Three normals are drawn from the point (3, 0) | vedclass

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(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Since it passes through $(3, 0)$,we have $

Vedclass · Accepted Answer

$A 	o P; B 	o S; C 	o R; D 	o Q$

Vedclass · Answer

(A) For the parabola $y^2 = 4ax$ (where $a=1$),the normal at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Since it passes through $(3, 0)$,we have $0 = -3t + 2t + t^3$,which simplifies to $t^3 - t = 0$. Thus,$t(t-1)(t+1) = 0$,giving $t_1 = 0, t_2 = 1, t_3 = -1$. The points are $P(0, 0), Q(1, 2), R(1, -2)$. $(A)$ Circumradius $R_{c} = \frac{abc}{4\Delta}$. Sides are $PQ = \sqrt{1^2 + 2^2} = \sqrt{5}$,$PR = \sqrt{5}$,$QR = 4$. Area $\Delta = \frac{1}{2} 	imes 4 	imes 1 = 2$. $R_{c} = \frac{\sqrt{5} \cdot \sqrt{5} \cdot 4}{4 \cdot 2} = \frac{5}{2}$. So $A 	o P$. $(B)$ Area of $\Delta PQR = 2$. So $B 	o S$. $(C)$ Centroid $G = (\frac{0+1+1}{3}, \frac{0+2-2}{3}) = (2/3, 0)$. So $C 	o R$. $(D)$ Circumcenter $O_{c}$ of $\Delta PQR$ with vertices $(0,0), (1,2), (1,-2)$ is $(5/2, 0)$. So $D 	o Q$. Therefore,$A 	o P, B 	o S, C 	o R, D 	o Q$.

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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