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

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

The Pointer
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3 Answers3

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

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

Ng Chung Tak
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  • Why did you get $(3k)^2 + 2(\frac{-k}{2})^2 + 3(k^2) = 1$? – The Pointer Sep 27 '16 at 07:07
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    Substitute $(X,Y,Z)=(3k,-\frac{k}{2},k)$ into the ellipsoid $x^2+2y^2+3z^2=1$ – Ng Chung Tak Sep 27 '16 at 07:11
  • And what's the purpose of that specific step? Is it to find the points on the ellipsoid that contains points corresponding to multiples of the normal vector of the plane? – The Pointer Sep 27 '16 at 07:35
  • To reiterate my post above, I do not understand why we have to perform that step. Please explain your reasoning. Thanks. – The Pointer Sep 27 '16 at 07:45
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    By comparing the coefficients of the given plane and tangent plane, you find the ratios among $X,Y,Z$. We still need to find the proportional constant $k$ for your point of contact. The point of contact is the intersection of the ellipsoid and the tangent plane. By substituting, $(X,Y,Z)$ into the ellipsoid, you can solve $k$ hence the value of $(X,Y,Z)$. – Ng Chung Tak Sep 27 '16 at 08:09
  • This doesn't make any sense to me. Why do we need a point of contact? If they're parallel they either never touch or touch at every single point (infinitely many times). – The Pointer Sep 27 '16 at 09:25
  • *That's on your question:* Find the points on the ellipsoid $x^2+2y^2+3z^2=1$ where the tangent plane is parallel to the plane $3x−y+3z=1$. – Ng Chung Tak Sep 27 '16 at 09:39
  • Why does the tangent plane equal to Xx+2Yy+3Zz=1? – StillFantasy May 02 '20 at 07:55
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    @StillFantasy, that's pole and polar relation. For a quadric $ax^2+by^2+cz^2=1$, the polar plane (or line for conic) w.r.t. a pole $(X,Y,Z)$ is $$aXx+bYy+cZz=1$$ See more on Joachimsthal's notations here. – Ng Chung Tak May 03 '20 at 09:19
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Let the equation of the ellipsoid be $f(x, y, z) = x^2 + 2y^2 + 3z^2 = 1$. The gradient is $\nabla f(a, b, c) = \langle2a, 4b, 6c\rangle$

We are looking for the points on the ellipsoid where the tangent plane is parallel to a provided plane, $ 3x - y + 3z = 1 $. In order for any plane to be parallel to another, the planes' normal vectors must be multiples of each other. The normal vector of the provided plane is $ \vec{n_1} = \langle3, -1, 3\rangle$. Gradients are normal to level surfaces, so the gradient $\nabla f(a, b, c) $ will be normal to the tangent planes to the ellipsoid. If $k$ is a scalar value, we thus need to find where $k\vec{n_1}=\nabla f(a, b, c)$

$k\langle3, -1, 3\rangle=\langle2a, 4b, 6c\rangle$

$a=\frac{3k}{2}\\ b=\frac{-k}{4}\\ c=\frac{k}{2}$

All points $\left(\frac{3k}{2}, \frac{-k}{4}, \frac{k}{2}\right)$ that lie on the ellipsoid will satisfy the question's requirements (parallel tangent planes). Plugging this into the ellipsoid equation to solve for valid k values, we get

$$\left(\frac{3}{2}k\right)^2 + 2\left(\frac{-1}{4}k\right)^2 + 3\left(\frac{1}{2}k\right)^2 = 1$$ Simplifying a bit: $$\frac{9}{4}k^2 + \frac{1}{8}k^2 + \frac{3}{4}k^2 = 1$$ $$\frac{25}{8}k^2 = 1$$ $$k = \pm\frac{2\sqrt2}{5}$$ Now, plug in $k$ to get the points on the ellipsoid where the tangent plane is parallel to the given plane: $$\langle\pm\frac{3\sqrt2}{5}, \mp\frac{\sqrt2}{10}, \pm\frac{\sqrt2}{5}\rangle$$

twiddler
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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 $\nabla f(p)$ 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 $\nabla f(p) = \lambda \langle 3,-1,3\rangle$.

Faraad Armwood
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  • Isn't it a level surface - not a level curve? Also, could you elaborate on your reasoning? I don't quite understand your proposed solution. – The Pointer Sep 27 '16 at 01:30
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    Eh, that was just a mistake by me. Let $p$ lie on the level curve given by $f(\textbf{x}) = k$. We can find a parametrization $c(t)$ who's image is a small section of this level curve with $c(0) = p$. Then $\frac{d}{dt} (k) = 0 = f(c(t))'|_{t=0} = \nabla f(p) \cdot c'(0)$ i.e the gradient vector at $p$ is perpendicular to the tangent vector at $p$ and so it is normal to the curve. Do you remember this result? – Faraad Armwood Sep 27 '16 at 01:37
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    The reasoning behind making that function $f$ was so that $f(c(t)) = 0$ where $c$ is described in my last comment. – Faraad Armwood Sep 27 '16 at 01:40