Parametric equations for Tangent line of an ellipse

[derek1999]

Homework Statement

The ellipsoid 4x^2+2y^2+z^2=16 intersects the plane y=2 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point (1,2,2)

Homework Equations

sin(t)^2 + cos(t)^2 = 1

The Attempt at a Solution

After plugging 2 in for y, I get 4x^2 + z^2 = 8
thus x^2/2 + z^2/8 = 1
if sin(t)^2 + cos(t)^2=1
then x^2/2=sin(t)^2
and x = root(2)sint(t)
z would = 2root(2)cos(t)
and thus the curve of
x=root(2)sint(t), y=2, z=2root(2)cos(t) would be the intersect of the plane and the ellipsoid.

to get a tangent of the line, it would be r'(t)/|r'(t)

at the point x = 1 = root(2)sint(t) -> t = 45degrees meaning that all cos(t) and sin(t) would be root(2)/2

however for r'(t), I get root(2)cos(t)i + 0j + -2root(2)cos(t) and plugging in root(2)/2, will yield a -1.
Finding the magnitude of the vector always gets me a root(5)

However the answer should be x = 1+t, y=2, z=2-2t

am I making this too complicated or do I not get how to parametrize the intersection of a surface and a plane?

 

[HallsofIvy]

You are being much too complicated!

First the problem asked for parametric equations for the tangent line. It is not necessary to get parametric equations for the ellipse,.

Yes, with y= 2, your equation reduces to $4x^2+ z^2= 8$. From that you can calculate immediately that $8x + 2z dz/dx= 0$ so that $dz/dx= -4x/z$. In particular, at x= 1, z= 2, $dz/dx= -4(1)/2= -2$. The tangent line through that point, in the y= 2 plane, is z= -2x+ 4.

Now, to get parametric equations of the line in three dimensions, take x= t as parameter:
x= t, y= 2, z= -2t+ 4.