Q: If two chords drawn from the point (4 , 4) to the parabola x^2 = 4y are divided by line y = m x in the ratio 1 : 2 , then
(A) m < – √3
(B) m < –√3 – 1
(C) m > √3
(D) m > √3– 1
Sol. Point (4, 4) lies on the parabola.

Let the point of intersection of the line y = m x with the chords be (a , ma), then
$\large \alpha = \frac{4+2x_1}{3}$
$\large x_1 = \frac{3\alpha – 4}{2}$
and , $\large m \alpha = \frac{4 + 2 y_1}{3}$
$\large y_1 = \frac{3m \alpha – 4}{2}$
as (x1, y1) lies on the curve
$\large (\frac{3\alpha – 4}{2})^2 = 4 (\frac{3m\alpha – 4}{2}) $
⇒ 9α2 + 16 – 24a = 8(3mα – 4)
Þ 9α2 – 24α(1 + m) + 48 = 0
Þ 3α2 – 8α(1 + m) + 16 = 0
two distinct chords are obtained
D > 0.
(8(1 + m))2 – 4 . 3 . 16 > 0
⇒ (1 + m)2 – 3 > 0
1 + m > √3 or 1 + m < –√3
⇒ m > √3– 1 or m < –( √3 + 1).
Hence (B), (C) and (D) are the correct answers.
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