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Page 1 Solved Examples on Hyperbola Q1: The equation of the conic with focus at (?? , -?? ), directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Sol: Here, focus (?? ) = (1, -1), eccentricity (?? ) = v 2 From definition, ???? = ?????? v(?? - 1) 2 + (?? + 1) 2 = v 2 · (?? - ?? + 1) v1 2 + 1 2 ? (?? - 1) 2 + (?? + 1) 2 = (?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2: If the foci of the ellipse ?? ?? ???? + ?? ?? ?? ?? = ?? and the hyperbola ?? ?? ?????? - ?? ?? ???? = ?? ???? coincide, then the value of ?? ?? is (a) 1 (b) 5 (c) 7 (d) 9 Ans: (c) Sol: For hyperbola, ?? 2 144 - ?? 2 81 = 1 25 ?? = v 144 25 , ?? = v 81 25 , ?? 1 = v 1 + ?? 2 ?? 2 = v 1 + 81 144 = v 225 144 = 5 4 Therefore foci = (±?? ?? 1 , 0) = (± 12 5 · 5 4 , 0) = (±3,0). Therefore foci of ellipse i.e., (±4?? , 0) = (±3,0) (For ellipse ?? = 4) ? ?? = 3 4 , Hence ?? 2 = 16(1 - 9 16 ) = 7. Page 2 Solved Examples on Hyperbola Q1: The equation of the conic with focus at (?? , -?? ), directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Sol: Here, focus (?? ) = (1, -1), eccentricity (?? ) = v 2 From definition, ???? = ?????? v(?? - 1) 2 + (?? + 1) 2 = v 2 · (?? - ?? + 1) v1 2 + 1 2 ? (?? - 1) 2 + (?? + 1) 2 = (?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2: If the foci of the ellipse ?? ?? ???? + ?? ?? ?? ?? = ?? and the hyperbola ?? ?? ?????? - ?? ?? ???? = ?? ???? coincide, then the value of ?? ?? is (a) 1 (b) 5 (c) 7 (d) 9 Ans: (c) Sol: For hyperbola, ?? 2 144 - ?? 2 81 = 1 25 ?? = v 144 25 , ?? = v 81 25 , ?? 1 = v 1 + ?? 2 ?? 2 = v 1 + 81 144 = v 225 144 = 5 4 Therefore foci = (±?? ?? 1 , 0) = (± 12 5 · 5 4 , 0) = (±3,0). Therefore foci of ellipse i.e., (±4?? , 0) = (±3,0) (For ellipse ?? = 4) ? ?? = 3 4 , Hence ?? 2 = 16(1 - 9 16 ) = 7. Q3: If ???? is a double ordinate of hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? such that ?????? is an equilateral triangle, ?? being the centre of the hyperbola. Then the eccentricity ?? of the hyperbola satisfies (a) ?? < ?? < ?? /v ?? (b) ?? = ?? /v ?? (c) ?? = v ?? /?? (d) ?? > ?? /v ?? Ans: (d) Sol: Let ?? (?? sec ?? , ?? tan ?? ); ?? (?? sec ?? , -?? tan ?? ) be end points of double ordinates and ?? (0,0) is the centre of the hyperbola Now ???? = 2?? tan ?? ; ???? = ???? = v?? 2 sec 2 ?? + ?? 2 tan 2 ?? Since ???? = ???? = ???? , ? 4?? 2 tan 2 ?? = ?? 2 sec 2 ?? + ?? 2 tan 2 ?? ? 3?? 2 tan 2 ?? = ?? 2 sec 2 ?? ? 3?? 2 sin 2 ?? = ?? 2 ? 3?? 2 (?? 2 - 1)sin 2 ?? = ?? 2 = 3(?? 2 - 1)sin 2 ?? = 1 ? 1 3(?? 2 - 1) = sin 2 ?? < 1 (? sin 2 ?? < 1) ? 1 ?? 2 - 1 < 3 ? ?? 2 - 1 > 1 3 = ?? 2 > 4 3 ? ?? > 2 v 3 Q4: The eccentricity of the conjugate hyperbola of the hyperbola ?? ?? - ?? ?? ?? = ?? , is (a) 2 (b) ?? v ?? (c) 4 Page 3 Solved Examples on Hyperbola Q1: The equation of the conic with focus at (?? , -?? ), directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Sol: Here, focus (?? ) = (1, -1), eccentricity (?? ) = v 2 From definition, ???? = ?????? v(?? - 1) 2 + (?? + 1) 2 = v 2 · (?? - ?? + 1) v1 2 + 1 2 ? (?? - 1) 2 + (?? + 1) 2 = (?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2: If the foci of the ellipse ?? ?? ???? + ?? ?? ?? ?? = ?? and the hyperbola ?? ?? ?????? - ?? ?? ???? = ?? ???? coincide, then the value of ?? ?? is (a) 1 (b) 5 (c) 7 (d) 9 Ans: (c) Sol: For hyperbola, ?? 2 144 - ?? 2 81 = 1 25 ?? = v 144 25 , ?? = v 81 25 , ?? 1 = v 1 + ?? 2 ?? 2 = v 1 + 81 144 = v 225 144 = 5 4 Therefore foci = (±?? ?? 1 , 0) = (± 12 5 · 5 4 , 0) = (±3,0). Therefore foci of ellipse i.e., (±4?? , 0) = (±3,0) (For ellipse ?? = 4) ? ?? = 3 4 , Hence ?? 2 = 16(1 - 9 16 ) = 7. Q3: If ???? is a double ordinate of hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? such that ?????? is an equilateral triangle, ?? being the centre of the hyperbola. Then the eccentricity ?? of the hyperbola satisfies (a) ?? < ?? < ?? /v ?? (b) ?? = ?? /v ?? (c) ?? = v ?? /?? (d) ?? > ?? /v ?? Ans: (d) Sol: Let ?? (?? sec ?? , ?? tan ?? ); ?? (?? sec ?? , -?? tan ?? ) be end points of double ordinates and ?? (0,0) is the centre of the hyperbola Now ???? = 2?? tan ?? ; ???? = ???? = v?? 2 sec 2 ?? + ?? 2 tan 2 ?? Since ???? = ???? = ???? , ? 4?? 2 tan 2 ?? = ?? 2 sec 2 ?? + ?? 2 tan 2 ?? ? 3?? 2 tan 2 ?? = ?? 2 sec 2 ?? ? 3?? 2 sin 2 ?? = ?? 2 ? 3?? 2 (?? 2 - 1)sin 2 ?? = ?? 2 = 3(?? 2 - 1)sin 2 ?? = 1 ? 1 3(?? 2 - 1) = sin 2 ?? < 1 (? sin 2 ?? < 1) ? 1 ?? 2 - 1 < 3 ? ?? 2 - 1 > 1 3 = ?? 2 > 4 3 ? ?? > 2 v 3 Q4: The eccentricity of the conjugate hyperbola of the hyperbola ?? ?? - ?? ?? ?? = ?? , is (a) 2 (b) ?? v ?? (c) 4 (d) ?? ?? Ans: (a) Sol: The given hyperbola is ?? 2 1 - ?? 2 1/3 = 1. Here ?? 2 = 1 and ?? 2 = 1 3 Since ?? 2 = ?? 2 (?? 2 - 1) ? 1 3 = 1(?? 2 - 1) ? ?? 2 = 4 3 ? ?? = 2 v 3 If ?? ' is the eccentricity of the conjugate hyperbola, then 1 ?? 2 + 1 ?? '2 = 1 ? 1 ?? '2 = 1 - 1 ?? 2 = 1 - 3 4 = 1 4 ? ?? ' = 2. Q5:The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: (c) Sol: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ? ?? 8 = sec ?? , ?? 8 = tan ?? ? sec 2 ?? - tan 2 ?? = 1 ? ?? 2 8 2 - ?? 2 8 2 = 1 Here ?? = 8, ?? = 8. Now ?? = v1 + ?? 2 ?? 2 = v1 + 8 2 8 2 = v 2 ? Distance between directrices = 2?? ?? = 2×8 v 2 = 8v 2. Q6: If ?? ?? and ?? ?? are the slopes of the tangents to the hyperbola ?? ?? ???? - ?? ?? ???? = ?? which pass through the point (?? , ?? ), then (a) ?? ?? + ?? ?? = ???? ???? (b) ?? ?? ?? ?? = ???? ???? (c) ?? ?? + ?? ?? = ???? ???? (d) ?? ?? ?? ?? = ???? ???? Ans: (a, b) Page 4 Solved Examples on Hyperbola Q1: The equation of the conic with focus at (?? , -?? ), directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Sol: Here, focus (?? ) = (1, -1), eccentricity (?? ) = v 2 From definition, ???? = ?????? v(?? - 1) 2 + (?? + 1) 2 = v 2 · (?? - ?? + 1) v1 2 + 1 2 ? (?? - 1) 2 + (?? + 1) 2 = (?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2: If the foci of the ellipse ?? ?? ???? + ?? ?? ?? ?? = ?? and the hyperbola ?? ?? ?????? - ?? ?? ???? = ?? ???? coincide, then the value of ?? ?? is (a) 1 (b) 5 (c) 7 (d) 9 Ans: (c) Sol: For hyperbola, ?? 2 144 - ?? 2 81 = 1 25 ?? = v 144 25 , ?? = v 81 25 , ?? 1 = v 1 + ?? 2 ?? 2 = v 1 + 81 144 = v 225 144 = 5 4 Therefore foci = (±?? ?? 1 , 0) = (± 12 5 · 5 4 , 0) = (±3,0). Therefore foci of ellipse i.e., (±4?? , 0) = (±3,0) (For ellipse ?? = 4) ? ?? = 3 4 , Hence ?? 2 = 16(1 - 9 16 ) = 7. Q3: If ???? is a double ordinate of hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? such that ?????? is an equilateral triangle, ?? being the centre of the hyperbola. Then the eccentricity ?? of the hyperbola satisfies (a) ?? < ?? < ?? /v ?? (b) ?? = ?? /v ?? (c) ?? = v ?? /?? (d) ?? > ?? /v ?? Ans: (d) Sol: Let ?? (?? sec ?? , ?? tan ?? ); ?? (?? sec ?? , -?? tan ?? ) be end points of double ordinates and ?? (0,0) is the centre of the hyperbola Now ???? = 2?? tan ?? ; ???? = ???? = v?? 2 sec 2 ?? + ?? 2 tan 2 ?? Since ???? = ???? = ???? , ? 4?? 2 tan 2 ?? = ?? 2 sec 2 ?? + ?? 2 tan 2 ?? ? 3?? 2 tan 2 ?? = ?? 2 sec 2 ?? ? 3?? 2 sin 2 ?? = ?? 2 ? 3?? 2 (?? 2 - 1)sin 2 ?? = ?? 2 = 3(?? 2 - 1)sin 2 ?? = 1 ? 1 3(?? 2 - 1) = sin 2 ?? < 1 (? sin 2 ?? < 1) ? 1 ?? 2 - 1 < 3 ? ?? 2 - 1 > 1 3 = ?? 2 > 4 3 ? ?? > 2 v 3 Q4: The eccentricity of the conjugate hyperbola of the hyperbola ?? ?? - ?? ?? ?? = ?? , is (a) 2 (b) ?? v ?? (c) 4 (d) ?? ?? Ans: (a) Sol: The given hyperbola is ?? 2 1 - ?? 2 1/3 = 1. Here ?? 2 = 1 and ?? 2 = 1 3 Since ?? 2 = ?? 2 (?? 2 - 1) ? 1 3 = 1(?? 2 - 1) ? ?? 2 = 4 3 ? ?? = 2 v 3 If ?? ' is the eccentricity of the conjugate hyperbola, then 1 ?? 2 + 1 ?? '2 = 1 ? 1 ?? '2 = 1 - 1 ?? 2 = 1 - 3 4 = 1 4 ? ?? ' = 2. Q5:The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: (c) Sol: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ? ?? 8 = sec ?? , ?? 8 = tan ?? ? sec 2 ?? - tan 2 ?? = 1 ? ?? 2 8 2 - ?? 2 8 2 = 1 Here ?? = 8, ?? = 8. Now ?? = v1 + ?? 2 ?? 2 = v1 + 8 2 8 2 = v 2 ? Distance between directrices = 2?? ?? = 2×8 v 2 = 8v 2. Q6: If ?? ?? and ?? ?? are the slopes of the tangents to the hyperbola ?? ?? ???? - ?? ?? ???? = ?? which pass through the point (?? , ?? ), then (a) ?? ?? + ?? ?? = ???? ???? (b) ?? ?? ?? ?? = ???? ???? (c) ?? ?? + ?? ?? = ???? ???? (d) ?? ?? ?? ?? = ???? ???? Ans: (a, b) Sol: The line through (6,2) is ?? - 2 = ?? (?? - 6) ? ?? = ???? + 2 - 6?? Now, from condition of tangency (2 - 6?? ) 2 = 25?? 2 - 16 ? 36?? 2 + 4 - 24?? - 25?? 2 + 16 = 0 ? 11?? 2 - 24?? + 20 = 0 Obviously, its roots are ?? 1 and ?? 2 , therefore ?? 1 + ?? 2 = 24 11 and ?? 1 ?? 2 = 20 11 Q7: If the tangent at the point (???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° (c) ???? ° (d) ???? ° Ans: (c) Sol: Here ?? = 2sec ?? and ?? = 3tan ?? Differentiating w.r.t. ?? ???? ???? = 2sec ?? tan ?? and ???? ???? = 3sec 2 ?? ? Gradient of tangent ???? ???? = ???? /???? ???? /???? = 3sec 2 ?? 2sec ?? tan ?? ; ? ???? ???? = 3 2 cosec ?? But tangent is parallel to 3?? - ?? + 4 = 0; ? Gradient ?? = 3 From (i) and (ii), 3 2 cosec ?? = 3 ? cosec ?? = 2, ? ?? = 30 ° Q8: The locus of the point of intersection of tangents to the hyperbola ?? ?? ?? - ?? ?? ?? = ???? which meet at a constant angle ?? /?? , is (a) (?? ?? + ?? ?? - ?? ) ?? = ?? (?? ?? ?? - ?? ?? ?? + ???? ) (b) (?? ?? + ?? ?? - ?? ) = ?? (?? ?? ?? - ?? ?? ?? + ???? ) (c) ?? (?? ?? + ?? ?? - ?? ) ?? = (?? ?? ?? - ?? ?? ?? + ???? ) (d) None of these Ans: (a) Sol: Let the point of intersection of tangents be ?? (?? 1 , ?? 1 ). Then the equation of pair of tangents from ?? (?? 1 , ?? 1 ) to the given hyperbola is (4?? 2 - 9?? 2 - 36)(4?? 1 2 - 9?? 1 2 - 36) = [4?? 1 ?? - 9?? 1 ?? - 36] 2 Page 5 Solved Examples on Hyperbola Q1: The equation of the conic with focus at (?? , -?? ), directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Sol: Here, focus (?? ) = (1, -1), eccentricity (?? ) = v 2 From definition, ???? = ?????? v(?? - 1) 2 + (?? + 1) 2 = v 2 · (?? - ?? + 1) v1 2 + 1 2 ? (?? - 1) 2 + (?? + 1) 2 = (?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2: If the foci of the ellipse ?? ?? ???? + ?? ?? ?? ?? = ?? and the hyperbola ?? ?? ?????? - ?? ?? ???? = ?? ???? coincide, then the value of ?? ?? is (a) 1 (b) 5 (c) 7 (d) 9 Ans: (c) Sol: For hyperbola, ?? 2 144 - ?? 2 81 = 1 25 ?? = v 144 25 , ?? = v 81 25 , ?? 1 = v 1 + ?? 2 ?? 2 = v 1 + 81 144 = v 225 144 = 5 4 Therefore foci = (±?? ?? 1 , 0) = (± 12 5 · 5 4 , 0) = (±3,0). Therefore foci of ellipse i.e., (±4?? , 0) = (±3,0) (For ellipse ?? = 4) ? ?? = 3 4 , Hence ?? 2 = 16(1 - 9 16 ) = 7. Q3: If ???? is a double ordinate of hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? such that ?????? is an equilateral triangle, ?? being the centre of the hyperbola. Then the eccentricity ?? of the hyperbola satisfies (a) ?? < ?? < ?? /v ?? (b) ?? = ?? /v ?? (c) ?? = v ?? /?? (d) ?? > ?? /v ?? Ans: (d) Sol: Let ?? (?? sec ?? , ?? tan ?? ); ?? (?? sec ?? , -?? tan ?? ) be end points of double ordinates and ?? (0,0) is the centre of the hyperbola Now ???? = 2?? tan ?? ; ???? = ???? = v?? 2 sec 2 ?? + ?? 2 tan 2 ?? Since ???? = ???? = ???? , ? 4?? 2 tan 2 ?? = ?? 2 sec 2 ?? + ?? 2 tan 2 ?? ? 3?? 2 tan 2 ?? = ?? 2 sec 2 ?? ? 3?? 2 sin 2 ?? = ?? 2 ? 3?? 2 (?? 2 - 1)sin 2 ?? = ?? 2 = 3(?? 2 - 1)sin 2 ?? = 1 ? 1 3(?? 2 - 1) = sin 2 ?? < 1 (? sin 2 ?? < 1) ? 1 ?? 2 - 1 < 3 ? ?? 2 - 1 > 1 3 = ?? 2 > 4 3 ? ?? > 2 v 3 Q4: The eccentricity of the conjugate hyperbola of the hyperbola ?? ?? - ?? ?? ?? = ?? , is (a) 2 (b) ?? v ?? (c) 4 (d) ?? ?? Ans: (a) Sol: The given hyperbola is ?? 2 1 - ?? 2 1/3 = 1. Here ?? 2 = 1 and ?? 2 = 1 3 Since ?? 2 = ?? 2 (?? 2 - 1) ? 1 3 = 1(?? 2 - 1) ? ?? 2 = 4 3 ? ?? = 2 v 3 If ?? ' is the eccentricity of the conjugate hyperbola, then 1 ?? 2 + 1 ?? '2 = 1 ? 1 ?? '2 = 1 - 1 ?? 2 = 1 - 3 4 = 1 4 ? ?? ' = 2. Q5:The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: (c) Sol: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ? ?? 8 = sec ?? , ?? 8 = tan ?? ? sec 2 ?? - tan 2 ?? = 1 ? ?? 2 8 2 - ?? 2 8 2 = 1 Here ?? = 8, ?? = 8. Now ?? = v1 + ?? 2 ?? 2 = v1 + 8 2 8 2 = v 2 ? Distance between directrices = 2?? ?? = 2×8 v 2 = 8v 2. Q6: If ?? ?? and ?? ?? are the slopes of the tangents to the hyperbola ?? ?? ???? - ?? ?? ???? = ?? which pass through the point (?? , ?? ), then (a) ?? ?? + ?? ?? = ???? ???? (b) ?? ?? ?? ?? = ???? ???? (c) ?? ?? + ?? ?? = ???? ???? (d) ?? ?? ?? ?? = ???? ???? Ans: (a, b) Sol: The line through (6,2) is ?? - 2 = ?? (?? - 6) ? ?? = ???? + 2 - 6?? Now, from condition of tangency (2 - 6?? ) 2 = 25?? 2 - 16 ? 36?? 2 + 4 - 24?? - 25?? 2 + 16 = 0 ? 11?? 2 - 24?? + 20 = 0 Obviously, its roots are ?? 1 and ?? 2 , therefore ?? 1 + ?? 2 = 24 11 and ?? 1 ?? 2 = 20 11 Q7: If the tangent at the point (???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° (c) ???? ° (d) ???? ° Ans: (c) Sol: Here ?? = 2sec ?? and ?? = 3tan ?? Differentiating w.r.t. ?? ???? ???? = 2sec ?? tan ?? and ???? ???? = 3sec 2 ?? ? Gradient of tangent ???? ???? = ???? /???? ???? /???? = 3sec 2 ?? 2sec ?? tan ?? ; ? ???? ???? = 3 2 cosec ?? But tangent is parallel to 3?? - ?? + 4 = 0; ? Gradient ?? = 3 From (i) and (ii), 3 2 cosec ?? = 3 ? cosec ?? = 2, ? ?? = 30 ° Q8: The locus of the point of intersection of tangents to the hyperbola ?? ?? ?? - ?? ?? ?? = ???? which meet at a constant angle ?? /?? , is (a) (?? ?? + ?? ?? - ?? ) ?? = ?? (?? ?? ?? - ?? ?? ?? + ???? ) (b) (?? ?? + ?? ?? - ?? ) = ?? (?? ?? ?? - ?? ?? ?? + ???? ) (c) ?? (?? ?? + ?? ?? - ?? ) ?? = (?? ?? ?? - ?? ?? ?? + ???? ) (d) None of these Ans: (a) Sol: Let the point of intersection of tangents be ?? (?? 1 , ?? 1 ). Then the equation of pair of tangents from ?? (?? 1 , ?? 1 ) to the given hyperbola is (4?? 2 - 9?? 2 - 36)(4?? 1 2 - 9?? 1 2 - 36) = [4?? 1 ?? - 9?? 1 ?? - 36] 2 From ?? ?? 1 = ?? 2 or ?? 2 (?? 1 2 + 4) + 2?? 1 ?? 1 ???? + ?? 2 (?? 1 2 - 9) + ? . = 0 Since angle between the tangents is ?? /4. ? tan (?? /4) = 2v[?? 1 2 ?? 1 2 -(?? 1 2 +4)(?? 1 2 -9)] ?? 1 2 +4+?? 1 2 -9 . Hence locus of ?? (?? 1 , ?? 1 ) is (?? 2 + ?? 2 - 5) 2 = 4(9?? 2 - 4?? 2 + 36). Q9: The locus of the mid-points of the chords of the circle ?? ?? + ?? ?? = ???? which are tangent to the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? is (a) (?? ?? + ?? ?? ) ?? = ???? ?? ?? - ?? ?? ?? (b) (?? ?? + ?? ?? ) ?? = ?? ?? ?? - ???? ?? ?? (c) (?? ?? - ?? ?? ) ?? = ???? ?? ?? - ?? ?? ?? (d) None of these Ans: (a) Sol: The given hyperbola is ?? 2 16 - ?? 2 9 = 1 Any tangent to (i) is ?? = ???? + v16?? 2 - 9 Let (?? 1 , ?? 1 ) be the mid point of the chord of the circle ?? 2 + ?? 2 = 16 Then equation of the chord is ?? = ?? 1 i.e., ?? ?? 1 + ?? ?? 1 - (?? 1 2 + ?? 1 2 ) = 0 Since (ii) and (iii) represent the same line. ? ?? ?? 1 = -1 ?? 1 = v16?? 2 - 9 -(?? 1 2 + ?? 1 2 ) ? ?? = - ?? 1 ?? 1 and (?? 1 2 + ?? 1 2 ) 2 = ?? 1 2 (16?? 2 - 9) ? (?? 1 2 + ?? 1 2 ) 2 = 16 · ?? 1 2 ?? 1 2 ?? 1 2 - 9?? 1 2 = 16?? 1 2 - 9?? 1 2 ? Locus of (?? 1 , ?? 1 ) is (?? 2 + ?? 2 ) 2 = 16?? 2 - 9?? 2 . Q10: From any point on the hyperbola, ?? ?? ?? ?? - ?? ?? ?? ?? = ?? tangents are drawn to the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? . The area cut-off by the chord of contact on the asymptotes is equal to (a) ???? ?? (b) ???? (c) ?? ????Read More
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