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(A) Let the parabola be $y^{2} = 4ax$. The center of the circle is $C(x_1, 0)$.Since the circle touches the parabola at $P$,the radius $CP$ is perpend

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(A) Let the parabola be $y^{2} = 4ax$. The center of the circle is $C(x_1, 0)$.Since the circle touches the parabola at $P$,the radius $CP$ is perpendicular to the tangent at $P$.The angle between $CP$ and the axis of the parabola is $120^{\circ}$,so the angle between $CP$ and the normal at $P$ is $30^{\circ}$ (or the slope of $CP$ is $	an(180^{\circ}-120^{\circ}) = 	an 60^{\circ} = \sqrt{3}$ is incorrect,rather the angle with the negative x-axis is $60^{\circ}$).Let $P = (x_0, y_0)$. The radius $r=2$. The coordinates of $P$ are $(x_1 - 2\cos 60^{\circ}, 2\sin 60^{\circ}) = (x_1 - 1, \sqrt{3})$.Since $P$ lies on $y^2 = 4ax$,we have $(\sqrt{3})^2 = 4a(x_1 - 1) \Rightarrow 3 = 4a(x_1 - 1)$.The slope of the tangent at $P$ is $m_t = \frac{2a}{y_0} = \frac{2a}{\sqrt{3}}$.The slope of the normal $CP$ is $m_n = \frac{\sqrt{3} - 0}{(x_1 - 1) - x_1} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$.Since the normal is perpendicular to the tangent,$m_t 	imes m_n = -1$.$\left(\frac{2a}{\sqrt{3}}ight)(-\sqrt{3}) = -1$ $\Rightarrow -2a = -1$ $\Rightarrow a = \frac{1}{2}$.The latus rectum is $4a = 4 	imes \frac{1}{2} = 2$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

$A$ circle has its center $C$ on the axis of a parabola and it touches the parabola at point $P$. The line segment $CP$ makes an angle of $120^{\circ}$ with the axis of the parabola. If the radius of the circle is $2$,then the latus rectum of the parabola is:

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