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Suppose a 4th grade curve meets a line in one point with multiplicity 4.

Example: the lemniscate $(x^2 + y^2)^2 = y^2 - x^2$ meets the line $x=y$ when the condition $x^4=0$ holds. This shows that line $x=y$ meets the lemniscate in one point $(0,0)$ with multiplicity 4. The line $x=y$ therefore is tangent to the curve in $(0,0)$ and also therefore $(0,0)$ is a node.

In an (old) article I read that in such a case obviously the tangents must be flexes, i.e. the nodes are also inflexion points.

Can anybody provide this "obvious" argument?

Gerard
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    Isn't a flex just any point where the tangent has a multiplicity of intersection $\geq3$? – AdLibitum Mar 28 '15 at 21:50
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    The curve $x^3=y^2$ and line $y=0$ has a multiplicity of intersection 3. But the origin $(0,0)$ is a cusp. – Gerard Mar 28 '15 at 22:14

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