Find the shortest distance?!?!?

Question

between this curve and the origin:

z^2=xy-1

Answer 1

(If the coords can be complex-valued), then z=i, x=y=0 actually gives a shorter distance than z=0, x=y=±1
This minimizes d at d=1 (instead of √2 for husoski)
Obviously z is complex-valued.
If z must be real then it's going to be x=y=±1, z=0
You didn't say which.

[Both the following methods prove there is no global minimum, thus the minimum d occurs at the endpoint.

Justification: it looks like symmetry requires the solution to have x=y. Then d² = x² + y² + z² = x² + y² + xy-1
becomes d² = 3x²-1
d = √(3x²-1)
∂d/∂x = 6x / 2√(3x²-1) = 3x / √(3x²-1)
∂d/∂x=0 => x=0
z² = xy-1 then means x=y=0

* Method of Lagrange multipliers, which locates the the global minimum as z=i, x=y=0, thus again there is no real-valued global minimum.

to prove this is

Note also we can eliminate variable z entirely as follows

Call the distance d(x,y,z)
We must have that d² = x² + y² + z²
and with the constraint z² = xy-1

=> d² = x² + y² + (xy-1)
d² = x² + y² + 2xy -xy -1
d² = (x+y)² -xy -1

d = √(x² + y² + (xy-1))

∇d = (∂d/∂x, ∂d/∂y) = (2x+y, 2y+x) / 2√(x² + y² + xy-1)

Use Lagrange multipliers:

d = √(x² + y² + xy-1) - λ(xy-1 -z²)

∇d = (∂d/∂x, ∂d/∂y) = (2x+y, 2y+x) / 2√(x² + y² + xy-1) - (λy, λx)
∂d/∂λ always just returns the constraint, so we ignore it.

Solve ∇d = 0 => solve ∂d/∂x =0 and note the local minimum (if it exists) must be symmetric in x and y:
(2x+y, 2y+x) / 2√(x² + y² + xy-1) = (λy, λx)

=> (2x+y) / 2√(x² + y² + xy-1) = λy
(2x+y) = 2λy √(x² + y² + xy-1)
(2x+y) = 2λy √(x² + y² + xy-1)

The only symmetric solution (∂d/∂x, ∂d/∂y) = (0,0)
must have x=y=0, (λ= don't-care)
This then gives z=1 (from the constraint). QED.

Answer 2

smci solution is wrong as the function does not exist at x=0 or y=0.
in fact, +-1 defines the lower limit for both variables.
i tried it also, but i realized it wrong as x,y cannot be 0.
z^2 implies non-negativity. xy>=1.

This is a system in 3 variables, which is most possibly a surface in 3 dimensional space.

In 3-dimensional space, the distance from the origin of any point x,y,z is defined by the formula
d = (x^2 + y^2 + z^2)^(1/2)

The distance from the origin of any point in the given function z^2 = xy - 1 can be computed thus.

Use y = (z^2 + 1)/x

d = (x^2 + (z^2 + 1)^2/x^2 + z^2)^(1/2)

To find the shortest distance, we now find if the distance formula has a minimum and this will give our
solution.

In a function of 2-variables, the critical pt (maximum, minimum, saddle pt) is found by finding
the pt where the first partial derivatives relative to each variable is equal to 0,
df/dx = 0, df/dz = 0. Also 2nd derivative are found to determine if it is a minimum, maximum, or a saddle point. delta = d2d/dx2 * d2d/dz2 - (d2d/dxdz)^2
*** df/dx here means partial derivative. i have yet to find/learn how to input these symbols***

To simplify the solution, we just instead use the square of the distance D = d^2.

D = (x^2 + (z^2 + 1)^2/x^2 + z^2)
dD/dx = 2x - 2(z^2 + 1)^2/x^3
dD/dz = 4z(z^2 + 1)/x^2 + 2z

solve dD/dx = 0:
2x -2(z^2 + 1)^2/x^3 = 0
x^2 = (z^2 + 1)

solve dD/dz = 0:
4z(z^2 + 1)/x^2 + 2z = 0
2z[2(z^2 + 1)/x^2 + 1] = 0
z=0 is a solution. 2(z^2 + 1)/x^2 + 1 =0 has no solution, the qty at left is always positive.
z=0 is a possible solution.

from dD/dx = 0, x^2 = (z^2 + 1), plugging in z=0, we have x=+-1

a possible solution is (z=0,x=1,y=1) (since it symmetric, x=-1 will yield the same solution)

To test if solution is a maximum or minimum, solve delta = d2D/dx2*d2D/dz2 - [d2D/dxdz]^2

d2D/dx2 = 2 + 6(z^2 + 1)^2/x^4
d2D/dz2 = [8z^2 + 4(z^2 + 1)]/x^2 + 2
d2D/dxdz = -8z(z^2 + 1)/x^3

given z=0, x=1
delta = 8*6 - 0 = 48
since delta > 0, D (as well as d) is minimum at z=0, x=1, y=1.
and the shortest distance d = (2)^(1/2)
The pt z=0, x=-1, y=-1 also gives the same value.

Answer 3

One equation in three unknowns...looks more like a surface than a curve. One curve I do see is the the curve defined by:
xy - 1 = 0 ...or... y = 1/x
which is the boundary line between those (x,y) pairs that have a defined value for z and those who don't.

It's clear to see that the pairs (1,1) and (-1,-1) are on this curve and tied for being closest to the origin. This leaves the points (1,1,0) and (-1,-1,0) are tied for being closest points on the surface to the origin.