If the normals drawn at the points Pleft(frac | vedclass

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(B) The equation of the parabola is $y^2 = 4ax$,where $4a = 3$,so $a = \frac{3}{4}$.For a point $(at^2, 2at)$ on the parabola,the normal at $t$ is $y

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$\left(\frac{27}{4}, -\frac{9}{2}\right)$

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(B) The equation of the parabola is $y^2 = 4ax$,where $4a = 3$,so $a = \frac{3}{4}$.For a point $(at^2, 2at)$ on the parabola,the normal at $t$ is $y = -tx + 2at + at^3$.For point $P\left(\frac{3}{4}, \frac{3}{2}ight)$,we have $at^2 = \frac{3}{4} \implies \frac{3}{4}t^2 = \frac{3}{4} \implies t_1 = 1$ (since $2at_1 = \frac{3}{2} \implies 2(\frac{3}{4})t_1 = \frac{3}{2} \implies t_1 = 1$).For point $Q(3,3)$,we have $at^2 = 3 \implies \frac{3}{4}t^2 = 3 \implies t^2 = 4 \implies t_2 = -2$ (since $2at_2 = 3 \implies 2(\frac{3}{4})t_2 = 3 \implies t_2 = 2$,but the normal at $t_2=2$ is $y = -2x + 2(\frac{3}{4})(2) + (\frac{3}{4})(8) = -2x + 3 + 6 = -2x + 9$. Checking $Q(3,3)$: $3 = -2(3) + 9 = 3$,so $t_2 = 2$).If normals at $t_1$ and $t_2$ meet at $R(t_3)$,then $t_3 = -(t_1 + t_2) = -(1 + 2) = -3$.The coordinates of $R$ are $(at_3^2, 2at_3) = (\frac{3}{4}(-3)^2, 2(\frac{3}{4})(-3)) = (\frac{3}{4} 	imes 9, -\frac{9}{2}) = (\frac{27}{4}, -\frac{9}{2})$.

If the normals drawn at the points $P\left(\frac{3}{4}, \frac{3}{2}\right)$ and $Q(3,3)$ on the parabola $y^2=3x$ intersect again on the parabola at $R$,then $R=$

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