JEE MAIN - Mathematics (2025 - 23rd January Morning Shift - No. 15)
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
$\frac{\pi}{2}-\tan ^{-1}\left(\frac{3}{2}\right)$
$\tan ^{-1}\left(\frac{9}{7}\right)$
$\tan ^{-1}\left(\frac{11}{9}\right)$
$\tan ^{-1}\left(\frac{4}{5}\right)$
Explanation
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$$\begin{aligned} & 3 x-2 y+12=0 \\ & 4 y=3 x^2 \\ & \therefore 2(3 x+12)=3 x^2 \\ & \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow x=-2,4 \\ & \mathrm{~m}_{\mathrm{OA}}=-3 / 2, \mathrm{~m}_{\mathrm{OB}}=3 \\ & \tan \theta=\left(\frac{\frac{-3}{2}-3}{1-\frac{9}{2}}\right)=\frac{9}{7} \\ & \theta=\tan ^{-1}\left(\frac{9}{7}\right) \text { (angle will be acute) } \end{aligned}$$
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