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

The given parabola is $$4y = 3x^2$$, which can be rewritten as $$y = \frac{3}{4}x^2$$. The vertex of this parabola is at the origin $$(0,0)$$ since it is in the standard form $$x^2 = \frac{4}{3}y$$, implying $$4a = \frac{4}{3}$$, so $$a = \frac{1}{3}$$.

The line is $$3x - 2y + 12 = 0$$, which can be rewritten as $$y = \frac{3}{2}x + 6$$.

To find the points of intersection A and B, substitute $$y = \frac{3}{4}x^2$$ into the line equation:

$$\frac{3}{4}x^2 = \frac{3}{2}x + 6$$

Multiply both sides by 4 to clear the denominator:

$$3x^2 = 6x + 24$$

Rearrange into standard quadratic form:

$$3x^2 - 6x - 24 = 0$$

Divide by 3:

$$x^2 - 2x - 8 = 0$$

Factorize:

$$(x - 4)(x + 2) = 0$$

So, $$x = 4$$ or $$x = -2$$.

Substitute these x-values into $$y = \frac{3}{2}x + 6$$:

For $$x = 4$$: $$y = \frac{3}{2}(4) + 6 = 6 + 6 = 12$$, so point A is $$(4, 12)$$.

For $$x = -2$$: $$y = \frac{3}{2}(-2) + 6 = -3 + 6 = 3$$, so point B is $$(-2, 3)$$.

The vertex is O$$(0,0)$$. The angle subtended by AB at O is the angle between the lines OA and OB.

The slope of OA is $$m_1 = \frac{12 - 0}{4 - 0} = 3$$.

The slope of OB is $$m_2 = \frac{3 - 0}{-2 - 0} = -\frac{3}{2}$$.

The formula for the angle $$\phi$$ between two lines with slopes $$m_1$$ and $$m_2$$ is:

$$\tan\phi = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substitute $$m_1 = 3$$ and $$m_2 = -\frac{3}{2}$$:

$$m_1 - m_2 = 3 - \left(-\frac{3}{2}\right) = 3 + \frac{3}{2} = \frac{9}{2}$$

$$1 + m_1 m_2 = 1 + (3) \left(-\frac{3}{2}\right) = 1 - \frac{9}{2} = -\frac{7}{2}$$

So,

$$\tan\phi = \left| \frac{\frac{9}{2}}{-\frac{7}{2}} \right| = \left| \frac{9}{2} \times -\frac{2}{7} \right| = \left| -\frac{9}{7} \right| = \frac{9}{7}$$

Thus, $$\phi = \tan^{-1}\left(\frac{9}{7}\right)$$.

The dot product of vectors $$\overrightarrow{OA} = (4, 12)$$ and $$\overrightarrow{OB} = (-2, 3)$$ is $$4 \times (-2) + 12 \times 3 = -8 + 36 = 28 > 0$$, confirming the angle is acute, so $$\tan\phi = \frac{9}{7}$$ is appropriate.

Comparing with the options, $$\tan^{-1}\left(\frac{9}{7}\right)$$ corresponds to option B.