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Range of t₁t₂ for Chord x-b+λy=0 of y²=4ax | JEE

JEE Maths question with a full step-by-step solution.

Question
The line xb+λy=0x - b + \lambda y = 0 cuts the parabola y2=4axy^2 = 4ax (a>0)(a > 0) at P(t1)P(t_1) and Q(t2)Q(t_2). If b[2a,4a]b \in [2a, 4a], then the range of t1t2t_1 t_2, where λR\lambda \in \mathbb{R}, is:
A[4,2][-4, -2]correct
B[2,4][2, 4]
C[4,16][4, 16]
D[16,4][-16, -4]
Solution
Step 1: The chord P(t1)Q(t2)P(t_1)Q(t_2) is xt1+t22y+at1t2=0x - \dfrac{t_1 + t_2}{2}y + at_1 t_2 = 0. Matching x+λyb=0x + \lambda y - b = 0 gives b=at1t2b = -at_1 t_2. Step 2: With b[2a,4a]b \in [2a, 4a] and a>0a > 0,
2t1t24t1t2[4,2].2 \le -t_1 t_2 \le 4 \Rightarrow t_1 t_2 \in [-4, -2].
Correct answer: (1)
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