# Multiple Tangents at a point: Intersection curve.

The article Wikipedia: multiple tangents at a point suggests that in order to find all the tangents at an implicit curve:

$(x^2+(y-1)^2-1)(x-y) = 0$

Which has a branch point on (0,0). To find its tangents We must exclude the lowest polynomial degree (2):

$-2y(x-y) = 0$

The solutions to these include the direction tangents to the point (IE, the vectors (c,0) and (c,c) where c is a constant).

I would like to know if the same approach can be applied in the case of intersection curve to multiple surfaces. For example:

$(x+y^2+z)(x^2+y^2+z) = 0$

$x^3+y^2+z(y-z) = 0$

Create an intersection curve with branch point occuring at (0,0,0). Can I find the tangents to this point by simply taking the lowest polynomial degree of every equation and solving for:

$z(x+z) = 0$

$y^2+z(y-z) = 0$

with let's say x = 1?

• I believe the tangent space you are referring to is called a tangent cone. For a hypersurface, taking the lowest degree term (assuming that we are computing the tangent cone at the origin) would suffice. But the tangent cone of a complete intersection is not a complete intersection in general. In other words, one may need more equations to define tangent cones. This is a special case of Grobner basis. Sometimes the defining ideal of a tangent cone is called as a leading ideal. Oct 28, 2017 at 5:54