What are the answers in this calculus problem?

You have the following curve in 3D-space:

x² - yz = 1

and want to know which points are closest to the origin.  We can use the distance equation for two points in 3D-space:

d = √[(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)²]

Using the known point (x, y, z) for one point and (0, 0, 0) for the other, we can simplify this to:

d = √(x² + y² + z²)

We can solve the first equation for x² in terms of y and z and substitute it into the second equation:

x² - yz = 1

x² = 1 + yz

That gives us:

d = √(1 + yz + y² + z²)

Now we have distance in terms of two unknowns.  We can solve for the zeroes of the first derivative.  We'll need to take partial derivatives with chain rules.  So next we get the derivatives in respect to both y and z, set them to zero, then see what we have left.  When taking the derivative in respect to y, treat z as a constant, and visa-versa:

Chain rule and two partial derivatives:

d = √u and u = 1 + yz + y² + z²

dd/du = 1 / (2√u) and du/dy = z + 2y and du/dz = y + 2z

dd/dy = dd/du * du/dy and dd/dz = dd/du * du/dz

dd/dy = 1 / (2√u) * (z + 2y) and dd/dz = 1 / (2√u) * (y + 2z)

dd/dy = (z + 2y) / (2√u)  and dd/dz = (y + 2z) / (2√u)

dd/dy = (z + 2y) / [2√(1 + yz + y² + z²)]  and dd/dz = (y + 2z) / [2√(1 + yz + y² + z²)]

Now we can set those to zero:

0 = (z + 2y) / [2√(1 + yz + y² + z²)]  and 0 = (y + 2z) / [2√(1 + yz + y² + z²)]

If we multiply both sides by the denominator this simplifies things and turns into a system of two equations and two unknowns that can be solved easily:

0 = z + 2y and 0 = y + 2z

-2y = z

0 = y + 2z

0 = y + 2(-2y)

0 = y - 4y

0 = -3y

0 = y

-2y = z

-2(0) = z

0 = z

Now we can solve for x:

x² = 1 + yz

x² = 1 + 0(0)

x² = 1 + 0

x² = 1

x = ±1

We do have two points after all.

The points are:

(-1, 0, 0) and (1, 0, 0)