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 (x_{1}, y_{1}) lies on the curve

$\large (\frac{3\alpha – 4}{2})^2 = 4 (\frac{3m\alpha – 4}{2}) $

⇒ 9α^{2} + 16 – 24a = 8(3mα – 4)

Þ 9α^{2} – 24α(1 + m) + 48 = 0

Þ 3α^{2} – 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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