If \( \Large t_{1}+t_{2}+t_{3}=-t_{1}t_{2}t_{3} \) then orthocentre of the triangle formed by the points \( \Large \left[ at_{1}t_{2},\ a \left(t_{1}+t_{2}\right) \right],\ \left[ at_{2}t_{3},\ a \left(t_{2}+t_{3}\right) \right]\ and\ \left[ at_{3}t_{1},\ a \left(t_{3}-t_{1}\right) \right] \) lies on;

Let \( \Large A\left[ at_{1}t_{2},\ a \left(t_{1}+t_{2}\right),\ B\left[ at_{2}t_{3},\ a \left(t_{2}+t_{3}\right) C \left[ at_{3}t_{1},\ a \left(t_{3}+t_{1}\right) \right] \right] \right] \)

Slope of \( \Large AB \left(m_{AB}\right) = \frac{a \left(t_{3}-t_{1}\right) }{at_{2} \left(t_{3}-t_{1}\right) } = \frac{1}{t_{2}} \)

Equation of line through C perpendicular to AB is

\( \Large y-a \left(t_{3}+t_{1}\right)=-t_{2} \left(x-at_{3}t_{1}\right) \)

=> \( \Large y-a \left(t_{3}+t_{1}\right)=-t_{2} x+at_{1}t_{2}t_{3} \) ...(i)

Similarly, equation of line through B perpendicular to CA is

\( \Large y-a \left(t_{2}+t_{3}\right)=-t_{1}x+at_{1}t_{2}t_{3} \) ...(ii)

Using \( \Large t_{1}t_{2}t_{3}=- \left(t_{1}+t_{2}+t_{3}\right) \) in equations (i) and (ii)

\( \Large y = -t_{2}x-at_{2} \)

and \( \Large y = -t_{1}x-at_{1} \)

=> \( \Large t_{2} \left(x+a\right) = t_{1} \left(x+a\right) \)

=> \( \Large x = -a,\ y = 0 \)