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Parametric equations for Tangent line of an ellipse
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Oct8-10, 03:31 AM
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
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?
Oct8-10, 11:16 AM
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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