Let P be a point on the parabola y^2 = 4ax,wh | vedclass

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(A) Let the point $P$ be $(at^2, 2at)$. The equation of the normal at $P$ is $y = -tx + 2at + at^3$.For the $x$-axis,set $y = 0$,which gives $0 = -tx

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$(2, 3)$

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(A) Let the point $P$ be $(at^2, 2at)$. The equation of the normal at $P$ is $y = -tx + 2at + at^3$.For the $x$-axis,set $y = 0$,which gives $0 = -tx + 2at + at^3$,so $x = 2a + at^2$. Thus,$Q$ is $(2a + at^2, 0)$.The focus $F$ is $(a, 0)$.The area of $	riangle PFQ$ is given by $\frac{1}{2} |x_P(y_F - y_Q) + x_F(y_Q - y_P) + x_Q(y_P - y_F)|$.Substituting the coordinates $P(at^2, 2at)$,$F(a, 0)$,and $Q(2a + at^2, 0)$:Area $= \frac{1}{2} |at^2(0 - 0) + a(0 - 2at) + (2a + at^2)(2at - 0)| = \frac{1}{2} |-2a^2t + 4a^2t + 2a^2t^3| = |a^2t + a^2t^3| = a^2|t|(1 + t^2)$.Given the slope of the normal $m = -t$,so $t = -m$. Since $m > 0$,$t$ is negative. Let $t = -m$,then $|t| = m$.Area $= a^2 m(1 + m^2) = 120$.For $a = 2$ and $m = 3$,Area $= 2^2 	imes 3(1 + 3^2) = 4 	imes 3(10) = 120$.Thus,the pair $(a, m)$ is $(2, 3)$.

Column-$I$	Column-$II$
$(A)$ Circumradius of $\Delta PQR$	$(P)$ $5/2$
$(B)$ Area of $\Delta PQR$	$(Q)$ $(5/2, 0)$
$(C)$ Centroid of $\Delta PQR$	$(R)$ $(2/3, 0)$
$(D)$ Circumcenter of $\Delta PQR$	$(S)$ $2$

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