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 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}

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

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$endgroup$

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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

Faraad ArmwoodFaraad Armwood

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