How do you solve this calculus problem? Please show all your steps?

Question

The path of a particle follows the expression x^2 - y^2=1. Find the position of the point on this path that is closest to the point A(0,2).

Answer 1

So you have to minimize the distance between some point on x^2 - y^2=1 and the point (0, 2). We can use the distance formula for that.
d = sqrt[(x-0)^2+ (y-2)^2] = sqrt[x^2+ (y-2)^2]

Now solve the original function for x^2.
x^2 - y^2=1
x^2 = y^2 + 1

Now substitute
d = sqrt[x^2+ (y-2)^2] = sqrt[y^2 + 1+ (y-2)^2]
= sqrt(y^2 + 1+ y^2 + 4 - 4y) = sqrt(2y^2 - 4y+ 5)

To minimize the distance, d, we have to take the derivative of this equation and set it equal to zero. (You could also graph it and find the x-coordinate of the graph's minimum for the value of y)
d = sqrt(2y^2 - 4y+ 5) = (2y^2 - 4y+ 5)^.5
d' = .5[(2y^2 - 4y+ 5)^-.5](4y-4) = 0
4y-4=0
y=1

Back to the original expression again.
x^2 - y^2=1
x^2 - 1^2=1
x^2 - 1=1
x^2 =2
abs(x) = sqrt(2)
x = +/- sqrt(2)
So there are two points, actually: (sqrt2, 1) and (-sqrt2, 1)

Answer 2

You can put is in this way
Minimize z=x^2 +(y-2)^2 with the condition x^2-y^2-1=0
using lagrangian multipliers
F=x^2+(y-2)^2+k(x^2-y^2-1)
Fx=2x+2kx =0
Fy= 2(y-2)-2ky=0
2x(1+k)=0 which gives x= 0 or k=-1
for x= 0 0^2-y^2=1 which cant be
so k=-1
and2y-4+2y=0 and 4y =4 so y=1 and x^2-2=0 so x=+-sqrt2