The point (2t2+2t+4,t2+t+1) lies on the line x + 2y = 1 for
The point (2t2+2t+4,t2+t+1) lies on the line
x + 2y = 1 if (2t2 + 2t + 4) + 2(t2 + t + 1) = 1
i.e. 4t2 + 4t + 5 = 0
Here, discriminant = 16 – 7.4.5 = –64 < 0
∴ No real value of t is possible
Hence, the given point cannot lie on the line.