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Find a plane $\pi$ which involves x-axis and its intersection line with $$\frac{x^2}{4}+y^2-z^2=1$$ is a circle.


Because the plane want to be find involves x-axis,so set as $By+Cz=0$,then I must to determine its value such that $$\begin{cases}By+Cz=0\\\frac{x^2}{4}+y^2-z^2=1\end{cases}$$ is a circle in 3-dimensional space,apparently,$B\not=0$,so I replace $y$ with $-\frac{C}{B}z$ in $\frac{x^2}{4}+y^2-z^2=1$,then i get a elliptic cylinder,at last ,i neet to determine $B$ and $C$ such that the intersection line between plane and elliptic cylinder is a circle.$$\begin{cases}By+Cz=0\\\frac{x^2}{4}+\left[(\frac{C}{B})^2-1\right]z^2=1\end{cases}$$ but i can't go any further.

thanks very much

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Assume $\pi$ has an equation like $$by+cz=0$$ for some unknown $c$ and $b$. If $b=0$ then $\pi: z=0$ and then we have an intersection like $x^2/4+y^2=1$ which is an ellipse not an circle. The same story would be for $c=0$. Let $c\neq0,b\neq0$ and so $z=-by/c$ so the intersection would be $$(1/4)x^2+y^2-b^2y^2/c^2 = 1$$ or $$(1/4)x^2+\left(1-b^2/c^2\right)y^2 = 1$$ if we want to have a circle as an intersection we need: $$b^2/c^2=3/4$$ For example set $b=\sqrt{3},c=2$ and look at the following picture:

enter image description here

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$\frac{b^2}{c^2}=\frac{\sqrt{3}}{4} \Rightarrow b=\sqrt{3}, c=2$ ? maybe a typo. –  Laura Jan 3 '13 at 12:20
    
@Tai: Ohhh sorry you're right friend. –  B. S. Jan 3 '13 at 12:25
    
your cylinder: $\frac{x^2}{4}+\frac{y^2}{4}=1$ is orthogonal to XOY plane, and intersection lines is a circle. but your plane $\pi$'s normal vector is $(0,\sqrt{3},2)$,it is not parallel to XOY plane. so the intersection line between $\pi$ and the cylinder maybe a ellipse,not a circle. and your figure looks like a ellipse :) –  Laura Jan 3 '13 at 13:31
    
@Tai: Yes. If your plane $\pi$ is parallel to $xy-$ plane and if you want $\pi$ contains the $x-$ axe so it would be $z=0$ and this is what you noted. It is ellipse not an circle. But we assume the equation as $by+cz=0$ and in this equation b and c is not zero at all. We just want an equation of a plane which contains $x-$ axe and intersect the surface. So if this plane is orthogonal or... let it be. –  B. S. Jan 3 '13 at 13:37
    
I don't want plane $\pi$ parallel to xy-plane. Because your cylinder is orthogonal to XOY plane, if $\pi$ don't parallel to XOY, the intersection curve will not be a circle, it maybe a ellipse. –  Laura Jan 3 '13 at 13:48
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