How to find the shortest distance to a hyperboloid, etc?

[RJLiberator]

Homework Statement

Consider the hyperboloid of two sheets: z^2=x^2+y^2+1
and a point P(0, 1, 0). Find the shortest distance between the hyperboloid and the point P. Also, find coordinates of all points on the surface for which this distance is attained.

Homework Equations

The Attempt at a Solution

So, I first used the distance function with the point P.
(x-0)^2+(y-1)^2+(z-0)^2
and this simplified (with the use of the equation z^2=x^2+y^2+1 to:
2x^2+2y^2-2y+3=f(x,y)
I then took the partial derivatives with respects to x and y
fx = 4x
fy = 4y-2

These equal 0 when x = 0 and y = 1/2. This is the ONLY critical point.
I then found the z value with original function so (0, 1/2, sqrt(5)/2) is the point.

I then found the distance to (0,1,0) through subtraction. (0, 1/2, -sqrt(5)/2) = Shortest distance from hyperboloid to point P.

Now I'm pretty sure that this is correct, but how do I go about finding the coordinates of all points on the surface for which the distance is attained? I'm struggling to start here. Any tips on this would be great.

 

[mfb]

This is the ONLY critical point.

I think you are done.

 

[RJLiberator]

Interesting. I had this hunch as well. It didn't make sense that there would be more coordinates if that was the only critical point.
In a sense, they were simply testing conceptual understanding with that add-on then.

Let me just clarify my understanding: Since there is only one critical point on the hyperboloid that minimizes the distance in such a way that it becomes the shortest possible distance from the hyperboloid and the point P, then that is all possible points on the surface for which the distance is attained.

Excellent.

Thank you.

 

[haruspex]

These equal 0 when x = 0 and y = 1/2. This is the ONLY critical point.
I then found the z value with original function so (0, 1/2, sqrt(5)/2) is the point.

Is that the only z value?

 

[RJLiberator]

Oi, perhaps +/- sqrt(5)/2 Eh? Yes, that exists because z^2. But then, because that exists that is a second point similar distance to point P.

 

[haruspex]

Oi, perhaps +/- sqrt(5)/2 Eh? Yes, that exists because z^2. But then, because that exists that is a second point similar distance to point P.

Yes. (Picturing it helps. You see the symmetry about the XY plane.)

 

[RJLiberator]

Yes, I was curious about this as I made a poster for hyperboloids with two sheets and noticed that there should be an opposite point that is symmetrical. Thank you.

 

[mfb]

Oops, good point, I missed the other sheet as well.