1. Write the parabola and identify its vertex

The parabola is $4y=3x^2 \implies y=\frac{3}{4}x^2.$ Its vertex is clearly $V=(0,0).$

2. Find the intersection points of the line with the parabola

The line is $3x-2y+12=0 \implies y=\frac{3x+12}{2}.$

Substitute into the parabola: $4\left(\frac{3x+12}{2}\right)=3x^2$ $2(3x+12)=3x^2$ $6x+24=3x^2$ $3x^2-6x-24=0$ $x^2-2x-8=0$ $(x-4)(x+2)=0.$

So the intersection points are at $x=4 \quad \text{and} \quad x=-2.$

Now find the corresponding $y$-coordinates using $y=\frac{3}{4}x^2$:

• For $x=4$: $y=\frac{3}{4}(16)=12,$ so $A=(4,12).$
• For $x=-2$: $y=\frac{3}{4}(4)=3,$ so $B=(-2,3).$

3. Find the angle subtended by chord $AB$ at the vertex

We need the angle between the vectors $\overrightarrow{VA}=(4,12), \qquad \overrightarrow{VB}=(-2,3).$

Let the required angle be $\theta$. Using the formula $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|,$ where $m_1,m_2$ are the slopes of $VA$ and $VB$.

Now, $m_1=\frac{12}{4}=3, \qquad m_2=\frac{3}{-2}=-\frac{3}{2}.$

Hence, $\tan\theta=\left|\frac{3-\left(-\frac{3}{2}\right)}{1+3\left(-\frac{3}{2}\right)}\right|$ $=\left|\frac{\frac{9}{2}}{1-\frac{9}{2}}\right|$ $=\left|\frac{\frac{9}{2}}{-\frac{7}{2}}\right|$ $=\frac{9}{7}.$

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

4. Match with the options

This corresponds to: $\boxed{\text{B }\tan^{-1}\left(\frac{9}{7}\right)}.$

5. Compare with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.