<s>[INST] Cross section of cylinder $5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$ Prove that
$$5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$$
represents a cylinder whose cross-section is an ellipse of eccentricity $\frac{1}{\sqrt 2}$.
I know how to find the plane of a cross section if I know the direction ratios of at least one generator of the cylinder, but I don't know how to find that out from just the equation of the cylinder alone.
 [/INST] The equation can be written $5(x+1)^2+5(y-1)^2+8z^2-2(x+1)(y-1)+8(y-1)z+8(x+1)z-6=0,$ and we can translate.
From @WillJagy's comment look at the plane $x+y-z=0.$ We're interested in the intersection $\langle 5x^2+5y^2+8z^2-2xy+8yz+8xz-6,x+y-z\rangle.$
Now rotate so that the normal vector $[\frac1{\sqrt{3}},\frac1{\sqrt{3}},-\frac1{\sqrt{3}}]$ goes to $[0,0,1],$ by way of the rotation matrix $\begin{pmatrix}\frac{1
 }{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\ -
 \frac{1}{\sqrt{2}\sqrt{3}}&-\frac{1}{\sqrt{2}
 \sqrt{3}}&-\frac{\sqrt{2}}{\sqrt{3}}\\ \frac{1
 }{\sqrt{3}}&\frac{1}{\sqrt{3}}&-
 \frac{1}{\sqrt{3}}\end{pmatrix}.$
This produces an equation in the new $XY$-plane: $6(X^2+2Y^2-1)=0,$ which has eccentricity $\frac1{\sqrt{2}}.$
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