1, 81 45, , [2], , 4x 2 4y 2, , 1, 81, 45, , [3], , x2 y2, , 1, 9, 5, , [4], , x2 y2, , 5, 81 45, , The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola, is, , [1], , Q.8, , x2, y2, , 1, 25 144, , [2], , (x  5)2, y2, x 2 (y  5)2, ,  1 [3], , 1, 25, 144, 25, 144, , [4], , (x  5)2 (y  5)2, , 1, 25, 144, , Equation of the hyprbola with eccentricity 3/2 and foci at (± 2, 0) is -, , [1], , x2 y2 4, , , 4, 5 9, , [2], , x2 y2 4, , , 9, 9 9, , [3], , x2 y2, , 1, 4, 9, , [4] none of these, , 157

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HYPERBOLA, , Q.9, , If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0) respectively, then the equation of the, hyperbola is [1] 4x2 – 5y2 = 8, , [2] 4x2 – 5y2 = 80, , [3] 5x2 – 4y2 = 80, , [4] 5x2 – 4y2 = 8, , 3, Q.10, , If the latus rectum of an hyperbola be 8 and eccentricity be, [1] 4x2 – 5y2 = 100, , Q.11, , Q.12, , [2] an ellipse if  > 0, , [3] a hyperbola if 8 <  < 12, , [4] none of these, , [2] ellipse, , [3] hyperbola, , [4] none of these, , The equation of the conic with focus at (1, – 1), directrix along x – y + 1 = 0 and with eccentricity, [2] xy = 1, , The eccentricity of the hyperbola –, , a2  b2, a2, , [2] e = +, , [3] 2xy – 4x + 4y + 1 = 0, , 2 is -, , [4] 2xy + 4x – 4y – 1 = 0, , x2 y2, ,  1 is given by a 2 b2, a2  b 2, a2, , [3] e = +, , b2  a2, a2, , [4] e = +, , a2  b2, b2, , If e and e' be the eccentricities of two conics S and S' such that e2 + e'2 = 3 then both S and S' are, [1] ellipse, , Q.16, , [4] 5x2 + 4y2 = 100, , x2, y2, ,  1, k > 1 represents The equation, 1 k 1 k, , [1] e = +, , Q.15, , [3] 4x2 + 5y2 = 100, , [1] a hyperbola if  < 8, , [1] x2 – y2 = 1, , Q.14, , , then the equation of the hyperbola is, , x2, y2, ,  1 represents The equation, 12   8  , , [1] circle, , Q.13, , [2] 5x2 – 4y2 = 100, , 5, , [2] Parabolas, , [3] hyperbolas, , [4] None of these, , x2 y2, The length of the latus rectum of the hyperbola 2  2 = – 1 is, a, b, 2a2, [1], b, , 2b2, [2], a, , b2, [3], a, , a2, [4], b, , 158

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HYPERBOLA, , Q.17, , The length of the transvere axis of a hyperbola is 7 and it passes through the point (5, – 2). The equation of the, hyperbola is, , Q.18, , [1], , 4 2 196 2, x –, y 1, 49, 51, , [2], , [3], , 4 2 51 2, x , y 1, 49, 196, , [4] none of these, , The locus of the point of intersect of the straight lines, [1] a circle, , Q.19, , Q.22, , Q.23, , [3] an ellipse, , [4] a hyperbola, , [2] 75x – 16y, , [3] 25x – 4y = 400, , [4] none of these, , The line y = x + 2 touches the hyperbola 5x2 – 9y2 = 45 at the point, [1] (0, 2), , Q.21, , [2] a parabola, , x y, x y 1,    and   ( is a variable) is, a b, a b , , Equation of the chord of the hyperbola 25x2 – 16y2 = 400 which is bisected at the point (6, 2) is, [1] 16x – 75y, , Q.20, , 49 2 51 2, x , y 1, 196, 196, , [2] (3, 1), , [3] (– 9/2, – 5/2), , Line lx + my + n = 0 is a tangent to the hyperbola, , x2, a, , 2, , , , y2, b2, , [4] none of these, ,  1 if, , [1] a2l2 – b2m2 = n2, , [2] a2l2 + m2 = n2b2, , [3] a2 + b2 = n2 (l2 + m2), , [4] a2l2 + b2m2 = n2, , The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is-, , [1], , ( x  1)2 ( y  5)2, , 1, 16, 9, , [2], , x2 y2, , 1, 16 9, , [3], , ( x  1)2 ( y  5 )2, ,  1, 16, 9, , [4] none of these, , The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance, between the foci, is-, , [1], Q.24, , 4, 3, , 3, , 2, [3], , 3, , [4] none of these, , The slope of the tangent to the hyperbola 2x2 – 3y2 = 6 at (3, 2) is[1] –1, , Q.25, , 4, [2], , [2] 1, , [3] 0, , [4] 2, , What will be the equation of the chord of hyperbola 25x2 – 16y2 = 400, whose mid point is (5, 3) [1] 115x – 117y = 17, , [2] 125x – 48y = 481, , [3] 127x + 33y = 341, , [4] 15x + 121y = 105, , 159

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HYPERBOLA, , EXERCISE # 3, Q.1, , 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-, , Q.2, , [1] 16x2 + 10xy + y2 = 2, , [2] 16x2 – 10xy + y2 = 2, , [3] 16x2 + 10xy + y2 = 4, , [4] none of these, , If a circle cuts the rectangular hyperbola xy = 1 in the points (xr, yr) where r = 1,2,3,4 then, [1] x1x2x3x4 = 2, , Q.3, , [IIT-97], , [2] x1x2x3x4 = 1, , [IIT-98], , [3] x1 + x2 + x3 + x4 = 0 [4] y1 + y2 + y3 + y4 = 0, , If x = 9 is the chord of contact of the hyperbola x2 – y2 = 9, then the equation of the corresponding pair of tangents, is-, , Q.4, , [IIT-99], , [1] 9x2 – 8y2 + 18x – 9 = 0, , [2] 9x2 – 8y2 – 18x + 9 = 0, , [3] 9x2 – 8y2 – 18x – 9 = 0, , [4] 9x2 – 8y2 + 18x + 9 = 0, , At which of the following points of the hyperbola x2 – y2 = 3, tangent is parallel to the line 2x + y + 8 = 0?[REE 99], [1] (2, 1), , Q.5, , [4] none of these, , x2 y2, , , be two points on the hyperbola 2  2  1., 2, a, b, If (h, k) is the point of intersection of the normals at P and Q, then K is equal to[IIT-2000], , a2  b2, a, , [2] , , a2  b2, a, , [3], , a2  b2, b, , [4] , , a2  b2, b, , The equation of common tangent of the curve y2 = 8x and xy = –1 is, [1] 3y = 9x + 2, , Q.7, , [3] (–2, –1), , Let P (a sec , b tan ) and Q(a sec , b tan ) where  +  =, , [1], Q.6, , [2] (2, –1), , In the hyperbola, , [2] y = x + 2, x2, cos 2 , , , , y2, sin 2 , , [3] 2y = x + 8, , [IIT-2002], [4] y = 2x + 1, ,  1 , corresponding to any chnage in , which of the following is not changed-, , [IIT-2003], , Q.8, , [1] abscissae of vertices, , [2] abscissae of foci, , [3] eccentricity, , [4] directrices, , The locus of a point P(a, b) moving under the condition that the line y = ax + b is a tangent to the hyperbola, , x 2 y2,   1 is, a 2 b2, , [AIEEE-2005], , [1] a hyperbola, , Q.9, , For the hyperbola, , [2] a parabola, , [3] a circle, , [4] an ellipse, , x2, y2, , 1 , which of the following remains constant when  varies? [AIEEE-2007], cos 2  sin 2 , , [1] Abscissae of vertices, , [2] Abscissae of foci, , [3] Eccentricity, , [4] Directrix, 162

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HYPERBOLA, , Q.10, , Match the conics in column-I with statements/expressions in column-II :, Column-I, , Column-II, , A. Circle, , p. The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4, , B. Parabola, , q. Points z in the complex plane satisfying |z + 2| – |z – 2| = ± 3, , C. Ellipse, ,  1  t2 , 2t, , y , r. Points of the conic have parametric representation x  3 , 2 , 1  t2, 1  t , , D. Hyperbola, , s. The eccentricity of the conic lies in the interval 1  x < , t. Points z in the complex plane satisfying Re(z + 1)2 = |z|2 + 1, , Codes :, , Q.11, , [IIT-JEE-2009], , [1] (A, p), (B, (s, t)), (C, r), (D, (q, s)), , [2] (A, p), (B, s), (C, r), (D, q), , [3] (A, q), (B, p), (C, s), (D, t), , [4] (A, (p, t)), (B, (s, t)), (C, q), (D, r), x2 y2, ,  1 intersect at points A and B. Equation of a common tangent, 9, 4, , The circle x2 + y2 – 8x = 0 and hyperbola, , with positive slope to the circle as well as to the hyperbola is :, [1] 2 x  y  20  0, Q.12, , [2] 2 x  5 y  4  0, , Let P(6, 3) be a point on the hyperbola, , [3] 3x  4 y  8  0, , [IIT-JEE-2010], [4] 4 x  3y  4  0, , x2 y2, , = 1. If the normal at the point P intersects the x-axis at (9, 0),, a2 b 2, , then the eccentricity of the hyperbola is :, [1], , Q.13, , 5, 2, , [2], , [IIT-JEE-2011], , 3, 2, , Let the eccentricity of the hyperbola, , [3], , 2, , [4], , 3, , x2 y2, , = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola, a2 b 2, , passes through a focus of the ellipse, then :, [1] the equation of the hyperbola is, , x2 y2, , 1, 3, 2, , [3] the eccentricity of the hyperbola is, , Q.14, , The circle x2 + y2 – 8x = 0 and hyperbola, as diameter is :, , 5, 3, , [IIT-JEE-2011], [2] a focus of the hyperbola is (2, 0), , [4] the equation of the hyperbola is x2 – 3y2 = 3, , x2 y2, ,  1 intersect at points A and B. Equation of the circle with AB, 9, 4, [IIT-JEE - 2010], , [1] x2 + y2 – 12x + 24 = 0 [2] x2 + y2 + 12x + 24 = 0 [3] x2 + y2 + 24x – 12 = 0 [4] x2 + y2 – 24x – 12 = 0, , 163

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HYPERBOLA, , ANSWER KEY, EXERCISE - 1, Que., Ans., , 1, 1, , 2, 3, , 3, 1, , 4, 1, , 5, 2, , 6, 2, , 7, 2, , 8, 1, , 9, 3, , 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25, 1, 3, 4, 3, 4, 3, 1, 3, 4, 2, 3, 1, 1, 3, 2, 2, , EXERCISE - 2, Q u e ., A n s., , 1, 2, , 2, 2, , 3, 3, , 4, 2, , 5, 1, , 6, 1, , 7, 2, , 8, 3, , 9, 2, , 10, 2, , 11, 2, , 12, 1, , 13, 4, , 14, 1, , EXERCISE - 3, , 164