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Page 1 Previous Year Questions (2021-24): Parabola Q1 - 2024 (27 Jan Shift 1) If the shortest distance of the parabola ?? 2 = 4?? from the centre of the circle ?? 2 + ?? 2 - 4?? - 16?? + 64 = 0 is d, then d 2 is equal to : (1) 16 (2) 24 (3) 20 (4) 36 Q2 - 2024 (29 Jan Shift 2) Let P( ?? , ?? ) be a point on the parabola y 2 = 4x. If P also lies on the chord of the parabola ?? 2 = 8?? whose mid point is ( 1, 5 4 ) . Then ( ?? - 28) ( ?? - 8) is equal to Q3 - 2024 (30 Jan Shift 1) The maximum area of a triangle whose one vertex is at ( 0,0) and the other two vertices lie on the curve ?? = -2?? 2 + 54 at points ( ?? , ?? ) and ( -?? , ?? ) where y > 0 is: (1) 88 (2) 122 (3) 92 (4) 108 Answer Key Q1 (3) Q2 (192) Q3 (4) Page 2 Previous Year Questions (2021-24): Parabola Q1 - 2024 (27 Jan Shift 1) If the shortest distance of the parabola ?? 2 = 4?? from the centre of the circle ?? 2 + ?? 2 - 4?? - 16?? + 64 = 0 is d, then d 2 is equal to : (1) 16 (2) 24 (3) 20 (4) 36 Q2 - 2024 (29 Jan Shift 2) Let P( ?? , ?? ) be a point on the parabola y 2 = 4x. If P also lies on the chord of the parabola ?? 2 = 8?? whose mid point is ( 1, 5 4 ) . Then ( ?? - 28) ( ?? - 8) is equal to Q3 - 2024 (30 Jan Shift 1) The maximum area of a triangle whose one vertex is at ( 0,0) and the other two vertices lie on the curve ?? = -2?? 2 + 54 at points ( ?? , ?? ) and ( -?? , ?? ) where y > 0 is: (1) 88 (2) 122 (3) 92 (4) 108 Answer Key Q1 (3) Q2 (192) Q3 (4) Solutions Q1 Equation of normal to parabola ?? = ???? - 2?? - ?? 3 this normal passing through center of circle ( 2,8) 8 = 2?? - 2?? - ?? 3 ?? = -2 So point P on parabola ? ( am 2 , -2am )= ( 4,4) And C = ( 2,8) PC = v4 + 16 = v20 d 2 = 20 Q2 Parabola is ?? 2 = 8?? Chord with mid point ( x 1 , y 1 ) is T = S 1 ? xx 1 - 4( y + y 1 )= x 1 2 - 8y 1 ? ( x 1 , y 1 )= ( 1, 5 4 ) ? x - 4 ( y + 5 4 )= 1 - 8 × 5 4 = -9 ? x - 4y + 4 = 0 … ( ?? , ?? ) lies on (i) \ & also on ?? 2 = 4?? ? ?? - 4?? + 4 = 0 … ( ???? ) &?? 2 = 4?? … (iii) Solving (ii) \ & (iii) Page 3 Previous Year Questions (2021-24): Parabola Q1 - 2024 (27 Jan Shift 1) If the shortest distance of the parabola ?? 2 = 4?? from the centre of the circle ?? 2 + ?? 2 - 4?? - 16?? + 64 = 0 is d, then d 2 is equal to : (1) 16 (2) 24 (3) 20 (4) 36 Q2 - 2024 (29 Jan Shift 2) Let P( ?? , ?? ) be a point on the parabola y 2 = 4x. If P also lies on the chord of the parabola ?? 2 = 8?? whose mid point is ( 1, 5 4 ) . Then ( ?? - 28) ( ?? - 8) is equal to Q3 - 2024 (30 Jan Shift 1) The maximum area of a triangle whose one vertex is at ( 0,0) and the other two vertices lie on the curve ?? = -2?? 2 + 54 at points ( ?? , ?? ) and ( -?? , ?? ) where y > 0 is: (1) 88 (2) 122 (3) 92 (4) 108 Answer Key Q1 (3) Q2 (192) Q3 (4) Solutions Q1 Equation of normal to parabola ?? = ???? - 2?? - ?? 3 this normal passing through center of circle ( 2,8) 8 = 2?? - 2?? - ?? 3 ?? = -2 So point P on parabola ? ( am 2 , -2am )= ( 4,4) And C = ( 2,8) PC = v4 + 16 = v20 d 2 = 20 Q2 Parabola is ?? 2 = 8?? Chord with mid point ( x 1 , y 1 ) is T = S 1 ? xx 1 - 4( y + y 1 )= x 1 2 - 8y 1 ? ( x 1 , y 1 )= ( 1, 5 4 ) ? x - 4 ( y + 5 4 )= 1 - 8 × 5 4 = -9 ? x - 4y + 4 = 0 … ( ?? , ?? ) lies on (i) \ & also on ?? 2 = 4?? ? ?? - 4?? + 4 = 0 … ( ???? ) &?? 2 = 4?? … (iii) Solving (ii) \ & (iii) ?? 2 = 4( 4?? - 4)? ?? 2 - 16?? + 16 = 0 matho ? ?? = 8 ± 4v3 and ?? = 4?? - 4 = 28 ± 16v3 ? ( ?? , ?? ) = ( 28 + 16v3, 8 + 4v3) & ( 28 - 16v3, 8 - 4v3) ? ( ?? - 28) ( ?? - 8)= ( ±16v3) ( ±4v3) matho = 192 Q3. Area of ? = 1 2 | 0 0 1 x y 1 -x y 1 | ? | 1 2 ( xy+ xy ) | = |xy | Area ( ?)= |xy | = |x( -2x 2 + 54) | d( ?) dx = |( -6x 2 + 54) | ? d? dx = 0 at x = 3 Area = 3( -2 × 9 + 54)= 108 January 2023 Q1 - 24 January - Shift 1 The equations of the sides AB and AC of a triangle ABC are ( ?? + 1) ?? + ???? = 4 and ???? + ( 1 - ?? ) ?? + ?? = 0 respectively. Its vertex ?? is on the ?? - axis and its orthocentre is ( 1,2) . The length of the tangent from the point ?? to the part of the parabola ?? 2 = 6?? in the first quadrant is Page 4 Previous Year Questions (2021-24): Parabola Q1 - 2024 (27 Jan Shift 1) If the shortest distance of the parabola ?? 2 = 4?? from the centre of the circle ?? 2 + ?? 2 - 4?? - 16?? + 64 = 0 is d, then d 2 is equal to : (1) 16 (2) 24 (3) 20 (4) 36 Q2 - 2024 (29 Jan Shift 2) Let P( ?? , ?? ) be a point on the parabola y 2 = 4x. If P also lies on the chord of the parabola ?? 2 = 8?? whose mid point is ( 1, 5 4 ) . Then ( ?? - 28) ( ?? - 8) is equal to Q3 - 2024 (30 Jan Shift 1) The maximum area of a triangle whose one vertex is at ( 0,0) and the other two vertices lie on the curve ?? = -2?? 2 + 54 at points ( ?? , ?? ) and ( -?? , ?? ) where y > 0 is: (1) 88 (2) 122 (3) 92 (4) 108 Answer Key Q1 (3) Q2 (192) Q3 (4) Solutions Q1 Equation of normal to parabola ?? = ???? - 2?? - ?? 3 this normal passing through center of circle ( 2,8) 8 = 2?? - 2?? - ?? 3 ?? = -2 So point P on parabola ? ( am 2 , -2am )= ( 4,4) And C = ( 2,8) PC = v4 + 16 = v20 d 2 = 20 Q2 Parabola is ?? 2 = 8?? Chord with mid point ( x 1 , y 1 ) is T = S 1 ? xx 1 - 4( y + y 1 )= x 1 2 - 8y 1 ? ( x 1 , y 1 )= ( 1, 5 4 ) ? x - 4 ( y + 5 4 )= 1 - 8 × 5 4 = -9 ? x - 4y + 4 = 0 … ( ?? , ?? ) lies on (i) \ & also on ?? 2 = 4?? ? ?? - 4?? + 4 = 0 … ( ???? ) &?? 2 = 4?? … (iii) Solving (ii) \ & (iii) ?? 2 = 4( 4?? - 4)? ?? 2 - 16?? + 16 = 0 matho ? ?? = 8 ± 4v3 and ?? = 4?? - 4 = 28 ± 16v3 ? ( ?? , ?? ) = ( 28 + 16v3, 8 + 4v3) & ( 28 - 16v3, 8 - 4v3) ? ( ?? - 28) ( ?? - 8)= ( ±16v3) ( ±4v3) matho = 192 Q3. Area of ? = 1 2 | 0 0 1 x y 1 -x y 1 | ? | 1 2 ( xy+ xy ) | = |xy | Area ( ?)= |xy | = |x( -2x 2 + 54) | d( ?) dx = |( -6x 2 + 54) | ? d? dx = 0 at x = 3 Area = 3( -2 × 9 + 54)= 108 January 2023 Q1 - 24 January - Shift 1 The equations of the sides AB and AC of a triangle ABC are ( ?? + 1) ?? + ???? = 4 and ???? + ( 1 - ?? ) ?? + ?? = 0 respectively. Its vertex ?? is on the ?? - axis and its orthocentre is ( 1,2) . The length of the tangent from the point ?? to the part of the parabola ?? 2 = 6?? in the first quadrant is (1) v 6 (2) 2v 2 (3) 2 (4) 4 Q2 - 24 January - Shift 2 The urns A, B and C contain 4 red, 6 black; 5 red, 5 black and ?? red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y 2 = ?? x with one vertex at the vertex of the parabola is Q3 - 24 January - Shift 2 The distance of the point ( 6, -2v 2) from the common tangent y = mx+ c, m > 0, of the curves x = 2y 2 and x = 1 + y 2 is (1) 1 3 (2) 5 (3) 14 3 (4) 5v 3 Q4 - 30 January - Shift 1 If P( h, k) be point on the parabola x = 4y 2 , which is nearest to the point Q( 0,33) , then the distance of P from the directrix of the parabola ?? 2 = 4( ?? + ?? ) is equal to : (1) 2 (2) 4 (3) 8 (4) 6 Q5 - 30 January - Shift 2 Let A be a point on the x-axis. Common tangents are drawn from A to the curves ?? 2 + ?? 2 = 8 and ?? 2 = 16x. If one of these tangents touches the two curves at ?? and ?? , then ( ???? ) 2 is equal to (1) 64 (2) 76 (3) 81 (4) 72 Page 5 Previous Year Questions (2021-24): Parabola Q1 - 2024 (27 Jan Shift 1) If the shortest distance of the parabola ?? 2 = 4?? from the centre of the circle ?? 2 + ?? 2 - 4?? - 16?? + 64 = 0 is d, then d 2 is equal to : (1) 16 (2) 24 (3) 20 (4) 36 Q2 - 2024 (29 Jan Shift 2) Let P( ?? , ?? ) be a point on the parabola y 2 = 4x. If P also lies on the chord of the parabola ?? 2 = 8?? whose mid point is ( 1, 5 4 ) . Then ( ?? - 28) ( ?? - 8) is equal to Q3 - 2024 (30 Jan Shift 1) The maximum area of a triangle whose one vertex is at ( 0,0) and the other two vertices lie on the curve ?? = -2?? 2 + 54 at points ( ?? , ?? ) and ( -?? , ?? ) where y > 0 is: (1) 88 (2) 122 (3) 92 (4) 108 Answer Key Q1 (3) Q2 (192) Q3 (4) Solutions Q1 Equation of normal to parabola ?? = ???? - 2?? - ?? 3 this normal passing through center of circle ( 2,8) 8 = 2?? - 2?? - ?? 3 ?? = -2 So point P on parabola ? ( am 2 , -2am )= ( 4,4) And C = ( 2,8) PC = v4 + 16 = v20 d 2 = 20 Q2 Parabola is ?? 2 = 8?? Chord with mid point ( x 1 , y 1 ) is T = S 1 ? xx 1 - 4( y + y 1 )= x 1 2 - 8y 1 ? ( x 1 , y 1 )= ( 1, 5 4 ) ? x - 4 ( y + 5 4 )= 1 - 8 × 5 4 = -9 ? x - 4y + 4 = 0 … ( ?? , ?? ) lies on (i) \ & also on ?? 2 = 4?? ? ?? - 4?? + 4 = 0 … ( ???? ) &?? 2 = 4?? … (iii) Solving (ii) \ & (iii) ?? 2 = 4( 4?? - 4)? ?? 2 - 16?? + 16 = 0 matho ? ?? = 8 ± 4v3 and ?? = 4?? - 4 = 28 ± 16v3 ? ( ?? , ?? ) = ( 28 + 16v3, 8 + 4v3) & ( 28 - 16v3, 8 - 4v3) ? ( ?? - 28) ( ?? - 8)= ( ±16v3) ( ±4v3) matho = 192 Q3. Area of ? = 1 2 | 0 0 1 x y 1 -x y 1 | ? | 1 2 ( xy+ xy ) | = |xy | Area ( ?)= |xy | = |x( -2x 2 + 54) | d( ?) dx = |( -6x 2 + 54) | ? d? dx = 0 at x = 3 Area = 3( -2 × 9 + 54)= 108 January 2023 Q1 - 24 January - Shift 1 The equations of the sides AB and AC of a triangle ABC are ( ?? + 1) ?? + ???? = 4 and ???? + ( 1 - ?? ) ?? + ?? = 0 respectively. Its vertex ?? is on the ?? - axis and its orthocentre is ( 1,2) . The length of the tangent from the point ?? to the part of the parabola ?? 2 = 6?? in the first quadrant is (1) v 6 (2) 2v 2 (3) 2 (4) 4 Q2 - 24 January - Shift 2 The urns A, B and C contain 4 red, 6 black; 5 red, 5 black and ?? red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y 2 = ?? x with one vertex at the vertex of the parabola is Q3 - 24 January - Shift 2 The distance of the point ( 6, -2v 2) from the common tangent y = mx+ c, m > 0, of the curves x = 2y 2 and x = 1 + y 2 is (1) 1 3 (2) 5 (3) 14 3 (4) 5v 3 Q4 - 30 January - Shift 1 If P( h, k) be point on the parabola x = 4y 2 , which is nearest to the point Q( 0,33) , then the distance of P from the directrix of the parabola ?? 2 = 4( ?? + ?? ) is equal to : (1) 2 (2) 4 (3) 8 (4) 6 Q5 - 30 January - Shift 2 Let A be a point on the x-axis. Common tangents are drawn from A to the curves ?? 2 + ?? 2 = 8 and ?? 2 = 16x. If one of these tangents touches the two curves at ?? and ?? , then ( ???? ) 2 is equal to (1) 64 (2) 76 (3) 81 (4) 72 Q6 - 01 February - Shift 1 Let ?? be the set of all ?? ? ?? such that the area of the triangle formed by the tangent at the point P( b, c) , b, c ? N, on the parabola y 2 = 2ax and the lines x = b, y = 0 is 16 unit 2 , then ? a?S ?a is equal to $ \qquad $ Q7 - 01 February - Shift 2 If the ?? -intercept of a focal chord of the parabola ?? 2 = 8?? + 4?? + 4 is 3 , then the length of this chord is equal to . Answer Key (As per Official NTA Key released on 2 Feb) Q1 (2) Q2 (432) Q3 (2) Q4 (4) Q5 (4) Q6 (146) Q7 (16) Solutions Q1 (2) AB : ( ?? + 1) x + ?? y = 4 ???? : ???? + ( 1 - ?? ) ?? + ?? = 0 Vertex ?? is on ?? -axis ? x = 0Read More
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FAQs on Parabola: JEE Mains Previous Year Questions (2021-2024) - Mathematics (Maths) for JEE Main & Advanced
| 1. What is a parabola and how is it defined in terms of its focus and directrix? |  |
Ans. A parabola is a conic section curve formed by the intersection of a right circular cone and a plane parallel to one of the cone's generating lines. It is defined as the set of all points in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix.
| 2. How can the equation of a parabola be written in standard form for a vertical or horizontal orientation? | |
Ans. For a parabola with a vertical orientation and vertex at (h, k), the equation in standard form is (x - h)^2 = 4p(y - k), where p is the distance between the vertex and the focus. For a parabola with a horizontal orientation and vertex at (h, k), the equation in standard form is (y - k)^2 = 4p(x - h).
| 3. What is the relationship between the focal length and the distance from the vertex to the focus in a parabola? | |
Ans. The focal length of a parabola is equal to the distance from the vertex to the focus. It is denoted by the letter 'p' in the standard form of the parabolic equation.
| 4. How can the focus and directrix of a parabola be determined from its equation in standard form? | |
Ans. For a parabola with equation (x - h)^2 = 4p(y - k) or (y - k)^2 = 4p(x - h), the focus is located at the point (h, k + p) and the directrix is the horizontal line y = k - p for a vertical parabola or the vertical line x = h - p for a horizontal parabola.
| 5. How can the axis of symmetry of a parabola be identified from its equation? | |
Ans. The axis of symmetry of a parabola is the vertical line passing through its vertex. For a parabola in standard form, the axis of symmetry is given by the equation x = h for a vertical parabola or y = k for a horizontal parabola, where (h, k) is the vertex of the parabola.
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