How do i find the point on a graph of y=(x^2-1)^(1/2) that is closest to the point (1,0)?

Answer 1

d^2 = (x-1)^2 + y^2
minimize d^2

d^2 = (x-1)^2 + (x^2-1)

d^2 = x^2 -2x +1 + x^2 -1

d^2 = 2 x^2 -2 x = 2 x (x-1)

note
d = 0 when x = 0 and when x = 1
you can not get closer than that
the solution when x = 0 is imaginary (y = sqrt(-1) )
so minimum when x = 1
in other words the curve you gave goes through the point (1,0)