Let and be points on the parabola, . Let be chosen on an arc of the parabola, where is the origin, such that the area of is maximum. Then, the area (in sq. units) of , is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
For a parabola y² = 4x, points are parametrized as (t², 2t). So A = (4,-4) gives t = -2, and B = (9,6) gives t = 3.
Area of triangle ACB with C = (t², 2t) is (1/2)|det of vectors|. Using the formula for area with vertices A(4,-4), B(9,6), C(t²,2t):
Area = (1/2)|( 4(6-2t) + 9(2t-(-4)) + t²(-4-6) )|
= (1/2)|24 - 8t + 18t + 36 - 10t²|
= (1/2)|60 + 10t - 10t²|
= 5|6 + t - t²|
Maximize f(t) = 6 + t - t² (for t between -2 and 3, arc AOB through origin means t ∈ (-2, 0) ∪ ... actually C is on arc AOB not containing the chord directly).
Since O is the origin (t=0) and the arc AOB passes through O (t=0), C must lie on arc from t=-2 to t=3 passing through t=0, so t ∈ (-2, 3).
Maximize f(t) = -t² + t + 6: df/dt = -2t + 1 = 0 → t = 1/2.
f(1/2) = -(1/4) + (1/2) + 6 = 6.25 = 25/4.
Maximum
Area of triangle ACB with C = (t², 2t) is (1/2)|det of vectors|. Using the formula for area with vertices A(4,-4), B(9,6), C(t²,2t):
Area = (1/2)|( 4(6-2t) + 9(2t-(-4)) + t²(-4-6) )|
= (1/2)|24 - 8t + 18t + 36 - 10t²|
= (1/2)|60 + 10t - 10t²|
= 5|6 + t - t²|
Maximize f(t) = 6 + t - t² (for t between -2 and 3, arc AOB through origin means t ∈ (-2, 0) ∪ ... actually C is on arc AOB not containing the chord directly).
Since O is the origin (t=0) and the arc AOB passes through O (t=0), C must lie on arc from t=-2 to t=3 passing through t=0, so t ∈ (-2, 3).
Maximize f(t) = -t² + t + 6: df/dt = -2t + 1 = 0 → t = 1/2.
f(1/2) = -(1/4) + (1/2) + 6 = 6.25 = 25/4.
Maximum
[JEE Main 2019 · 9 Jan · Shift II]