text: To be the closest point, the line from (3,0) to the point must be normal to the hyperbola.  The slope of the normal is the negative reciprocal of the slope of the curve.  \n\nThe curve is: y = 8/x\nThe curve slope is dy/dx = -8/x^2\nThe slope of the line from (3,0) to the cloest point is: m = (x^2)/8\n\nThe distance from (3,0) to any point (x, y) is D = sqrt((x-3)^2 + y^2)\n\nTo be along the line y = 8/x, the the distance is:\n\nD = sqrt((x-3)^2 + (8/x)^2)\n\nLook at the derivative dD/dx which is how the distance changes with x:\n\ndD/dx =(x - 3 - (64/x^3))/sqrt((x-3)^2 + 64/x^2)\n\nFor the distance to be minimum, the numerator must be zero:\n\n0 =  x - 3 - (64/x^3)\n\nBy inspection, this is solved at x = 4.  The closest point is then (4, 2)