Parabola: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Parabola  
 
(January 2026) 
Q1: Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the 
parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 
4/3), then (BC)² is equal to: 
A: 89 
B: 80 
C: 32 
D: 41 
Answer: B 
Explanation: 
 
Identify the Parabola Parameters The parabola is y² = 8x. Comparing this with the 
standard form y² = 4ax, we find 4a = 8, so a = 2. 
Focus A : The focus is (a, 0), which is (2, 0). 
Define Parametric Points for B and C: Let the two distinct points of intersection be B 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Parabola  
 
(January 2026) 
Q1: Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the 
parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 
4/3), then (BC)² is equal to: 
A: 89 
B: 80 
C: 32 
D: 41 
Answer: B 
Explanation: 
 
Identify the Parabola Parameters The parabola is y² = 8x. Comparing this with the 
standard form y² = 4ax, we find 4a = 8, so a = 2. 
Focus A : The focus is (a, 0), which is (2, 0). 
Define Parametric Points for B and C: Let the two distinct points of intersection be B 
a C. Using parametric coordinates (at², 2at): 
 
 
 
Q2: Let the image of parabola x² = 4y, in the line x - y = 1 be (y + a)² = b(x - c), a, 
b, c ? N. Then a + b + c is equal to 
A: 12 
B: 8 
C: 6 
D: 4 
Answer: C 
Explanation: 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Parabola  
 
(January 2026) 
Q1: Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the 
parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 
4/3), then (BC)² is equal to: 
A: 89 
B: 80 
C: 32 
D: 41 
Answer: B 
Explanation: 
 
Identify the Parabola Parameters The parabola is y² = 8x. Comparing this with the 
standard form y² = 4ax, we find 4a = 8, so a = 2. 
Focus A : The focus is (a, 0), which is (2, 0). 
Define Parametric Points for B and C: Let the two distinct points of intersection be B 
a C. Using parametric coordinates (at², 2at): 
 
 
 
Q2: Let the image of parabola x² = 4y, in the line x - y = 1 be (y + a)² = b(x - c), a, 
b, c ? N. Then a + b + c is equal to 
A: 12 
B: 8 
C: 6 
D: 4 
Answer: C 
Explanation: 
 
from (1) and (2) eliminate parameter t to get 
image of curve. 
so, from (2) put t = y+1/2 in equation (1). 
 
Compare with given image of curve (y + a)² = b(x - c) 
to find value of a, b, &c. 
a = 1, b = 4, c = 1  
a + b + c = 1 + 4 + 1 = 6. 
 
Q3: An equilateral triangle OAB is inscribed in the parabola y² = 4x with the vertex 
O at the vertex of the parabola. Then the minimum distance of the circle having 
AB as a diameter from the origin is 
A: 2(8 - 3v3) 
B: 4(6 + v3) 
C: 4(3 - v3) 
D: 2(3 + v3) 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Parabola  
 
(January 2026) 
Q1: Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the 
parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 
4/3), then (BC)² is equal to: 
A: 89 
B: 80 
C: 32 
D: 41 
Answer: B 
Explanation: 
 
Identify the Parabola Parameters The parabola is y² = 8x. Comparing this with the 
standard form y² = 4ax, we find 4a = 8, so a = 2. 
Focus A : The focus is (a, 0), which is (2, 0). 
Define Parametric Points for B and C: Let the two distinct points of intersection be B 
a C. Using parametric coordinates (at², 2at): 
 
 
 
Q2: Let the image of parabola x² = 4y, in the line x - y = 1 be (y + a)² = b(x - c), a, 
b, c ? N. Then a + b + c is equal to 
A: 12 
B: 8 
C: 6 
D: 4 
Answer: C 
Explanation: 
 
from (1) and (2) eliminate parameter t to get 
image of curve. 
so, from (2) put t = y+1/2 in equation (1). 
 
Compare with given image of curve (y + a)² = b(x - c) 
to find value of a, b, &c. 
a = 1, b = 4, c = 1  
a + b + c = 1 + 4 + 1 = 6. 
 
Q3: An equilateral triangle OAB is inscribed in the parabola y² = 4x with the vertex 
O at the vertex of the parabola. Then the minimum distance of the circle having 
AB as a diameter from the origin is 
A: 2(8 - 3v3) 
B: 4(6 + v3) 
C: 4(3 - v3) 
D: 2(3 + v3) 
Answer: C 
Explanation: 
 
y² = 4x ? a = 1 
Let A = (t², 2t) and B = (t², -2t) due to symmetry. 
Triangle OAB is equilateral: 
Slope of OA = tan(30°) = 1/v3 
 
Coordinates: 
A = (12, 4v3) B = (12, -4v3) 
Circle with diameter AB: 
 
Minimum distance from origin O(0, 0) : 
d = |OC - r|  
 
d = 12 - 4v3 = 4(3 - v3) 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Parabola  
 
(January 2026) 
Q1: Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the 
parabola at two distinct points B and C. If the centroid of the triangle ABC is (7/3, 
4/3), then (BC)² is equal to: 
A: 89 
B: 80 
C: 32 
D: 41 
Answer: B 
Explanation: 
 
Identify the Parabola Parameters The parabola is y² = 8x. Comparing this with the 
standard form y² = 4ax, we find 4a = 8, so a = 2. 
Focus A : The focus is (a, 0), which is (2, 0). 
Define Parametric Points for B and C: Let the two distinct points of intersection be B 
a C. Using parametric coordinates (at², 2at): 
 
 
 
Q2: Let the image of parabola x² = 4y, in the line x - y = 1 be (y + a)² = b(x - c), a, 
b, c ? N. Then a + b + c is equal to 
A: 12 
B: 8 
C: 6 
D: 4 
Answer: C 
Explanation: 
 
from (1) and (2) eliminate parameter t to get 
image of curve. 
so, from (2) put t = y+1/2 in equation (1). 
 
Compare with given image of curve (y + a)² = b(x - c) 
to find value of a, b, &c. 
a = 1, b = 4, c = 1  
a + b + c = 1 + 4 + 1 = 6. 
 
Q3: An equilateral triangle OAB is inscribed in the parabola y² = 4x with the vertex 
O at the vertex of the parabola. Then the minimum distance of the circle having 
AB as a diameter from the origin is 
A: 2(8 - 3v3) 
B: 4(6 + v3) 
C: 4(3 - v3) 
D: 2(3 + v3) 
Answer: C 
Explanation: 
 
y² = 4x ? a = 1 
Let A = (t², 2t) and B = (t², -2t) due to symmetry. 
Triangle OAB is equilateral: 
Slope of OA = tan(30°) = 1/v3 
 
Coordinates: 
A = (12, 4v3) B = (12, -4v3) 
Circle with diameter AB: 
 
Minimum distance from origin O(0, 0) : 
d = |OC - r|  
 
d = 12 - 4v3 = 4(3 - v3) 
 
Q4: Let the locus of the mid-point of the chord through the origin O of the 
parabola y² = 4x be the curve S. Let P be any point on S. Then the locus of the 
point, which internally divides OP in the ratio 3 : 1, is: 
A 2x² = 3y 
B 2y² = 3x 
C 3y² = 2x 
D 3x² = 2y 
Answer: B 
Explanation: 
Let M(x 1, y1) be the locus of the midpoint of the chord of the parabola y² = 4x 
? Equation of the chord is S 1 = S 11 ? yy1 - 2(x + x1) = y1² - 4x 1 
It passes through O(0, 0) 
? -2x 1 = y1² - 4x 1 
Locus of M is y² = 2x 
 
 
 
 
Q5: If the chord joining the points P 1(x 1, y 1) and P 2(x 2, y 2) on the parabola y² = 12x 
subtends a right angle at the vertex of the parabola, the x 1x 2 - y 1y 2 is equal to 
A: 280 
B: 288 
C: 292 
D: 284 
Answer: B 
Explanation: