Tangent plane to ellipsoid

Oct 2008
9
2
How can I find a tangent plane to an ellipsoid when the plane has to contain a given line?

I found the general equation of a tangent plane to the ellipsoid and the equation of the bundle of planes passing through the line. I can't seem to get anywhere from there. I would like some ideas on how to solve the problem. Please Help.

The ellipsoid given in the problem is:
x^2/16 + y^/12 + z^2/4 = 1

And the line:

(x-2)/2 = (y-3)/-1 = (z-2)/0

Thanks
 
Oct 2008
9
2
Well you can define a line in space that way.

Then you can convert it to a parametric form so z-2/0 = t

z = 2

The line will then be contained in the z = 2 plane.
 

dwsmith

MHF Hall of Honor
Mar 2010
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Well you can define a line in space that way.

Then you can convert it to a parametric form so z-2/0 = t

z = 2

The line will then be contained in the z = 2 plane.
If \(\displaystyle z=2\), then \(\displaystyle \frac{z-2}{0}\to\frac{z}{0}-\frac{2}{0}\to\frac{0}{0}-\frac{2}{0}=t\).
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Equation of a line in space

The vector (2,-1,0) represents the direction of the line in space.
If you have a point and vector, you can form the plane.

\(\displaystyle <2,-1,0>\ (2,3,2)\)

\(\displaystyle a(x-x_1)+b(y-y_1)+c(z-z_1)=0\) will be the equation of the plane.

\(\displaystyle 2(x-2)-(y-3)=0\to 2x-y=1\)

parametric:

\(\displaystyle x=2+2t\)
\(\displaystyle y=-1+3t\)
\(\displaystyle z=2t\)

Solve for t

\(\displaystyle \frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{2}\)