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The line x - b +lambda y = 0 cuts the pa...

The line `x - b +lambda y = 0` cuts the parabola `y^(2) = 4ax (a gt 0)` at `P(t_(1))` and `Q(t_(2))`. If `b in [2a, 4a]` then range of `t_(1)t_(2)` where `lambda in R` is

A

`[-4,-2]`

B

`[2,4]`

C

`[4,16]`

D

`[-16,-4]`

Text Solution

Verified by Experts

The correct Answer is:
A
Line PQ is `(t_(1)+t_(2)) y = 2(x+a t_(1)t_(2))` (1)
`lambda y =- x +b` (2)
`(t_(1)+t_(2))/(lambda) = (2)/(-1) =(2a t_(1)t_(2))/(b)`
`:. B =- a t_(1)t_(2)` and `2a le -a t_(1)t_(2) le 4a`
`:. -4 le t_(1)t_(2) le -2`
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