Parametric equations for Tangent line of an ellipse

by derek1999
Tags: ellipse, equations, line, parametric, tangent
derek1999 is offline
Oct8-10, 03:31 AM
P: 1
1. The problem statement, all variables and given/known data

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)

2. Relevant equations
sin(t)^2 + cos(t)^2 = 1

3. 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?
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HallsofIvy is offline
Oct8-10, 11:16 AM
Sci Advisor
PF Gold
P: 38,877
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 [itex]4x^2+ z^2= 8[/itex]. From that you can calculate immediately that [itex]8x + 2z dz/dx= 0[/itex] so that [itex]dz/dx= -4x/z[/itex]. In particular, at x= 1, z= 2, [itex]dz/dx= -4(1)/2= -2[/itex]. 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.

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