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

Step-by-step solution

First, we find the equation of the normal to the parabola y^2 = 4x at the point (1, 2) . Differentiating y^2 = 4x with respect to x : 2y dy dx = 4 dy dx = 2 y At (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) y = -x + 3 x = 3 - y The region is bounded by the parabola x = y^2 4 , the normal x = 3 - y , and the x-axis ( y = 0 ). To find the area A , it is convenient to integrate with respect to y from y = 0 to the i