The vertices of a triangle are `[a t_1t_2,a(t_1 +t_2)]`, `[a t_2t_3,a(t_2 +t_3)]`, `[a t_3t_1,a(t_3 +t_1)]` Then the orthocenter of the triangle is (a) `(-a, a(t_1+t_2+t_3)-at_1t_2t_3)` (b) `(-a, a(t_1+t_2+t_3)+at_1t_2t_3)` (c) `(a, a(t_1+t_2+t_3)+at_1t_2t_3)` (d) `(a, a(t_1+t_2+t_3)-at_1t_2t_3)`

Explanation

Let the given points be `{A}, {B}` and `{C}` respectively.


Slope of BC by `frac{y_{2}-y_{1}}{x_{2}-x_{1}}` is `frac{1}{t_{3}}`


Hence the line through A perpendicular to `B C` is



{y}-{a}({t}_{1}+{t}_{2})=-{t}_{3}({x}-{at}_{1} {t}_{2})



Similarly the line through `{B} perp` to `{CA}` is


`y-a(t_{2}+t_{3})=-t_{3}(x-a t_{2} t_{3}) ldots(2)` Subtracting, we get `x=-a` and hence `y=`


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