Find the point on the hyperbola xy8 that is close Find the p

Solution

let us assume the closest to point given is (3,0) because we need a point so that we can say that the point in hyperbola is close to (3,0)
this point will be given to you always:

The distance between two points in the xy plane is:
D = [(x2-x1)^2 + (y2-y1)^2]^0.5
Let (x1,y1) be the point (3,0)
D = [(x-3)^2 + y^2]^0.5
y = 8/x, so D = [(x-3)^2 + 64x^-2]^0.5
dD/dx = 0.5 [(x-3)^2 + 64x^-2]^-0.5 * [2(x-3) - 128x^-3]

D is at the minimum when dD/dx = 0, so solve for:

[2(x-3) - 128x^-3] = 0
x = 4, so y = 2.
The hyperbola is closest to (3, 0) at (4, 2) where the distance is the square root of 5