The area (in sq. units) of the region bounded by the parabola y^2 = 4x and its normal chord drawn at the point (1, 2) is

Step-by-step solution

The equation of the parabola is y^2 = 4x . Differentiating with respect to x , we get 2y dy dx = 4 dy dx = 2 y . At the point (1, 2) , the slope of the tangent is 2 2 = 1 . Therefore, the slope of the normal is -1 . The equation of the normal at (1, 2) is: y - 2 = -1(x - 1) x = 3 - y To find the points of intersection of the normal chord and the parabola, substitute x into the parabola's equation: y^2 = 4(3 - y) y^2 + 4y - 12 = 0 (y + 6)(y - 2) = 0 Thus, the y -coordinates of the intersection points are y = -6 and