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Parametric equations for Tangent line of an ellipse

by derek1999
Tags: ellipse, equations, line, parametric, tangent
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derek1999
#1
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
#2
Oct8-10, 11:16 AM
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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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