If the normal drawn from the point (t_1^2, 2t | vedclass

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(D) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.Here,$a = 1$,so the normal at $(t_1^2,

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$t_2 = -t_1 - \frac{2}{t_1}$

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(D) The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.Here,$a = 1$,so the normal at $(t_1^2, 2t_1)$ is $y = -t_1x + 2t_1 + t_1^3$.Since this normal passes through $(t_2^2, 2t_2)$,we substitute these coordinates into the equation:$2t_2 = -t_1(t_2^2) + 2t_1 + t_1^3$.Rearranging the terms:$2(t_2 - t_1) = -t_1(t_2^2 - t_1^2)$.$2(t_2 - t_1) = -t_1(t_2 - t_1)(t_2 + t_1)$.Since $t_1 
eq t_2$ for the normal to intersect at a different point,we can divide by $(t_2 - t_1)$:$2 = -t_1(t_2 + t_1)$.$2 = -t_1t_2 - t_1^2$.$t_1t_2 = -(t_1^2 + 2)$.

If the normal drawn from the point $(t_1^2, 2t_1)$ to the parabola $y^2 = 4x$ intersects the parabola again at the point $(t_2^2, 2t_2)$,then -

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