# At What Point On The Ellipsoid Is The Tangent Plane Parallel

Find the points on the ellipsoid \$x^2 + 2y^2 + 3z^2 = 1\$, where the tangent plane is parallel to the plane \$3x – y + 3z = 1\$.

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I”m not sure how to go about solving this. I”d appreciate some help.

Let the contact point be \$(X,Y,Z)\$, now the tangent plane is

\$\$Xx+2Yy+3Zz=1\$\$

Comparing coefficients, \$\$X:2Y:3Z=3:-1:3\$\$

That is \$\$frac{X}{3}=frac{2Y}{-1}=frac{3Z}{3}=k\$\$

Now substitute \$(X,Y,Z)=left( 3k,-dfrac{k}{2},k
ight)\$ into \$x^{2}+2y^{2}+3z^{2}=1\$

We haveegin{align*} (3k)^{2}+2left( -frac{k}{2}
ight)^{2}+3(k)^{2} &= 1 \ frac{25}{2}k^{2} &= 1 \ k &= pm frac{sqrt{2}}{5} \ (X,Y,Z) &= left( pm frac{3sqrt{2}}{5}, mp frac{1}{5sqrt{2}}, pm frac{sqrt{2}}{5}
ight)end{align*}

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edited Sep 27 “16 at 8:18
answered Sep 27 “16 at 6:52

Ng Chung TakNg Chung Tak
\$endgroup\$
9
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\$egingroup\$
Consider the function \$f(x,y,z) = x^2+2y^2+3z^2-1\$ then the \$E\$=ellipsoid is given by \$f^{-1}(0)\$ i.e a level surface. Let \$p in E\$ then \$
abla f(p)\$ is a normal vector for the tangent plane. For the tangent plane to be parallel to \$3x-y+3z=1\$, you need \$p in E\$ such that \$
abla f(p) = lambda langle 3,-1,3
angle\$.

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edited Sep 27 “16 at 1:32
answered Sep 27 “16 at 1:18

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