Tangent and normal are drawn at on the parabola , which intersect the axis of the parabola at and , respectively. If is the centre of the circle through the points , and and , then a value of is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
For y² = 16x, we have 4a = 16, so a = 4. Point P(16,16) corresponds to parameter t = 2 (since x = at², y = 2at gives t = 2).
Tangent at P: ty = x + at² → 2y = x + 16. Setting y = 0: x = -16, so A = (-16, 0).
Normal at P: y - 16 = -t(x - 16) → y - 16 = -2(x - 16) → y = -2x + 48. Setting y = 0: x = 24, so B = (24, 0).
C is the circumcenter of triangle PAB. Since AB is along the x-axis and the circle passes through P, A, B, the center C lies on the perpendicular bisector of AB (x = 4) and the perpendicular bisector of PB.
Midpoint of AB = (4, 0), so C has x = 4.
Midpoint of PB = (20, 8), slope PB = (16-0)/(16-24) = -2, so perpendicular bisector has slope 1/2: y - 8 = (1/2)(x - 20) → at x = 4: y - 8 = -8, y = 0.
So C = (4, 0).
Now find ∠CPB: vectors from P(16,16) to C(4,0) and from P(16,16) to B(24,0).
PC = (4-16, 0-16) = (-12, -16)
PB = (
Tangent at P: ty = x + at² → 2y = x + 16. Setting y = 0: x = -16, so A = (-16, 0).
Normal at P: y - 16 = -t(x - 16) → y - 16 = -2(x - 16) → y = -2x + 48. Setting y = 0: x = 24, so B = (24, 0).
C is the circumcenter of triangle PAB. Since AB is along the x-axis and the circle passes through P, A, B, the center C lies on the perpendicular bisector of AB (x = 4) and the perpendicular bisector of PB.
Midpoint of AB = (4, 0), so C has x = 4.
Midpoint of PB = (20, 8), slope PB = (16-0)/(16-24) = -2, so perpendicular bisector has slope 1/2: y - 8 = (1/2)(x - 20) → at x = 4: y - 8 = -8, y = 0.
So C = (4, 0).
Now find ∠CPB: vectors from P(16,16) to C(4,0) and from P(16,16) to B(24,0).
PC = (4-16, 0-16) = (-12, -16)
PB = (
[JEE Main 2018]