Let A be the area of the region bounded by the parabola y^2 = 2(x-2) and the normal to this parabola at the point (4,2) . The value of 12A is

Step-by-step solution

The equation of the given parabola is y^2 = 2(x-2) . Differentiating with respect to x , we get: 2y dy dx = 2 dy dx = 1 y At the point (4,2) , the slope of the tangent is m_t = 1 2 . Therefore, the slope of the normal at (4,2) is m_n = -2 . The equation of the normal is: y - 2 = -2(x - 4) y - 2 = -2x + 8 x = 10-y 2 = 5 - y 2 To find the points of intersection of the normal and the parabola, substitute x from the parabola's equation x = y^2 2 + 2 into the normal's equation: y^2 2 + 2 = 5 - y 2 y^2 + y - 6 = 0 (y+3)(