Step 1: Recall how the three centres line up.
For any triangle the orthocentre $H$, centroid $G$ and circumcentre $O$ lie on one line, and $\vec{OH}=3\,\vec{OG}$. Here $O$ is the origin, so simply $H=3G$.
Step 2: Find the centroid $G$.
$G$ is the midpoint of $(a^2+1,a^2+1)$ and $(2a,-2a)$: \[ G\left(\frac{a^2+1+2a}{2},\frac{a^2+1-2a}{2}\right). \]
Step 3: Use perfect squares to tidy $G$.
Since $a^2+2a+1=(a+1)^2$ and $a^2-2a+1=(a-1)^2$, \[ G\left(\frac{(a+1)^2}{2},\frac{(a-1)^2}{2}\right). \]
Step 4: Get the orthocentre.
Multiply $G$ by $3$: \[ H\left(\frac{3(a+1)^2}{2},\frac{3(a-1)^2}{2}\right). \]
Step 5: Find the line from origin through $H$.
A curve through the origin and $H$ has slope \[ m=\frac{3(a-1)^2/2}{3(a+1)^2/2}=\frac{(a-1)^2}{(a+1)^2}. \] So $y=\dfrac{(a-1)^2}{(a+1)^2}x$.
Step 6: Clear the fraction.
Cross multiply: $(a+1)^2y=(a-1)^2x$, which rearranges to the answer. \[ \boxed{(a-1)^2x-(a+1)^2y=0} \]