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Homesingular points of plane curve
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singular points of plane curve
The points of a plane curve
$x\;=\;x(t),\quad y\;=\;y(t)$ 
in which both derivatives $x^{{\prime}}(t)$ and $y^{{\prime}}(t)$ vanish, are in general singular points of this curve. For studying such points we suppose that
$x^{{\prime}}(t_{0})\;=\;y^{{\prime}}(t_{0})\;=\;0$ 
and that $x(t)$ and $y(t)$ have in a neighbourhood of $t_{0}$ the derivatives of all orders. Thus we have the Taylor expansions
$\displaystyle\begin{cases}x(t)\;=\;x_{0}+a_{2}(t\!\!t_{0})^{2}+a_{3}(t\!\!t_% {0})^{3}+\ldots\\ y(t)\;=\;y_{0}+b_{2}(t\!\!t_{0})^{2}+b_{3}(t\!\!t_{0})^{3}+\ldots,\end{cases}$  (1) 
where $x_{0}=x(t_{0}),\;\,y_{0}=y(t_{0})$.
Assume now that $a_{2}$ and $b_{2}$ are not both 0. Then the slope of the chord (secant line) between the points $(x_{0},\,y_{0})$ and $(x,\,y)$ is $m=(yy_{0})\!:\!(xx_{0})$ and by (1) its limit as $t\to t_{0}$ equals $\frac{b_{2}}{a_{2}}$ (or the limit of $1/m$ is $\frac{a_{2}}{b_{2}}$). Accordingly, the curve has in the point $(x_{0},\,y_{0})$ a definite tangent line (which may be vertical if $a_{2}=0$). From the expression
$y\!\!y_{0}\;=\;b_{2}(t\!\!t_{0})^{2}+b_{3}(t\!\!t_{0})^{3}+\ldots$ 
one sees that when $t\!\!t_{0}$ is sufficiently small, the difference $y\!\!y_{0}$ of the ordinates has the same sign as $b_{2}$, i.e. the sign is the same on both sides of $t_{0}$. This means that the curve has a cusp at the point.
If the slope angle of the tangent line is $\alpha$, we have
$\sin\alpha\;=\;\frac{b_{2}}{\sqrt{a_{2}^{2}\!+\!b_{2}^{2}}},\quad\cos\alpha\;=% \;\frac{a_{2}}{\sqrt{a_{2}^{2}\!+\!b_{2}^{2}}}.$ 
We can form the projection of the chord to the normal line of the tangent, obtaining
$\displaystyle(x\!\!x_{0})\cos(\alpha\!+\!\frac{\pi}{2})+(y\!\!y_{0})\sin(% \alpha\!+\!\frac{\pi}{2})\;=\;\frac{(a_{2}b_{3}\!\!a_{3}b_{2})(t\!\!t_{0})^{% 3}\!+\!(a_{2}b_{4}\!\!a_{4}b_{2})(t\!\!t_{0})^{4}\!+\ldots}{\sqrt{a_{2}^{2}% \!+\!b_{2}^{2}}}.$  (2) 

The case $a_{2}b_{3}\!\!a_{3}b_{2}\neq 0$. When $t\!\!t_{0}$ is sufficiently small, the expression (2) of the projection changes its sign at the same time as $t\!\!t_{0}$ (due to the third power). Thus the two branches of the curve are on different sides of the tangent line. One speaks of a ordinary cusp.

The case $a_{2}b_{3}\!\!a_{3}b_{2}=0$ but $a_{2}b_{4}\!\!a_{4}b_{2}\neq 0$. The expansion (2) begins with the term with the even power $(t\!\!t_{0})^{4}$, the projection keeps its sign as $t$ passes through $t_{0}$. Therefore the both branches are on the same side of the tangent line. Now there is a ramphoid cusp (in German die Schnabelspitze) on the curve.
Example. Examine the singular points of the algebraic curve
$(y\!\!x^{2})^{2}\;=\;x^{5}.$ 
Let us take the ratio $\displaystyle\frac{y}{x^{2}}=:t$ as the parametre. This yields first $(x^{2}t\!\!x^{2})^{2}=x^{5}$; dividing by $x^{4}$ gives the parametric presentation
$\displaystyle\begin{cases}x\;=\;(t\!\!1)^{2},\\ y=(t\!\!1)^{4}t.\end{cases}$ 
The derivatives $\frac{dx}{dt}=2(t\!\!1)$ and $\frac{dy}{dt}=4(t\!\!1)^{3}t+(t\!\!1)^{4}$ have the common zero $t=1$, whence there is a cusp in the point $(0,\,0)$. Now the Taylor expansions in $t=1$ are the polynomials
$\displaystyle\begin{cases}x\;=\;(t\!\!1)^{2},\\ y\;=\;(t\!\!1)^{4}(1\!+\!(t\!\!1))\;=\;(t\!\!1)^{4}+(t\!\!1)^{5}.\end{cases}$ 
Thus $a_{2}=b_{4}=b_{5}=1,\;\,a_{3}=a_{4}=b_{2}=b_{3}=0$, and accordingly $a_{2}b_{3}\!\!a_{3}b_{2}=0$, $a_{2}b_{4}\!\!a_{4}b_{2}=1\neq 0$. It is a question of a ramphoid cusp. Both branches start from the origin to the right, their common tangent is the $x$axis. Note that the curve may be given in the form $y=x^{2}(1\pm\sqrt{x})$.
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