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.

  • $\begingroup$ the axis is parallel to $(-1,-1,1)$ which is the null direction of the Hessian matrix. The axes of the elliptical cross section are in directions $(-1,1,0)$ and $(1,1,2)$ $\endgroup$
    – Will Jagy
    Nov 24, 2020 at 16:50
  • 1
    $\begingroup$ Geogebra says the eccentricity of $\langle 5x^2 + 5y^2 + 8z^2 - 2xy + 8yz + 8zx + 12x - 12y + 6,x+y-z\rangle$ is $\frac1{\sqrt{2}}.$ $\endgroup$ Nov 24, 2020 at 19:30
  • $\begingroup$ My apologies. I've corrected the eccentricity. $\endgroup$ Nov 24, 2020 at 21:51

2 Answers 2


Let $$f(x,y,z)=5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6$$ The axis of the cylinder can be obtained via $f’_x=f’_y=f’_z=0$, which leads to its equation $$(x,y,z)=(0,2,-1)+t(1,1,-1)$$ The elliptical axes of the cross section containing the center $(0,2,-1)$ can be obtained via $$f’_x:f’_y:f’_z=(x-0):(y-2):(z+1)$$ which leads to the respective equations for the major and minor axes $$(0,2,-1)+t(1,-1,0),\>\>\>\>\>\>\>(0,2,-1)+t(1,1,2)$$ Then, substitute them into $f(x,y,z)= 0$ to get the major vertexes $(\pm\frac1{\sqrt2},2\mp\frac1{\sqrt2},-1)$ and the minor vertexes $(\pm\frac1{2\sqrt3},2\pm\frac1{2\sqrt3},-1\pm\frac1{\sqrt3})$, and in turn the lengths of the respective major and minor axes $a=2$ and $b={\sqrt2}$. Thus, the eccentricity is $\frac1{\sqrt2}$.


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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