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(B) The equation of the given parabola is $y^2=4ax$,which has its vertex at $V(0,0)$.Let a point on the parabola be $P(at^2, 2at)$.The slope of the li

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$\frac{4a \cos 	heta}{\sin^2 	heta}$

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(B) The equation of the given parabola is $y^2=4ax$,which has its vertex at $V(0,0)$.Let a point on the parabola be $P(at^2, 2at)$.The slope of the line segment joining $V(0,0)$ and $P(at^2, 2at)$ is given by $\frac{2at-0}{at^2-0} = 	an 	heta$.Simplifying this,we get $\frac{2}{t} = 	an 	heta$,which implies $t = 2 \cot 	heta$.The length of the line segment $VP$ is given by $\sqrt{(at^2-0)^2 + (2at-0)^2} = \sqrt{a^2t^4 + 4a^2t^2} = a|t|\sqrt{t^2+4}$.Substituting $t = 2 \cot 	heta$:$VP = a(2 \cot 	heta) \sqrt{4 \cot^2 	heta + 4} = 2a \cot 	heta \cdot 2 \sqrt{\cot^2 	heta + 1} = 4a \cot 	heta \operatorname{cosec} 	heta$.$VP = 4a \left(\frac{\cos 	heta}{\sin 	heta}ight) \left(\frac{1}{\sin 	heta}ight) = \frac{4a \cos 	heta}{\sin^2 	heta}$.Thus,option $B$ is correct.

If the line segment joining the vertex of the parabola $y^2=4ax$ and a point on the parabola makes an angle $\theta$ with the positive $X$-axis,then the length of that line segment is

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