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Math Help - Do I need to change this into parametric form?

  1. #1
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    Do I need to change this into parametric form?

    I need to find the points on:

    x^2+\frac{y^2}{4}+\frac{z^2}{9}=1

    where the tangent line is parallel to:

    x+2y+z=0


    This is what I've done:

    \langle 2x, \frac{y}{4}, \frac{2z}{9} \rangle \cdot \langle 1,2,1 \rangle =0

    =2x+y+\frac{2z}{9}=0

    Not sure what to do now... do I have to change this to parametric form? Should I have changed it first?
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  2. #2
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    Quote Originally Posted by MathSucker View Post
    I need to find the points on:

    x^2+\frac{y^2}{4}+\frac{z^2}{9}=1

    where the tangent line is parallel to:

    x+2y+z=0


    This is what I've done:

    \langle 2x, \frac{{\color{red}2}y}{4}, \frac{2z}{9} \rangle \cdot \langle 1,2,1 \rangle =0
    If the planes are parallel then their normal vectors should also be parallel, not perpendicular. So your equation should be \langle 2x, \tfrac{y}{2}, \tfrac{2z}{9} \rangle = \lambda \langle 1,2,1 \rangle. Then \langle x,y,z\rangle = \langle\tfrac\lambda2,4\lambda,\tfrac{9\lambda}2\r  angle. Substitute those values for x, y, z into the equation x^2+\frac{y^2}{4}+\frac{z^2}{9}=1 to find the two values of \lambda for which that point lies on the ellipsoid.
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  3. #3
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    So if I get \lambda= \pm \frac {2}{\sqrt{13}},

    are the two points:

     \pm (\frac{2}{\sqrt{13}}, \frac{4}{\sqrt{13}}, \frac{2}{\sqrt{13}})
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