Given:f(x,y) = x^2 + 4y^2...Find an equation for the tangent line to the curve which its intersection of the surface z=(x,y) and the plane y=1 through the point (-1,1,5)

Question

anonymous · Answer

can you do in non parametric?

dumbcow · Answer

how do you write a line in 3-d thats not parametric

dumbcow · Answer

also i didn't quite understand this part "which its intersection of the surface z=(x,y) and the plane y=1"

anonymous · Answer

which is the intersetion*

dumbcow · Answer

hmm i guess i did it wrong
so when f(x,y) intersects the plane y=1 the result is a parabola in the xz plane
z = x^2 +4
dz/dx = 2x
evaluate tangent line at point (-1,1,5)
slope = -2
z-5 = -2(x+1)
z = -2x +3

anonymous · Answer

hmmm i look all over the book and internet yet you and another guy gave me the same answer.  Are you sure?

anonymous · Answer

I cant find how you did that anywhere or proof anywhere

dumbcow · Answer

which answer? the one from my 1st post

dumbcow · Answer

here is the curve formed from intersection http://www.wolframalpha.com/input/?i=z%3D+x^2+%2B4y^2%2C+y%3D1 then i found the tangent line going through point (-1,5) in xz plane