Find the shortest distance from the point (1,1,0) to the surface x^2+3y^2-z^2=1/3

Question

We are asked to find the shortest distance from 3 points but that's all

Answer 1

Let's say (a,b,c) is the closest point on the curve, The line from (a,b,c) to (1,1,0) is (a-1,b-1,c)

The normal to the surface at (a,b,c) is ∇S(at a,b,c) = (2a, 6b, -2c). these lines must be the same, so

2a/(a-1) = 6b/(b-1) = -2c/c = -2 Solve for a and b --> (1/2, 1/4, c)

You can now find c by solving c = ±sqrt(a²+3b² - 1/3) (the sign doesn't matter here)

You can then calculate d = sqrt((a-1)²+(b-1)² + c²) should be around .957