Let A be the area of the region bounded by the parabola y^2 = 4x and its normal drawn at the point (1, 2) . Then the value of 3A is

Step-by-step solution

The given curve is y^2 = 4x . Differentiating with respect to x , we get 2y dy dx = 4 , which implies dy dx = 2 y . At the point (1, 2) , the slope of the tangent is 2 2 = 1 . The slope of the normal at this point is -1 . The equation of the normal at (1, 2) is: y - 2 = -1(x - 1) y = -x + 3 x = 3 - y To find the points of intersection of the normal and the parabola, substitute x = 3 - y into y^2 = 4x : y^2 = 4(3 - y) y^2 + 4y - 12 = 0 (y + 6)(y - 2) = 0 The points of intersection correspond to y = 2 and y = -6 . Th