Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).

Answer:(4, 4)Step-by-step explanation:There are a couple of ways to go at this:Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.Find the equation of a normal line to the parabola that goes through the given point.1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of xDifferentiating with respect to x and setting dd/dx=0, we have ...... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)We can factor 2 from this to get... 0 = x -2 +(y -8)(dy/dx)Differentiating the parabola's equation, we find ...... 2y(dy/dx) = 4... dy/dx = 2/ySubstituting for x (=y²/4) and dy/dx into our derivative equation above, we get... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y... 64 = y³ . . . . . . multiply by 4y, add 64... 4 = y . . . . . . . . cube root... y²/4 = 16/4 = x = 4_____2. The derivative above tells us the slope at point (x, y) on the parabola is ...... dy/dx = 2/yThen the slope of the normal line at that point is ... ... -1/(dy/dx) = -y/2The normal line through the point (2, 8) will have equation (in point-slope form) ...... y - 8 = (-y/2)(x -2)Substituting for x using the equation of the parabola, we get... y - 8 = (-y/2)(y²/4 -2)Multiplying by 8 gives ...... 8y -64 = -y³ +8y... y³ = 64 . . . . subtract 8y, multiply by -1... y = 4 . . . . . . cube root... x = y²/4 = 4The point on the parabola that is closest to the point (2, 8) is (4, 4).