Question

{"given":"Parabola: $y^2 = 12x$, so $4a = 12 \\Rightarrow a = 3$. Parametric form: any point on the parabola is $(3t^2,\\ 6t)$. Let $P_1 = (3t_1^2,\\ 6t_1)$ and $P_2 = (3t_2^2,\\ 6t_2)$.","key_observation":"Two lines through the origin are perpendicular if and only if the product of their slopes equals $-1$. Using the parametric slopes, this gives $t_1 t_2 = -4$, which is used to compute both $x_1 x_2$ and $y_1 y_2$.","option_analysis":[{"label":"(A)","text":"$288$","verdict":"correct","explanation":"Slope of $OP_1 = \\frac{2}{t_1}$ and slope of $OP_2 = \\frac{2}{t_2}$. Perpendicularity gives $t_1 t_2 = -4$. Then $x_1 x_2 = 9(t_1 t_2)^2 = 144$ and $y_1 y_2 = 36 t_1 t_2 = -144$, so $x_1 x_2 - y_1 y_2 = 144 - (-144) = 288$."},{"label":"(B)","text":"$280$","verdict":"incorrect","explanation":"This value does not follow from the correct parametric computation with $t_1 t_2 = -4$; it is a distractor."},{"label":"(C)","text":"$284$","verdict":"incorrect","explanation":"This value does not follow from the correct parametric computation with $t_1 t_2 = -4$; it is a distractor."},{"label":"(D)","text":"$292$","verdict":"incorrect","explanation":"This value does not follow from the correct parametric computation with $t_1 t_2 = -4$; it is a distractor."}],"answer":"(A)","formula_steps":[]}