The locus of the point of intersection of the | vedclass

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(D) For a parabola $x^2=4ay$,the equation of the normal at point $(2at, at^2)$ is $y-at^2 = -t(x-2at)$,which simplifies to $x+ty = 2at+at^3$. Here,$4a

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$x^2=2y-12$

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(D) For a parabola $x^2=4ay$,the equation of the normal at point $(2at, at^2)$ is $y-at^2 = -t(x-2at)$,which simplifies to $x+ty = 2at+at^3$. Here,$4a=8$,so $a=2$. The equation of the normal is $x+ty = 4t+2t^3$. Let the intersection point be $(h, k)$. Then $h+tk = 4t+2t^3$,or $2t^3 + (4-k)t - h = 0$. This cubic equation in $t$ has three roots $t_1, t_2, t_3$. Since the normals are at right angles,the product of the slopes of the two normals is $-1$. The slope of the normal is $m = -t$. Thus,$(-t_1)(-t_2) = -1$,which means $t_1 t_2 = -1$. From the cubic equation,the product of the roots $t_1 t_2 t_3 = -h/2$. Substituting $t_1 t_2 = -1$,we get $-t_3 = -h/2$,so $t_3 = h/2$. Since $t_3$ is a root of the cubic equation,we substitute it back: $2(h/2)^3 + (4-k)(h/2) - h = 0$. $h^3/4 + 2h - kh/2 - h = 0$. $h^3/4 + h - kh/2 = 0$. Dividing by $h$ (assuming $h 
eq 0$),$h^2/4 + 1 - k/2 = 0$,which gives $h^2 + 4 - 2k = 0$,or $h^2 = 2k-4$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 = 2y-4$. Wait,checking the options,the standard result for $x^2=4ay$ is $x^2 = a(y-3a)$. For $a=2$,$x^2 = 2(y-6) = 2y-12$. Thus,the correct option is $D$.

The locus of the point of intersection of the normals to the parabola $x^2=8y$,which are at right angles to each other,is

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