Let $C$ be the curve of intersection of the ellipsoid $x^2+2y^2+3z^2=39$ and the plane $3x+y-7z=0$. Find the parametric equations for the tangent line to $C$ at $(5,-1,2)$.

I don't know how to find the curve $C$.. I know the normal vector of the plane is $(3,1,-7$), but what can I do with the ellipsoid?


The tangent plane to the ellipsoid $r(y,z)=(\sqrt{39-2y^2-3z^2},y,z)$ has the basis $$u=r_y=(\frac{-2y}{\sqrt{39-2y^2-3z^2}},1,0)=(2/5,1,0) \\v=r_z=(\frac{-3z}{\sqrt{39-2y^2-3z^2}},0,1)=(-6/5,0,1)$$ Any tangent vector to the ellipsoid is a linear combination of $u,v$. We need one of these vectors perpendicular to $(3,1,-7)$, so we solve for $w=au+bv$ where $$w\cdot (3,1,-7)=0 \\ ((2a-6b)/5,a,b)\cdot(3,1,-7)=0 \\(6a-18b)/5+a-7b=0 \\11a-53b=0$$ Now we can take any value $b=1\implies a=53/11$ and so $w=(53/11)u+v$(evaluate it). Finally, the equation of the tangent line is $$\alpha(t)=(5,-1,2)+tw$$

  • $\begingroup$ could you please tell me how you get the u and v? $\endgroup$ – Gavin Z. Feb 18 '14 at 23:11
  • $\begingroup$ @GavinZ. the tangent vectors to a surface $r(u,v)$ are $\frac{\partial r}{\partial u},\frac{\partial r}{\partial v}$ $\endgroup$ – Semsem Feb 18 '14 at 23:23
  • $\begingroup$ @GavinZ. then i used the given point $\endgroup$ – Semsem Feb 18 '14 at 23:25

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