The vertices of a triangle are (at_1t_2, a(t_ | vedclass

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(B) Let the vertices be $A(at_1t_2, a(t_1 + t_2))$,$B(at_2t_3, a(t_2 + t_3))$,and $C(at_3t_1, a(t_3 + t_1))$.The slope of $BC$ is $m_{BC} = \frac{a(t_

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$(-a, a(t_1 + t_2 + t_3 + t_1t_2t_3))$

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(B) Let the vertices be $A(at_1t_2, a(t_1 + t_2))$,$B(at_2t_3, a(t_2 + t_3))$,and $C(at_3t_1, a(t_3 + t_1))$.The slope of $BC$ is $m_{BC} = \frac{a(t_2 + t_3) - a(t_3 + t_1)}{at_2t_3 - at_3t_1} = \frac{a(t_2 - t_1)}{at_3(t_2 - t_1)} = \frac{1}{t_3}$.The slope of the altitude from $A$ to $BC$ is $-t_3$. The equation of this altitude is $y - a(t_1 + t_2) = -t_3(x - at_1t_2)$,which simplifies to $t_3x + y = a(t_1 + t_2 + t_1t_2t_3)$.Similarly,the equation of the altitude from $B$ to $AC$ is $t_1x + y = a(t_2 + t_3 + t_1t_2t_3)$.Subtracting the two equations: $(t_3 - t_1)x = a(t_1 - t_3)$,so $x = -a$.Substituting $x = -a$ into the first altitude equation: $-at_3 + y = a(t_1 + t_2 + t_1t_2t_3)$,which gives $y = a(t_1 + t_2 + t_3 + t_1t_2t_3)$.Thus,the orthocentre is $(-a, a(t_1 + t_2 + t_3 + t_1t_2t_3))$.

The vertices of a triangle are $(at_1t_2, a(t_1 + t_2))$,$(at_2t_3, a(t_2 + t_3))$,and $(at_3t_1, a(t_3 + t_1))$. Find the coordinates of its orthocentre.

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