If the line 3x - 2y + 12 = 0 intersects the p | vedclass

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(D) The given line is $3x - 2y + 12 = 0$ and the parabola is $4y = 3x^2$.From the line equation,$2y = 3x + 12$.Substituting this into the parabola equ

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$	an^{-1}\left(\frac{9}{7}ight)$

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(D) The given line is $3x - 2y + 12 = 0$ and the parabola is $4y = 3x^2$.From the line equation,$2y = 3x + 12$.Substituting this into the parabola equation: $2(3x + 12) = 3x^2$.$3x^2 - 6x - 24 = 0 \Rightarrow x^2 - 2x - 8 = 0$.$(x - 4)(x + 2) = 0$,so $x = 4$ or $x = -2$.If $x = 4$,$4y = 3(16) = 48 \Rightarrow y = 12$. Point $B = (4, 12)$.If $x = -2$,$4y = 3(4) = 12 \Rightarrow y = 3$. Point $A = (-2, 3)$.The vertex of the parabola $4y = 3x^2$ is $O(0, 0)$.The slope of $OA$ is $m_1 = \frac{3 - 0}{-2 - 0} = -\frac{3}{2}$.The slope of $OB$ is $m_2 = \frac{12 - 0}{4 - 0} = 3$.The angle $	heta$ subtended by $AB$ at the vertex $O$ is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{3 - (-3/2)}{1 + (3)(-3/2)} ight| = \left| \frac{9/2}{1 - 9/2} ight| = \left| \frac{9/2}{-7/2} ight| = \frac{9}{7}$.Therefore,$	heta = 	an^{-1}\left(\frac{9}{7}ight)$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

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

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