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


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 haveegin{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}

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


Ng Chung TakNg Chung Tak
17.4k44 gold badges1616 silver badges4242 bronze badges
| Show 4 more comments
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

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


Faraad ArmwoodFaraad Armwood
8,46422 gold badges99 silver badges2121 bronze badges
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