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) = 1i.e., 4t2 + 4t + 5 = 0Here, discriminent = 16 – 4 × 4 × 5 = –64 < 0.∴No real value of t is possible.Hence, the given point cannot lie on the line.