Question

If the line 3x - 2y + 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 

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

Answer 1

Correct option is (4) \(\tan ^{-1}\left(\frac{9}{7}\right)\)  

3x - 2y + 12 = 0

\(4y = 3 x^{2}\)

\(\therefore 2(3 \mathrm{x}+12)=3 \mathrm{x}^{2}\)

\(\Rightarrow \mathrm{x}^{2}-2 \mathrm{x}-8=0\)

\(\Rightarrow \mathrm{x}=-2,4\)

\(\mathrm{m}_{\mathrm{OA}}=-3 / 2, \mathrm{~m}_{\mathrm{OB}}=3\)

\(\tan \theta=\left(\frac{\frac{-3}{2}-3}{1-\frac{9}{2}}\right)=\frac{9}{7}\)

\(\theta=\tan ^{-1}\left(\frac{9}{7}\right)\) (angle will be acute)