Given curve is y2 = 4x + 5 on differentiating, we get 2y dxdy = 4 ⇒ dxdy = y2 Given line is 2x - y + 5 = 0 ⇒ y = 2x + 5 slope of line is 2. Therefore, y2 = 2 ⇒ y = 1 put y = 1 in the equation of curve, we get 1 = 4x + 5 x = - 1 Hence, point of contact is (- 1, 1)