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Page 1 JEE Mains Previous Year Questions (2021-2024): Hyperbola 2024 Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : A. ?? ?? B. ?? ?? ???? C. ?? ?? D. ?? ?? [JEE Main 2024 (Online) 1st February Morning Shift] Ans: (c) Explanation: Currently no explanation available Q2: If the foci of a hyperbola are same as that of the ellipse ?? ?? ?? + ?? ?? ???? = ?? and the eccentricity of the hyperbola is ???? ?? times the eccentricity of the ellipse, then the smaller focal distance of the point ( v ?? , ???? ?? v ?? ?? ) on the hyperbola, is equal to A. ???? v ?? ?? - ?? ?? B. ?? v ?? ?? + ?? ?? C. ?? v ?? ?? - ?? ?? D. ???? v ?? ?? - ???? ?? [JEE Main 2024 (Online) 31st January Morning Shift] Ans: (c) x 2 9 + y 2 25 = 1 a = 3 , b = 5 e = v 1 - 9 25 = 4 5 ? foci = ( 0 , ± be ) = ( 0 , ± 4 ) ? ? e H = 4 5 × 15 8 = 3 2 Let equation hyperbola Page 2 JEE Mains Previous Year Questions (2021-2024): Hyperbola 2024 Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : A. ?? ?? B. ?? ?? ???? C. ?? ?? D. ?? ?? [JEE Main 2024 (Online) 1st February Morning Shift] Ans: (c) Explanation: Currently no explanation available Q2: If the foci of a hyperbola are same as that of the ellipse ?? ?? ?? + ?? ?? ???? = ?? and the eccentricity of the hyperbola is ???? ?? times the eccentricity of the ellipse, then the smaller focal distance of the point ( v ?? , ???? ?? v ?? ?? ) on the hyperbola, is equal to A. ???? v ?? ?? - ?? ?? B. ?? v ?? ?? + ?? ?? C. ?? v ?? ?? - ?? ?? D. ???? v ?? ?? - ???? ?? [JEE Main 2024 (Online) 31st January Morning Shift] Ans: (c) x 2 9 + y 2 25 = 1 a = 3 , b = 5 e = v 1 - 9 25 = 4 5 ? foci = ( 0 , ± be ) = ( 0 , ± 4 ) ? ? e H = 4 5 × 15 8 = 3 2 Let equation hyperbola x 2 A 2 - y 2 B 2 = - 1 ? ? B · e H = 4 ? ? B = 8 3 ? ? A 2 = B 2 ( e H 2 - 1 ) = 64 9 ( 9 4 - 1 ) ? A 2 = 80 9 ? ? x 2 80 9 - y 2 64 9 = - 1 Directrix : y = ± B e H = ± 16 9 PS = e · PM = 3 2 | 14 3 · v 2 5 - 16 9 | ? = 7 v 2 5 - 8 3 Q3: Let ?? be a point on the hyperbola ?? : ?? ?? ?? - ?? ?? ?? = ?? , in the first quadrant such that the area of triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the origin is A. 26 B. 22 C. 20 D. 18 [JEE Main 2024 (Online) 30th January Evening Shift] Ans: (b) Page 3 JEE Mains Previous Year Questions (2021-2024): Hyperbola 2024 Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : A. ?? ?? B. ?? ?? ???? C. ?? ?? D. ?? ?? [JEE Main 2024 (Online) 1st February Morning Shift] Ans: (c) Explanation: Currently no explanation available Q2: If the foci of a hyperbola are same as that of the ellipse ?? ?? ?? + ?? ?? ???? = ?? and the eccentricity of the hyperbola is ???? ?? times the eccentricity of the ellipse, then the smaller focal distance of the point ( v ?? , ???? ?? v ?? ?? ) on the hyperbola, is equal to A. ???? v ?? ?? - ?? ?? B. ?? v ?? ?? + ?? ?? C. ?? v ?? ?? - ?? ?? D. ???? v ?? ?? - ???? ?? [JEE Main 2024 (Online) 31st January Morning Shift] Ans: (c) x 2 9 + y 2 25 = 1 a = 3 , b = 5 e = v 1 - 9 25 = 4 5 ? foci = ( 0 , ± be ) = ( 0 , ± 4 ) ? ? e H = 4 5 × 15 8 = 3 2 Let equation hyperbola x 2 A 2 - y 2 B 2 = - 1 ? ? B · e H = 4 ? ? B = 8 3 ? ? A 2 = B 2 ( e H 2 - 1 ) = 64 9 ( 9 4 - 1 ) ? A 2 = 80 9 ? ? x 2 80 9 - y 2 64 9 = - 1 Directrix : y = ± B e H = ± 16 9 PS = e · PM = 3 2 | 14 3 · v 2 5 - 16 9 | ? = 7 v 2 5 - 8 3 Q3: Let ?? be a point on the hyperbola ?? : ?? ?? ?? - ?? ?? ?? = ?? , in the first quadrant such that the area of triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the origin is A. 26 B. 22 C. 20 D. 18 [JEE Main 2024 (Online) 30th January Evening Shift] Ans: (b) ?? 2 9 - ?? 2 4 = 1 ?? 2 = 9 , ?? 2 = 4 ?? 2 = ?? 2 ( ?? 2 - 1 ) ? ?? 2 = 1 + ?? 2 ?? 2 ?? 2 = 1 + 4 9 = 13 9 ?? = v 13 3 ? ?? 1 ?? 2 = 2ae = 2 × 3 × v 13 3 = 2 v 13 Area of ? PS 1 S 2 = 1 2 × ?? × S 1 S 2 = 2 v 13 ? ? 1 2 × ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2 ?? 2 9 - ?? 2 4 = 1 ? ?? 2 9 - 1 = 1 ? ?? 2 = 18 ? ?? = 3 v 2 Distance of P from origin = v ?? 2 + ?? 2 = v 18 + 4 = v 22 Q4: 4Let ?? ?? be the eccentricity of the hyperbola ?? ?? ???? - ?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : A. ?? v ?? ?? B. ?? v ?? C. ?? v ?? D. ???? v ?? ?? [JEE Main 2024 (Online) 27th January Evening Shift] Ans: (d) Explanation: Currently no explanation available Q5: Let the foci and length of the latus rectum of an ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , respectively. Then, the square of the eccentricity of the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? ?? ?? = ?? equals [JEE Main 2024 (Online) 31st January Morning Shift] Ans: 51 Page 4 JEE Mains Previous Year Questions (2021-2024): Hyperbola 2024 Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : A. ?? ?? B. ?? ?? ???? C. ?? ?? D. ?? ?? [JEE Main 2024 (Online) 1st February Morning Shift] Ans: (c) Explanation: Currently no explanation available Q2: If the foci of a hyperbola are same as that of the ellipse ?? ?? ?? + ?? ?? ???? = ?? and the eccentricity of the hyperbola is ???? ?? times the eccentricity of the ellipse, then the smaller focal distance of the point ( v ?? , ???? ?? v ?? ?? ) on the hyperbola, is equal to A. ???? v ?? ?? - ?? ?? B. ?? v ?? ?? + ?? ?? C. ?? v ?? ?? - ?? ?? D. ???? v ?? ?? - ???? ?? [JEE Main 2024 (Online) 31st January Morning Shift] Ans: (c) x 2 9 + y 2 25 = 1 a = 3 , b = 5 e = v 1 - 9 25 = 4 5 ? foci = ( 0 , ± be ) = ( 0 , ± 4 ) ? ? e H = 4 5 × 15 8 = 3 2 Let equation hyperbola x 2 A 2 - y 2 B 2 = - 1 ? ? B · e H = 4 ? ? B = 8 3 ? ? A 2 = B 2 ( e H 2 - 1 ) = 64 9 ( 9 4 - 1 ) ? A 2 = 80 9 ? ? x 2 80 9 - y 2 64 9 = - 1 Directrix : y = ± B e H = ± 16 9 PS = e · PM = 3 2 | 14 3 · v 2 5 - 16 9 | ? = 7 v 2 5 - 8 3 Q3: Let ?? be a point on the hyperbola ?? : ?? ?? ?? - ?? ?? ?? = ?? , in the first quadrant such that the area of triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the origin is A. 26 B. 22 C. 20 D. 18 [JEE Main 2024 (Online) 30th January Evening Shift] Ans: (b) ?? 2 9 - ?? 2 4 = 1 ?? 2 = 9 , ?? 2 = 4 ?? 2 = ?? 2 ( ?? 2 - 1 ) ? ?? 2 = 1 + ?? 2 ?? 2 ?? 2 = 1 + 4 9 = 13 9 ?? = v 13 3 ? ?? 1 ?? 2 = 2ae = 2 × 3 × v 13 3 = 2 v 13 Area of ? PS 1 S 2 = 1 2 × ?? × S 1 S 2 = 2 v 13 ? ? 1 2 × ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2 ?? 2 9 - ?? 2 4 = 1 ? ?? 2 9 - 1 = 1 ? ?? 2 = 18 ? ?? = 3 v 2 Distance of P from origin = v ?? 2 + ?? 2 = v 18 + 4 = v 22 Q4: 4Let ?? ?? be the eccentricity of the hyperbola ?? ?? ???? - ?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : A. ?? v ?? ?? B. ?? v ?? C. ?? v ?? D. ???? v ?? ?? [JEE Main 2024 (Online) 27th January Evening Shift] Ans: (d) Explanation: Currently no explanation available Q5: Let the foci and length of the latus rectum of an ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , respectively. Then, the square of the eccentricity of the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? ?? ?? = ?? equals [JEE Main 2024 (Online) 31st January Morning Shift] Ans: 51 focii = ( ± 5 , 0 ) ; 2 ?? 2 ?? = v 50 ?? = 5 ? ?? 2 = 5 v 2 ?? 2 ?? 2 = ?? 2 ( 1 - ?? 2 ) = 5 v 2 ?? 2 ? ? a ( 1 - e 2 ) = 5 v 2 2 ? ? 5 e ( 1 - e 2 ) = 5 v 2 2 ? ? v 2 - v 2 e 2 = e ? ? v 2 e 2 + e - v 2 = 0 ? ? v 2 e 2 + 2e - e - v 2 = 0 ? ? v 2 e ( e + v 2 ) - 1 ( 1 + v 2 ) = 0 ? ? ( e + v 2 ) ( v 2 e - 1 ) = 0 ? ? e ? - v 2 ; e = 1 v 2 x 2 b 2 - y 2 a 2 b 2 = 1 ? a = 5 v 2 ?? = 5 ?? 2 ?? 2 = ?? 2 ( ?? 1 2 - 1 ) ? ?? 1 2 = 51 Q6: Let the latus rectum of the hyperbola ?? ?? ?? - ?? ?? ?? ?? = ?? subtend an angle of ?? ?? at the centre of the hyperbola. If ?? ?? is equal to ?? ?? ( ?? + v ?? ) , where ?? and ?? are coprime numbers, then ?? ?? + ?? ?? + ?? ?? is equal to [JEE Main 2024 (Online) 30th January Morning Shift] Ans: 182 LR subtends 60 ° at centre Page 5 JEE Mains Previous Year Questions (2021-2024): Hyperbola 2024 Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : A. ?? ?? B. ?? ?? ???? C. ?? ?? D. ?? ?? [JEE Main 2024 (Online) 1st February Morning Shift] Ans: (c) Explanation: Currently no explanation available Q2: If the foci of a hyperbola are same as that of the ellipse ?? ?? ?? + ?? ?? ???? = ?? and the eccentricity of the hyperbola is ???? ?? times the eccentricity of the ellipse, then the smaller focal distance of the point ( v ?? , ???? ?? v ?? ?? ) on the hyperbola, is equal to A. ???? v ?? ?? - ?? ?? B. ?? v ?? ?? + ?? ?? C. ?? v ?? ?? - ?? ?? D. ???? v ?? ?? - ???? ?? [JEE Main 2024 (Online) 31st January Morning Shift] Ans: (c) x 2 9 + y 2 25 = 1 a = 3 , b = 5 e = v 1 - 9 25 = 4 5 ? foci = ( 0 , ± be ) = ( 0 , ± 4 ) ? ? e H = 4 5 × 15 8 = 3 2 Let equation hyperbola x 2 A 2 - y 2 B 2 = - 1 ? ? B · e H = 4 ? ? B = 8 3 ? ? A 2 = B 2 ( e H 2 - 1 ) = 64 9 ( 9 4 - 1 ) ? A 2 = 80 9 ? ? x 2 80 9 - y 2 64 9 = - 1 Directrix : y = ± B e H = ± 16 9 PS = e · PM = 3 2 | 14 3 · v 2 5 - 16 9 | ? = 7 v 2 5 - 8 3 Q3: Let ?? be a point on the hyperbola ?? : ?? ?? ?? - ?? ?? ?? = ?? , in the first quadrant such that the area of triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the origin is A. 26 B. 22 C. 20 D. 18 [JEE Main 2024 (Online) 30th January Evening Shift] Ans: (b) ?? 2 9 - ?? 2 4 = 1 ?? 2 = 9 , ?? 2 = 4 ?? 2 = ?? 2 ( ?? 2 - 1 ) ? ?? 2 = 1 + ?? 2 ?? 2 ?? 2 = 1 + 4 9 = 13 9 ?? = v 13 3 ? ?? 1 ?? 2 = 2ae = 2 × 3 × v 13 3 = 2 v 13 Area of ? PS 1 S 2 = 1 2 × ?? × S 1 S 2 = 2 v 13 ? ? 1 2 × ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2 ?? 2 9 - ?? 2 4 = 1 ? ?? 2 9 - 1 = 1 ? ?? 2 = 18 ? ?? = 3 v 2 Distance of P from origin = v ?? 2 + ?? 2 = v 18 + 4 = v 22 Q4: 4Let ?? ?? be the eccentricity of the hyperbola ?? ?? ???? - ?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : A. ?? v ?? ?? B. ?? v ?? C. ?? v ?? D. ???? v ?? ?? [JEE Main 2024 (Online) 27th January Evening Shift] Ans: (d) Explanation: Currently no explanation available Q5: Let the foci and length of the latus rectum of an ellipse ?? ?? ?? ?? + ?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , respectively. Then, the square of the eccentricity of the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? ?? ?? = ?? equals [JEE Main 2024 (Online) 31st January Morning Shift] Ans: 51 focii = ( ± 5 , 0 ) ; 2 ?? 2 ?? = v 50 ?? = 5 ? ?? 2 = 5 v 2 ?? 2 ?? 2 = ?? 2 ( 1 - ?? 2 ) = 5 v 2 ?? 2 ? ? a ( 1 - e 2 ) = 5 v 2 2 ? ? 5 e ( 1 - e 2 ) = 5 v 2 2 ? ? v 2 - v 2 e 2 = e ? ? v 2 e 2 + e - v 2 = 0 ? ? v 2 e 2 + 2e - e - v 2 = 0 ? ? v 2 e ( e + v 2 ) - 1 ( 1 + v 2 ) = 0 ? ? ( e + v 2 ) ( v 2 e - 1 ) = 0 ? ? e ? - v 2 ; e = 1 v 2 x 2 b 2 - y 2 a 2 b 2 = 1 ? a = 5 v 2 ?? = 5 ?? 2 ?? 2 = ?? 2 ( ?? 1 2 - 1 ) ? ?? 1 2 = 51 Q6: Let the latus rectum of the hyperbola ?? ?? ?? - ?? ?? ?? ?? = ?? subtend an angle of ?? ?? at the centre of the hyperbola. If ?? ?? is equal to ?? ?? ( ?? + v ?? ) , where ?? and ?? are coprime numbers, then ?? ?? + ?? ?? + ?? ?? is equal to [JEE Main 2024 (Online) 30th January Morning Shift] Ans: 182 LR subtends 60 ° at centre ? ? t a n ? 30 ° = b 2 / a ae = b 2 a 2 e = 1 v 3 ? ? e = v 3 b 2 9 Also, e 2 = 1 + b 2 9 ? 1 + b 2 9 = 3 b 4 81 ? ? ?? 4 = 3 ?? 2 + 27 ? ? ?? 4 - 3 ?? 2 - 27 = 0 ? ? ?? 2 = 3 2 ( 1 + v 13 ) ? ? l = 3 , m = 2 , n = 13 ? ? l 2 + m 2 + n 2 = 182Read More
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