If the normal at the point t_1 (i.e.,at (at_1 | vedclass

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at point $t_1$ is given by $y + t_1x = 2at_1 + at_1^3$. Since this normal meets the parabol

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$-2 - t_1^2$

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(A) The equation of the normal to the parabola $y^2 = 4ax$ at point $t_1$ is given by $y + t_1x = 2at_1 + at_1^3$. Since this normal meets the parabola again at point $t_2 = (at_2^2, 2at_2)$,the point $(at_2^2, 2at_2)$ must satisfy the equation of the normal. Substituting $x = at_2^2$ and $y = 2at_2$ into the normal equation: $2at_2 + t_1(at_2^2) = 2at_1 + at_1^3$. Dividing by $a$ (assuming $a 
eq 0$): $2t_2 + t_1t_2^2 = 2t_1 + t_1^3$. Rearranging the terms: $t_1(t_2^2 - t_1^2) + 2(t_2 - t_1) = 0$. $t_1(t_2 - t_1)(t_2 + t_1) + 2(t_2 - t_1) = 0$. Since $t_1 
eq t_2$ for the normal to meet the parabola at a distinct second point,we can divide by $(t_2 - t_1)$: $t_1(t_2 + t_1) + 2 = 0$. $t_1t_2 + t_1^2 + 2 = 0$. Therefore,$t_1t_2 = -2 - t_1^2$.

If the normal at the point $t_1$ (i.e.,at $(at_1^2, 2at_1)$) on the parabola $y^2 = 4ax$ meets the parabola again at the point $t_2$,then $t_1t_2$ is equal to:

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