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Page 1 Solved Examples on Hyperbola JEE Mains Q1. The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 From definition, ???? = ?????? v( ?? - 1) 2 + ( ?? + 1) 2 = v 2 · ( ?? - ?? + 1) v 1 2 + 1 2 ? ( ?? - 1) 2 + ( ?? + 1) 2 = ( ?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2. The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are (a) ( ±?? , ?? ) (b) ( ?? , ±?? ) (c) ( ±?? , ?? ) (d) ( ?? , ±?? ) Ans: (c) The equation of hyperbola is ?? 2 16 - ?? 2 9 = 1. Now, ?? 2 = ?? 2 ( ?? 2 - 1) ? 9 = 16( ?? 2 - 1) ? ?? = 5 4 . Hence foci are ( ±???? , 0) = (±4. 5 4 , 0) i.e., ( ±5,0) Q3. 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) 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 JEE Mains Q1. The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 From definition, ???? = ?????? v( ?? - 1) 2 + ( ?? + 1) 2 = v 2 · ( ?? - ?? + 1) v 1 2 + 1 2 ? ( ?? - 1) 2 + ( ?? + 1) 2 = ( ?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2. The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are (a) ( ±?? , ?? ) (b) ( ?? , ±?? ) (c) ( ±?? , ?? ) (d) ( ?? , ±?? ) Ans: (c) The equation of hyperbola is ?? 2 16 - ?? 2 9 = 1. Now, ?? 2 = ?? 2 ( ?? 2 - 1) ? 9 = 16( ?? 2 - 1) ? ?? = 5 4 . Hence foci are ( ±???? , 0) = (±4. 5 4 , 0) i.e., ( ±5,0) Q3. 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) 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. Q4. The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are (a) ?? = ±?? (b) ?? = ±?? (c) ?? = ±v ?? (d) ?? = ±v ?? Ans: (c) ( ?? + 1) 2 - ?? 2 - 1 + 5 = 0 ? - ( ?? +1) 2 4 + ?? 2 4 = 1 Equation of directrices of ?? 2 ?? 2 - ?? 2 ?? 2 = 1 are ?? = ± ?? ?? Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ± 2 v 2 ? ?? = ±v 2. Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: 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. The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is (a) ( ?? , ?? ) (b) ( ?? , ?? ) (c) ( ?? , -?? ) (d) None of these Ans: (a) The equation of line and hyperbola are ?? = ?? - 1 ....(i) and 3?? 2 - 4?? 2 = 12 .....(ii) From (i) and (ii), we get 3?? 2 - 4( ?? - 1) 2 = 12 ? 3?? 2 - 4( ?? 2 - 2?? + 1) = 12 or ?? 2 - 8?? + 16 = 0 ? ?? = 4 From (i), ?? = 3 so points of contact is ( 4,3) Trick: Points of contact are (± ?? 2 ?? v?? 2 ?? 2 -?? 2 , ± ?? 2 v?? 2 ?? 2 -?? 2 ). Here ?? 2 = 4, ?? 2 = 3 and ?? = 1. So the required points of contact is ( 4,3) . Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° Page 3 Solved Examples on Hyperbola JEE Mains Q1. The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 From definition, ???? = ?????? v( ?? - 1) 2 + ( ?? + 1) 2 = v 2 · ( ?? - ?? + 1) v 1 2 + 1 2 ? ( ?? - 1) 2 + ( ?? + 1) 2 = ( ?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2. The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are (a) ( ±?? , ?? ) (b) ( ?? , ±?? ) (c) ( ±?? , ?? ) (d) ( ?? , ±?? ) Ans: (c) The equation of hyperbola is ?? 2 16 - ?? 2 9 = 1. Now, ?? 2 = ?? 2 ( ?? 2 - 1) ? 9 = 16( ?? 2 - 1) ? ?? = 5 4 . Hence foci are ( ±???? , 0) = (±4. 5 4 , 0) i.e., ( ±5,0) Q3. 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) 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. Q4. The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are (a) ?? = ±?? (b) ?? = ±?? (c) ?? = ±v ?? (d) ?? = ±v ?? Ans: (c) ( ?? + 1) 2 - ?? 2 - 1 + 5 = 0 ? - ( ?? +1) 2 4 + ?? 2 4 = 1 Equation of directrices of ?? 2 ?? 2 - ?? 2 ?? 2 = 1 are ?? = ± ?? ?? Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ± 2 v 2 ? ?? = ±v 2. Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: 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. The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is (a) ( ?? , ?? ) (b) ( ?? , ?? ) (c) ( ?? , -?? ) (d) None of these Ans: (a) The equation of line and hyperbola are ?? = ?? - 1 ....(i) and 3?? 2 - 4?? 2 = 12 .....(ii) From (i) and (ii), we get 3?? 2 - 4( ?? - 1) 2 = 12 ? 3?? 2 - 4( ?? 2 - 2?? + 1) = 12 or ?? 2 - 8?? + 16 = 0 ? ?? = 4 From (i), ?? = 3 so points of contact is ( 4,3) Trick: Points of contact are (± ?? 2 ?? v?? 2 ?? 2 -?? 2 , ± ?? 2 v?? 2 ?? 2 -?? 2 ). Here ?? 2 = 4, ?? 2 = 3 and ?? = 1. So the required points of contact is ( 4,3) . Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° (c) ???? ° (d) ???? ° Ans: (c) 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. If the normal at ' ?? ' on the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? meets transverse axis at ?? , then ???? · ?? ' ?? = (Where ?? and ?? ' are the vertices of the hyperbola) (a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) (b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) (c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) (d) None of these Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? + ???? cot ?? = ( ?? 2 + ?? 2 ) This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are (( ?? 2 +?? 2 ?? )sec ?? , 0) and the co-ordinates of the vertices ?? and ?? ' are ?? ( ?? , 0) and ?? ' ( -?? , 0) respectively. ? ???? · ?? ' ?? = (-?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) (?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) = ( ?? 2 + ?? 2 ?? ) 2 sec 2 ?? - ?? 2 = ( ?? ?? 2 ) 2 sec 2 ?? - ?? 2 = ?? 2 ( ?? 4 sec 2 ?? - 1) Q9. What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose mid point is ( ?? , ?? ) (a) ?????? ?? - ?????? ?? = ???? (b) ?????? ?? - ???? ?? = ?????? (c) ?????? ?? + ???? ?? = ?????? (d) ???? ?? + ?????? ?? = ?????? Ans: (b) According to question, ?? = 25?? 2 - 16?? 2 - 400 = 0 Equation of required chord is ?? 1 = ?? Here ?? 1 = 25( 5) 2 - 16( 3) 2 - 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1 - 16?? ?? 1 - 400, where ?? 1 = 5, ?? 1 = 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. Q10. If the polar of a point w.r.t. ?? ?? ?? ?? + ?? ?? ?? ?? = ?? touches the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? , then the locus of the point is (a) Given hyperbola (b) Ellipse (c) Circle (d) None of these Page 4 Solved Examples on Hyperbola JEE Mains Q1. The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 From definition, ???? = ?????? v( ?? - 1) 2 + ( ?? + 1) 2 = v 2 · ( ?? - ?? + 1) v 1 2 + 1 2 ? ( ?? - 1) 2 + ( ?? + 1) 2 = ( ?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2. The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are (a) ( ±?? , ?? ) (b) ( ?? , ±?? ) (c) ( ±?? , ?? ) (d) ( ?? , ±?? ) Ans: (c) The equation of hyperbola is ?? 2 16 - ?? 2 9 = 1. Now, ?? 2 = ?? 2 ( ?? 2 - 1) ? 9 = 16( ?? 2 - 1) ? ?? = 5 4 . Hence foci are ( ±???? , 0) = (±4. 5 4 , 0) i.e., ( ±5,0) Q3. 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) 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. Q4. The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are (a) ?? = ±?? (b) ?? = ±?? (c) ?? = ±v ?? (d) ?? = ±v ?? Ans: (c) ( ?? + 1) 2 - ?? 2 - 1 + 5 = 0 ? - ( ?? +1) 2 4 + ?? 2 4 = 1 Equation of directrices of ?? 2 ?? 2 - ?? 2 ?? 2 = 1 are ?? = ± ?? ?? Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ± 2 v 2 ? ?? = ±v 2. Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: 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. The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is (a) ( ?? , ?? ) (b) ( ?? , ?? ) (c) ( ?? , -?? ) (d) None of these Ans: (a) The equation of line and hyperbola are ?? = ?? - 1 ....(i) and 3?? 2 - 4?? 2 = 12 .....(ii) From (i) and (ii), we get 3?? 2 - 4( ?? - 1) 2 = 12 ? 3?? 2 - 4( ?? 2 - 2?? + 1) = 12 or ?? 2 - 8?? + 16 = 0 ? ?? = 4 From (i), ?? = 3 so points of contact is ( 4,3) Trick: Points of contact are (± ?? 2 ?? v?? 2 ?? 2 -?? 2 , ± ?? 2 v?? 2 ?? 2 -?? 2 ). Here ?? 2 = 4, ?? 2 = 3 and ?? = 1. So the required points of contact is ( 4,3) . Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° (c) ???? ° (d) ???? ° Ans: (c) 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. If the normal at ' ?? ' on the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? meets transverse axis at ?? , then ???? · ?? ' ?? = (Where ?? and ?? ' are the vertices of the hyperbola) (a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) (b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) (c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) (d) None of these Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? + ???? cot ?? = ( ?? 2 + ?? 2 ) This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are (( ?? 2 +?? 2 ?? )sec ?? , 0) and the co-ordinates of the vertices ?? and ?? ' are ?? ( ?? , 0) and ?? ' ( -?? , 0) respectively. ? ???? · ?? ' ?? = (-?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) (?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) = ( ?? 2 + ?? 2 ?? ) 2 sec 2 ?? - ?? 2 = ( ?? ?? 2 ) 2 sec 2 ?? - ?? 2 = ?? 2 ( ?? 4 sec 2 ?? - 1) Q9. What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose mid point is ( ?? , ?? ) (a) ?????? ?? - ?????? ?? = ???? (b) ?????? ?? - ???? ?? = ?????? (c) ?????? ?? + ???? ?? = ?????? (d) ???? ?? + ?????? ?? = ?????? Ans: (b) According to question, ?? = 25?? 2 - 16?? 2 - 400 = 0 Equation of required chord is ?? 1 = ?? Here ?? 1 = 25( 5) 2 - 16( 3) 2 - 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1 - 16?? ?? 1 - 400, where ?? 1 = 5, ?? 1 = 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. Q10. If the polar of a point w.r.t. ?? ?? ?? ?? + ?? ?? ?? ?? = ?? touches the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? , then the locus of the point is (a) Given hyperbola (b) Ellipse (c) Circle (d) None of these Ans: (a) Let ( ?? 1 , ?? 1 ) be the given point. Its polar w.r.t. ?? 2 ?? 2 + ?? 2 ?? 2 = 1 is ?? ?? 1 ?? 2 + ?? ?? 1 ?? 2 = 1 i.e., ?? = ?? 2 ?? 1 (1 - ?? ?? 1 ?? 2 ) = - ?? 2 ?? 1 ?? 2 ?? 1 ?? + ?? 2 ?? 1 This touches ?? 2 ?? 2 - ?? 2 ?? 2 = 1 if ( ?? 2 ?? 1 ) 2 = ?? 2 · ( ?? 2 ?? 1 ?? 2 ?? 1 ) - ?? 2 ? ?? 4 ?? 1 2 = ?? 2 ?? 4 ?? 1 2 ?? 4 ?? 1 2 - ?? 2 ? ?? 2 ?? 1 2 = ?? 2 ?? 1 2 ?? 2 ?? 1 2 - 1 ? ?? 1 2 ?? 2 - ?? 1 2 ?? 2 = 1 ? Locus of ( ?? 1 , ?? 1 ) is ?? 2 ?? 2 - ?? 2 ?? 2 = 1. Which is the same hyperbola. Q11. The combined equation of the asymptotes of the hyperbola ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? = ?? (a) ?? ?? ?? + ?? ???? + ?? ?? ?? = ?? (b) ?? ?? ?? + ?? ???? + ?? ?? ?? - ?? ?? + ?? ?? + ?? = ?? = ?? (c) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? - ?? = ?? (d) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? + ?? = ?? Ans: (d) Given, equation of hyperbola 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? = 0 and equation of asymptotes 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? + ?? = 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 ?? ?? 2 + 2h???? + ?? ?? 2 + 2???? + 2???? + ?? = 0 Comparing equation (i) with standard equation, we get ?? = 2, ?? = 2, h = 5 2 , ?? = 2, ?? = 5 2 and ?? = ?? . We also know that the condition for a pair of straight lines is ?????? + 2???? h - ?? ?? 2 - ?? ?? 2 - ?? h 2 = 0. Therefore, 4?? + 25 - 25 2 - 8 - 25 4 ?? = 0 or -9?? 4 + 9 2 = 0 or ?? = 2 Substituting value of ?? in equation (i), we get 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? + 2 = 0. Q12. If the normal at (???? , ?? ?? ) on the curve ???? = ?? ?? meets the curve again in ?? ' , then (a) ?? ' = - ?? ?? ?? (b) ?? ' = - ?? ?? (c) ?? ' = ?? ?? ?? (d) ?? '?? = - ?? ?? ?? Ans: (a) The equation of the tangent at (???? , ?? ?? ) is ???? = ?? 3 ?? - ?? ?? 4 + ?? If it passes through (?? ?? ' , ?? ?? ' ) then ? ???? ?? ' = ?? 3 ?? ?? ' - ?? ?? 4 + ?? ? ?? = ?? 3 ?? '2 - ?? 4 ?? ' + ?? ' ? ?? - ?? ' = ?? 3 ?? ' ( ?? ' - ?? ) ? ?? ' = - 1 ?? 3 Q13. A variable straight line of slope 4 intersects the hyperbola ???? = ?? at two points. The locus of the point which divides the line segment between these two points in the ratio ?? : ?? is (a) ???? ?? ?? + ???? ???? + ?? ?? = ?? (b) ???? ?? ?? - ???? ???? + ?? ?? = ?? (c) ???? ?? ?? + ???? ???? + ?? ?? = ?? (d) None of these Page 5 Solved Examples on Hyperbola JEE Mains Q1. The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? and with eccentricity v ?? is (a) ?? ?? - ?? ?? = ?? (b) ???? = ?? (c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? (d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 From definition, ???? = ?????? v( ?? - 1) 2 + ( ?? + 1) 2 = v 2 · ( ?? - ?? + 1) v 1 2 + 1 2 ? ( ?? - 1) 2 + ( ?? + 1) 2 = ( ?? - ?? + 1) 2 ? 2???? - 4?? + 4?? + 1 = 0, which is the required equation of conic (Rectangular hyperbola) Q2. The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are (a) ( ±?? , ?? ) (b) ( ?? , ±?? ) (c) ( ±?? , ?? ) (d) ( ?? , ±?? ) Ans: (c) The equation of hyperbola is ?? 2 16 - ?? 2 9 = 1. Now, ?? 2 = ?? 2 ( ?? 2 - 1) ? 9 = 16( ?? 2 - 1) ? ?? = 5 4 . Hence foci are ( ±???? , 0) = (±4. 5 4 , 0) i.e., ( ±5,0) Q3. 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) 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. Q4. The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are (a) ?? = ±?? (b) ?? = ±?? (c) ?? = ±v ?? (d) ?? = ±v ?? Ans: (c) ( ?? + 1) 2 - ?? 2 - 1 + 5 = 0 ? - ( ?? +1) 2 4 + ?? 2 4 = 1 Equation of directrices of ?? 2 ?? 2 - ?? 2 ?? 2 = 1 are ?? = ± ?? ?? Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ± 2 v 2 ? ?? = ±v 2. Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? is (a) ???? v ?? (b) v ?? (c) ?? v ?? (d) ?? , v ?? Ans: 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. The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is (a) ( ?? , ?? ) (b) ( ?? , ?? ) (c) ( ?? , -?? ) (d) None of these Ans: (a) The equation of line and hyperbola are ?? = ?? - 1 ....(i) and 3?? 2 - 4?? 2 = 12 .....(ii) From (i) and (ii), we get 3?? 2 - 4( ?? - 1) 2 = 12 ? 3?? 2 - 4( ?? 2 - 2?? + 1) = 12 or ?? 2 - 8?? + 16 = 0 ? ?? = 4 From (i), ?? = 3 so points of contact is ( 4,3) Trick: Points of contact are (± ?? 2 ?? v?? 2 ?? 2 -?? 2 , ± ?? 2 v?? 2 ?? 2 -?? 2 ). Here ?? 2 = 4, ?? 2 = 3 and ?? = 1. So the required points of contact is ( 4,3) . Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola ?? ?? ?? - ?? ?? ?? = ?? is parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is (a) ???? ° (b) ???? ° (c) ???? ° (d) ???? ° Ans: (c) 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. If the normal at ' ?? ' on the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? meets transverse axis at ?? , then ???? · ?? ' ?? = (Where ?? and ?? ' are the vertices of the hyperbola) (a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) (b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) (c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) (d) None of these Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? + ???? cot ?? = ( ?? 2 + ?? 2 ) This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are (( ?? 2 +?? 2 ?? )sec ?? , 0) and the co-ordinates of the vertices ?? and ?? ' are ?? ( ?? , 0) and ?? ' ( -?? , 0) respectively. ? ???? · ?? ' ?? = (-?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) (?? + ( ?? 2 + ?? 2 ?? ) sec ?? ) = ( ?? 2 + ?? 2 ?? ) 2 sec 2 ?? - ?? 2 = ( ?? ?? 2 ) 2 sec 2 ?? - ?? 2 = ?? 2 ( ?? 4 sec 2 ?? - 1) Q9. What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose mid point is ( ?? , ?? ) (a) ?????? ?? - ?????? ?? = ???? (b) ?????? ?? - ???? ?? = ?????? (c) ?????? ?? + ???? ?? = ?????? (d) ???? ?? + ?????? ?? = ?????? Ans: (b) According to question, ?? = 25?? 2 - 16?? 2 - 400 = 0 Equation of required chord is ?? 1 = ?? Here ?? 1 = 25( 5) 2 - 16( 3) 2 - 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1 - 16?? ?? 1 - 400, where ?? 1 = 5, ?? 1 = 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. Q10. If the polar of a point w.r.t. ?? ?? ?? ?? + ?? ?? ?? ?? = ?? touches the hyperbola ?? ?? ?? ?? - ?? ?? ?? ?? = ?? , then the locus of the point is (a) Given hyperbola (b) Ellipse (c) Circle (d) None of these Ans: (a) Let ( ?? 1 , ?? 1 ) be the given point. Its polar w.r.t. ?? 2 ?? 2 + ?? 2 ?? 2 = 1 is ?? ?? 1 ?? 2 + ?? ?? 1 ?? 2 = 1 i.e., ?? = ?? 2 ?? 1 (1 - ?? ?? 1 ?? 2 ) = - ?? 2 ?? 1 ?? 2 ?? 1 ?? + ?? 2 ?? 1 This touches ?? 2 ?? 2 - ?? 2 ?? 2 = 1 if ( ?? 2 ?? 1 ) 2 = ?? 2 · ( ?? 2 ?? 1 ?? 2 ?? 1 ) - ?? 2 ? ?? 4 ?? 1 2 = ?? 2 ?? 4 ?? 1 2 ?? 4 ?? 1 2 - ?? 2 ? ?? 2 ?? 1 2 = ?? 2 ?? 1 2 ?? 2 ?? 1 2 - 1 ? ?? 1 2 ?? 2 - ?? 1 2 ?? 2 = 1 ? Locus of ( ?? 1 , ?? 1 ) is ?? 2 ?? 2 - ?? 2 ?? 2 = 1. Which is the same hyperbola. Q11. The combined equation of the asymptotes of the hyperbola ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? = ?? (a) ?? ?? ?? + ?? ???? + ?? ?? ?? = ?? (b) ?? ?? ?? + ?? ???? + ?? ?? ?? - ?? ?? + ?? ?? + ?? = ?? = ?? (c) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? - ?? = ?? (d) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? + ?? = ?? Ans: (d) Given, equation of hyperbola 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? = 0 and equation of asymptotes 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? + ?? = 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 ?? ?? 2 + 2h???? + ?? ?? 2 + 2???? + 2???? + ?? = 0 Comparing equation (i) with standard equation, we get ?? = 2, ?? = 2, h = 5 2 , ?? = 2, ?? = 5 2 and ?? = ?? . We also know that the condition for a pair of straight lines is ?????? + 2???? h - ?? ?? 2 - ?? ?? 2 - ?? h 2 = 0. Therefore, 4?? + 25 - 25 2 - 8 - 25 4 ?? = 0 or -9?? 4 + 9 2 = 0 or ?? = 2 Substituting value of ?? in equation (i), we get 2?? 2 + 5???? + 2?? 2 + 4?? + 5?? + 2 = 0. Q12. If the normal at (???? , ?? ?? ) on the curve ???? = ?? ?? meets the curve again in ?? ' , then (a) ?? ' = - ?? ?? ?? (b) ?? ' = - ?? ?? (c) ?? ' = ?? ?? ?? (d) ?? '?? = - ?? ?? ?? Ans: (a) The equation of the tangent at (???? , ?? ?? ) is ???? = ?? 3 ?? - ?? ?? 4 + ?? If it passes through (?? ?? ' , ?? ?? ' ) then ? ???? ?? ' = ?? 3 ?? ?? ' - ?? ?? 4 + ?? ? ?? = ?? 3 ?? '2 - ?? 4 ?? ' + ?? ' ? ?? - ?? ' = ?? 3 ?? ' ( ?? ' - ?? ) ? ?? ' = - 1 ?? 3 Q13. A variable straight line of slope 4 intersects the hyperbola ???? = ?? at two points. The locus of the point which divides the line segment between these two points in the ratio ?? : ?? is (a) ???? ?? ?? + ???? ???? + ?? ?? = ?? (b) ???? ?? ?? - ???? ???? + ?? ?? = ?? (c) ???? ?? ?? + ???? ???? + ?? ?? = ?? (d) None of these Ans: (a) Let ?? ( h, ?? ) be any point on the locus. Equation of the line through ?? and having slope 4 is ?? - ?? = 4( ?? - h) Suppose this meets ???? = 1 ..(ii) in ?? ( ?? 1 , ?? 1 ) and ?? ( ?? 2 , ?? 2 ) Eliminating ?? between (i) and (ii), we get 1 ?? - ?? = 4( ?? - h) ? 1 - ???? = 4?? 2 - 4h?? ? 4?? 2 - ( 4h - ?? ) ?? - 1 = 0 This has two roots say ?? 1 , ?? 2 ; ?? 1 + ?? 2 = 4h-?? 4 (iv) and ?? 1 ?? 2 = - 1 4 Also, 2?? 1 +?? 2 3 = h [? ?? divides ???? in the ratio 1: 2] i.e., 2?? 1 + ?? 2 = 3h (vi) - (iv) gives, ?? 1 = 3h - 4h-?? 4 = 8h+?? 4 and ?? 2 = 3h - 2 · 8h+?? 4 = - 2h+?? 2 Putting in (v), we get 8h+?? 4 (- 2h+?? 2 ) = - 1 4 ? ( 8h + ?? ) ( 2h + ?? ) = 2 ? 16h 2 + 10h?? + ?? 2 = 2 ? Required locus of ?? ( h, ?? ) is 16?? 2 + 10???? + ?? 2 = 2. Q14. ???? and ???? are two perpendicular chords of the rectangular hyperbola ???? = ?? ?? . If ?? is the centre of the rectangular hyperbola, then the product of the slopes of ???? , ???? , ???? and ???? is equal to (a) -1 (b) 1 (c) 0 (d) None of these Ans: (b) Let ?? 1 , ?? 2 , ?? 3 , ?? 4 be the parameters of the points ?? , ?? , ?? and ?? respectively. Then, the coordinates of ?? , ?? , ?? and ?? are (?? ?? 1 , ?? ?? 1 ), (?? ?? 2 , ?? ?? 2 ), (?? ?? 3 , ?? ?? 3 ) and (?? ?? 4 , ?? ?? 4 ) respectively. Now, ???? ? ???? ? ?? ?? 2 - ?? ?? 1 ?? ?? 2 -?? ?? 1 × ?? ?? 4 - ?? ?? 3 ?? ?? 4 -?? ?? 3 = -1 ? - 1 ?? 1 ?? 2 × - 1 ?? 3 ?? 4 = -1 ? ?? 1 ?? 2 ?? 3 ?? 4 = -1 ? Product of the slopes of ???? , ???? , ???? and ???? 1 ?? 1 2 × 1 ?? 2 2 × 1 ?? 3 2 × 1 ?? 4 2 = 1 ?? 1 2 ?? 2 2 ?? 3 2 ?? 4 2 = 1 [Using (i)] Q15. If the circle ?? ?? + ?? ?? = ?? ?? intersects the hyperbola ???? = ?? ?? in four points ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) then (a) ?? ?? + ?? ?? + ?? ?? + ?? ?? = ?? (b) ?? ?? + ?? ?? + ?? ?? + ?? ?? = ?? (c) ?? ?? ?? ?? ?? ?? ?? ?? = ?? ?? (d) ?? ?? ?? ?? ?? ?? ?? ?? = ?? ?? Ans: (a,b,c,d) Given, circle is ?? 2 + ?? 2 = ?? 2 (i) and hyperbola be ???? = ?? 2 from (ii) ?? = ?? 2 ?? . Putting in (i), we get ?? 2 + ?? 4 ?? 2 = ?? 2 ? ?? 4 - ?? 2 ?? 2 + ?? 4 = 0 ? ?? 1 + ?? 2 + ?? 3 + ?? 4 = 0, ?? 1 ?? 2 ?? 3 ?? 4 = ?? 4 Since both the curves are symmetric in ?? and ?? , ? ?? 1 + ?? 2 + ?? 3 + ?? 4 = 0; ?? 1 ?? 2 ?? 3 ?? 4 = ?? 4 .Read More
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