Question ID: #224
If the line $3x – 2y + 12 = 0$ intersects the parabola $4y = 3x^2$ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to
- (1) $\tan^{-1}\left(\frac{11}{9}\right)$
- (2) $\frac{\pi}{2} – \tan^{-1}\left(\frac{3}{2}\right)$
- (3) $\tan^{-1}\left(\frac{4}{5}\right)$
- (4) $\tan^{-1}\left(\frac{9}{7}\right)$
Solution:
Parabola: $3x^2 = 4y$; Line: $2y = 3x+12$.
Substitute $2y$:
$$3x^2 = 2(3x+12) \Rightarrow x^2 – 2x – 8 = 0 \Rightarrow (x-4)(x+2) = 0$$
Points of intersection:
1. $x = -2 \Rightarrow y = 3$. Point A $(-2, 3)$.
2. $x = 4 \Rightarrow y = 12$. Point B $(4, 12)$.
Slopes connecting vertex O(0,0) to A and B:
$$m_1 = \frac{3}{-2} = -\frac{3}{2}, \quad m_2 = \frac{12}{4} = 3$$
Angle $\theta$:
$$\tan \theta = \left| \frac{m_2 – m_1}{1 + m_1 m_2} \right| = \left| \frac{3 – (-\frac{3}{2})}{1 + 3(-\frac{3}{2})} \right| = \left| \frac{\frac{9}{2}}{1 – \frac{9}{2}} \right| = \left| \frac{\frac{9}{2}}{-\frac{7}{2}} \right| = \frac{9}{7}$$
$$\theta = \tan^{-1}\left(\frac{9}{7}\right)$$
Ans.(4)
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