If  ${\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3=-{\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3$ then the orthocentre of the triangle formed by the points  $\mathrm{A}\left({\mathrm{at}}_1,\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\right)\mathrm{B}\left({\mathrm{at}}_2{\mathrm{t}}_3,\mathrm{a}\left({\mathrm{t}}_2+{\mathrm{t}}_3\right)\right),\mathrm{C}\left({\mathrm{at}}_3{\mathrm{t}}_1,\mathrm{a}\left({\mathrm{t}}_3+{\mathrm{t}}_1\right)\right)$ , lies on

Equation of altitude from the point $\mathrm{A}\left({\mathrm{at}}_1{\mathrm{t}}_2,\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\right)$ is  $\mathrm{y}-\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)=-{\mathrm{t}}_3\left(\mathrm{x}-{\mathrm{at}}_1{\mathrm{t}}_2\right)$
Re write this equation as 
    ${\mathrm{t}}_3\mathrm{x}+\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)+{\mathrm{at}}_1{\mathrm{t}}_2{\mathrm{t}}_3$   …… (1) 
Equation of altitude from the point $\mathrm{B}\left({\mathrm{at}}_2{\mathrm{t}}_3,\mathrm{a}\left({\mathrm{t}}_2+{\mathrm{t}}_3\right)\right)$ is 
    ${\mathrm{t}}_1\mathrm{x}+\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_2+{\mathrm{t}}_3\right)+{\mathrm{at}}_1{\mathrm{t}}_2{\mathrm{t}}_3$   …… (2) 
The ortho center is the point of intersection of the above two lines 
Hence, subtract equation (1) from (2) 
 $\mathrm{x}\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)=\mathrm{a}\left({\mathrm{t}}_1-{\mathrm{t}}_3\right)\Rightarrow \mathrm{x}=-\mathrm{a}$
Substitute x=-a  in the equation ${\mathrm{t}}_3\mathrm{x}+\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)+{\mathrm{at}}_1{\mathrm{t}}_2{\mathrm{t}}_3$
$\begin{array}{c}{\mathrm{t}}_3(-\mathrm{a})+\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)+{\mathrm{at}}_1{\mathrm{t}}_2{\mathrm{t}}_3\\ \mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3\right)+{\mathrm{at}}_1{\mathrm{t}}_2{\mathrm{t}}_3\\ =0\end{array}$

Therefore, the ortho center is $\left(-a,0\right)$  it lies on x- axis.