If $t_{1} + t_{2} + t_{3} = - t_{1} t_{2} t_{3}$ , then the orthocentre of the triangle formed by the points $(a t_{1} t_{2}, a (t_{1} + t_{2}))$ , $(a t_{2} t_{3}, a (t_{2} + t_{3}))$ and $(a t_{3} t_{1}, a (t_{3} + t_{1}))$ lies on:

x -

axis

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y -

axis

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y = x

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x = a

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Solution

The correct option is B $x -$ axis
Let ABC be triangle whose vertices are $A (a t_{1} t_{2}, a (t_{1} + t_{2})), B (a t_{2} t_{3}, a (t_{2} + t_{3}))$ and $C (a t_{1} t_{3}, a (t_{1} + t_{3}))$
Then slope of BC = $\frac{a (t_{2} + t_{3}) - a (t_{1} + t_{3})}{a t_{1} t_{3} - a t_{1} t_{2}} = \frac{1}{t_{3}}$
slope of AC= $\frac{a (t_{1} + t_{3}) - a (t_{1} + t_{2})}{a t_{1} t_{3} - a t_{1} t_{2}} = \frac{1}{t_{1}}$
So, the equation of aline through A perpendicular to BC is
$y - a (t_{1} + t_{2}) = - t_{3} (x - a t_{1} t_{2})$ ...(1)
And equation of a line through B perpendicular to AC is
$y - a (t_{3} + t_{2}) = - t_{1} (x - a t_{3} t_{2})$ ...(2)
The point of intersection of (1) and (2) is the orthocentre
Solving (1) and (2) we get coordinates of orthocentre are
$(- a, a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3})) = (- a, 0)$

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