If the ordinates of points P and Q on the par | vedclass

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(A) Given the parabola $y^2=12x$,we have $4a=12$,so $a=3$. Let the parameters of points $P$ and $Q$ be $t_1$ and $t_2$ respectively. The ordinates are

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$y+18\left(\frac{x-6}{21}\right)^{3/2}=0$

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(A) Given the parabola $y^2=12x$,we have $4a=12$,so $a=3$. Let the parameters of points $P$ and $Q$ be $t_1$ and $t_2$ respectively. The ordinates are $y_1=2at_1$ and $y_2=2at_2$. Given $y_1:y_2=1:2$,we have $t_1:t_2=1:2$,so $t_2=2t_1$. Let $t_1=t$,then $t_2=2t$.The point of intersection $(x, y)$ of the normals at $t_1$ and $t_2$ is given by:$x = 2a + a(t_1^2 + t_2^2 + t_1t_2) = 2(3) + 3(t^2 + 4t^2 + 2t^2) = 6 + 21t^2$$y = -at_1t_2(t_1+t_2) = -3(t)(2t)(t+2t) = -6t^2(3t) = -18t^3$From $x=6+21t^2$,we get $t^2 = \frac{x-6}{21}$,so $t = \pm \left(\frac{x-6}{21}\right)^{1/2}$.Substituting $t$ into $y=-18t^3$,we get $y = -18 \left(\frac{x-6}{21}\right)^{3/2}$,which simplifies to $y+18\left(\frac{x-6}{21}\right)^{3/2}=0$.

If the ordinates of points $P$ and $Q$ on the parabola $y^2=12x$ are in the ratio $1:2$,then the locus of the point of intersection of the normals to the parabola at $P$ and $Q$ is

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