The area enclosed by the parabola y^2 = 4x and its normal 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 m_T = 2 2 = 1 . The slope of the normal is m_N = -1 . The equation of the normal at (1, 2) is: y - 2 = -1(x - 1) x + y = 3 x = 3 - y To find the intersection of the normal with 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 y = -6, y = 2 The area enclosed between the parabola and the normal is obtaine