# parameterization of the intersection's curve of two surfaces

I'm stuck on this problem and i'm not really sure how to proceed to resolve it. I have a first curve, parameterized by:

$$z = \sqrt{x+y}$$

And a second one, parameterized by:

$$z= 1+y$$

I need to find the curve for the intersection of those two surfaces, but i'm not sure even where to begin and what I should be doing...

• You mean "I have a first surface", right? Mar 10, 2018 at 21:31

from the second equation,

$$y=z-1$$ and from the first

$$x=z^2-y=z^2-z+1$$

• ...so that $t \mapsto (t^2 - t - 1, t-1,t)$ is a parameterization of the intersection curve, valid only for $z(t) \ge 0$, hence $t \ge 0$. Mar 10, 2018 at 21:37
• @JohnHughes Yes {}{}{}{}. Mar 10, 2018 at 21:39

$$z = \sqrt{x+y}$$

and $$z= 1+y$$

Thus $$\sqrt{x+y}=1+y$$

$$x+y=y^2+2y+1$$

$$x=y^2+y+1$$

The parametrization could be $$x=t^2+t+1, y=t, z=1+t$$ ,

we get the following equation $$1+y=\sqrt{x+y}$$ by squaring we get $$y^2+y+x-1=0$$ can you solve this equation? $$y_{1,2}=-\frac{1}{2}\pm\sqrt{\frac{1}{4}-x+1}$$