The mid-point of a line segment joining the points (x1,y1) and (x2,y2) is (2x1+x2,2y1+y2)
Given, the centroid G is the mid-point of the line segment joining the points (a2+1,a2+1) and (2a,−2a), thus G≡(2a2+1+2a,2a2+1−2a)
⇒G≡(2(a+1)2,2(a−1)2).
Also, given circumcentre C is origin ⇒C≡(0,0).
We know that, for a triangle circumcentre, orthocentre and centroid are collinear.
Thus, C,G and orthocentre H are collinear.
Again, we know that the equation of a line passing through the points (x1,y1) and (x2,y2) is y−y1=(x2−x1y2−y1)(x−x1)
Thus, the orthocentre lies on the line joining the points (0,0) and (2(a+1)2,2(a−1)2) and its equation, is
y−0=(2(a+1)2−0)(2(a−1)2−0)(x−0)
⇒y=(a+1)2(a−1)2(x)
⇒(a−1)2x−(a+1)2y=0.