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

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

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

(B) $\frac{\pi}{2}-\tan ^{-1}\left(\frac{3}{2}\right)$

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

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Parabola

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

Answer: (D)

Substitute

y = \frac{3 x + 12}{2}

(from the line) into

4 y = 3 x^{2}

2 (3 x + 12) = 3 x^{2} ⟹ x^{2} - 2 x - 8 = 0 ⟹ (x - 4) (x + 2) = 0

x = - 2

gives

A = (- 2, 3)

, and

x = 4

gives

B = (4, 12)

.

Slopes from the vertex

O (0, 0)

m_{O A} = \frac{3}{- 2} = - \frac{3}{2}, m_{O B} = \frac{12}{4} = 3

tan θ = \frac{m _{O A} - m _{O B}}{1 + m _{O A} m _{O B}} = \frac{- \frac{3}{2} - 3}{1 - \frac{9}{2}} = \frac{- \frac{9}{2}}{- \frac{7}{2}} = \frac{9}{7}

θ = tan^{- 1} \frac{9}{7}

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option