In our class, we proved that the orthocentre of the triangle formed by intersection of 3 tangents to a parabola, $y^2=4ax$ always lies on its directrix, $x=-a$. I wanted to find the condition for which the circumcentre of this triangle also lies on the directrix. To do the same, I thought of finding the condition for the centroid to lie on the directrix(which implies that the circumcentre also lies on the same, Euler line).
My attempt:
Let the points of contact of the tangents be $A(t_1) , B(t_2) , C(t_3)$ where $t_1,t_2,t_3$ are parameters for the parabola. Let $P,Q,R$ be their points of intersection. $P,Q,R \equiv (at_it_j,a(t_i+t_j))$ $\Rightarrow$ x coordinate of centroid, $$G_x=\frac{a(t_1t_2+t_2t_3+t_3t_1)}{3}$$ For $G_x=-a$,
$\Rightarrow t_1t_2+t_2t_3+t_1t_3=-3$. which is of the form $$xy+yz+zx=-3$$. $\Rightarrow (x+y+z)^2 - x^2-y^2-z^2=-6$
My question is, how many real solutions of $(x,y,z)$ does the given equation have?
I have been able to deduce only one condition.
1)$x\neq y\neq z$ because the coordinates have to be distinct, for a triangle to be formed.
And with your answer, could you also please tell me how to approach equations like the said equation when you have only one relation between $(x,y,z)$ to deduce the number of solutions (finite or infinite)?