- #1

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## Homework Statement

Consider the curve of intersection of the surface x

^{2}+ y

^{2}+ 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-x

^{2}-y

^{2}-x = x-2

3-y

^{2}= x

^{2}+2x

-y

^{2}= x

^{2}+2x-3

let x = t to parameterize

so r(t) = [t , sqrt(-t

^{2}-2t+3) , t-2]

r'(t) = (1 , 0.5(-t

^{2}-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?