text: closest proximity of a point on a hyperbola? find the point on the hyperbola xy=8 that is closest to the point (3,0) 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.    the curve is: y = 8/x the curve slope is dy/dx = -8/x^2 the slope of the line from (3,0) to the cloest point is: m = (x^2)/8  the distance from (3,0) to any point (x, y) is d = sqrt((x-3)^2 + y^2)  to be along the line y = 8/x, the the distance is:  d = sqrt((x-3)^2 + (8/x)^2)  look at the derivative dd/dx which is how the distance changes with x:  dd/dx =(x - 3 - (64/x^3))/sqrt((x-3)^2 + 64/x^2)  for the distance to be minimum, the numerator must be zero:  0 =  x - 3 - (64/x^3)  by inspection, this is solved at x = 4.  the closest point is then (4, 2)