Find the point on the hyperbola xy=8 that is closest to the point (3, 0).?

Please show work =)

Answer 1

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), then D is:

[(x2-3)^2 + (y2)^2]^0.5, or to save clutter:

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.

Leave a Comment