Consider the curve of intersection of the surface x2 + y2 + x + z = 1 and
the plane z - x + 2 = 0. Find parametric equations for the tangent line to the intersection curve at the point (1; 0;-1).
The Attempt at a Solution
i solve for intersection curve by getting rid of z
i get 1-x2-y2-x = x-2
3-y2 = x2+2x
-y2 = x2+2x-3
let x = t to parameterize
so r(t) = [t , sqrt(-t2-2t+3) , t-2]
r'(t) = (1 , 0.5(-t2-2t+3)-0.5(-2t-2) , 1)
at point (1,0,-1) my t = 1
therefore r'(t) = (1 , 0 , 1) this is a vector parallel to the line right?
therefore the line parameterization at (1,0,-1) is
x = 1 + t
y = 0
z = -1 + t
is this correct?