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Q.

If the line $3 x - 2 y + 12 = 0$ intersects the parabola $4 y = 3 x^{2}$ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

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a

$\frac{π}{2} - \tan^{- 1} (\frac{3}{2})$

b

$\tan^{- 1} (\frac{4}{5})$

c

$\tan^{- 1} (\frac{9}{7})$

d

$\tan^{- 1} (\frac{11}{9})$

answer is D.

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Detailed Solution

$\begin{aligned} 3 x - 2 y + 12 = 0 \\ 4 y = 3 x^{2} \\ ∴ 2 (3 x + 12) = 3 x^{2} \\ \Rightarrow x^{2} - 2 x - 8 = 0 \\ \Rightarrow x = - 2, 4 \\ m_{OA} = - 3 / 2, m_{OB} = 3 \\ \tan θ = (\frac{\frac{- 3}{2} - 3}{1 - \frac{9}{2}}) = \frac{9}{7} \end{aligned}$

$θ = \tan^{- 1} (\frac{9}{7})$

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If the line 3x – 2y + 12 = 0 intersects the parabola 4y = 3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to