The area of the region bounded by the parabola y^2 = 8x , the normal to the parabola drawn at the point (2, 4) , and the x -axis in the first quadrant is equal to

Step-by-step solution

The equation of the parabola is y^2 = 8x . Differentiating with respect to x , we get: 2y dy dx = 8 dy dx = 4 y At the point (2, 4) , the slope of the tangent is m_T = 4 4 = 1 . Thus, the slope of the normal is m_N = -1 . The equation of the normal at (2, 4) is: y - 4 = -1(x - 2) x + y = 6 x = 6 - y The region is bounded by the parabola x = y^2 8 , the normal x = 6 - y , and the x -axis ( y = 0 ) in the first quadrant. To find the upper limit for y , we find the intersection of the parabola and the normal in the fi