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How to find tangent cone in singular point of surface? For example, considering surface in $\mathbb{R}^3$ given by equation $x^2z=y^2$, what is it's tangent cone in the origin? enter image description here

UPD:By tangent cone in the point $p$ I mean set of equivalence classes of smooth maps from $[0,\varepsilon]$ to the surface, such that image of $0$ is $p$, there maps $f,g$ are called equivalent iff $f-g=o(1)$.(analog of tangent space for singular points)

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  • $\begingroup$ What definition of tangent cone do you mean?<a href="en.wikipedia.org/wiki/… one?</a> $\endgroup$ – cws Oct 14 '14 at 17:20
  • $\begingroup$ @cws added it to the question $\endgroup$ – user165101 Oct 14 '14 at 17:27
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Each equivalence class can be identified with the initial velocity vector $f'(0+)$. Since $f$ makes into the surface, the vector $v=f'(0+)$ has the property that $\operatorname{dist} (p+tv,S)=o(t)$ as $t\to 0$; here $S$ is the surface and $p$ is the point at which the tangent cone is considered.

Within $\epsilon$-neighborhood of the origin, the surface in your example is squeezed between the planes $y=\pm \sqrt{\epsilon} x$. This rules out any vector with nonzero $y$-component.

Also, since the surface is contained in the halfspace $z\ge 0$, any vector with negative $z$-component is out.

The above leaves the vectors $(a,0,c)$ with $c\ge 0$. I claim all these are in the tangent cone, i.e., can be realized as initial velocity vectors of curves beginning at $(0,0)$. This is not hard to check: $f(t)=(at,a\sqrt{c}t^{3/2},ct)$ does the job. (I put linear functions for $x,z$ and then found what $y$ needs to be for this to work.)

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