Given the paraboloid z = 6 - x - x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}-2y^{2}and the plane x = 1, find curve of intersection and the parametric equations of the tangent line to this curve at point (1,2,-4).

So I plugged x=1 into the paraboloid equation and got z = 4-2y^{2}.

Then I take the derivative of the curve of intersection:

dz/dy = -4y

Then I plug in 2 for t since y=t=2 at the given point and get the slope -8.

The parametric equation of a line is r = r0 +tv

I have r0 = (1,2,-4) and I have a slope -8. I'm not sure how to get a vector.

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# Finding the tangent line of the curve of intersection

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