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Homedetermining envelope

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# determining envelope

Theorem. Let $c$ be the parameter of the family $F(x,\,y,\,c)=0$ of curves and suppose that the function $F$ has the partial derivatives $F^{{\prime}}_{x}$, $F^{{\prime}}_{y}$ and $F^{{\prime}}_{c}$ in a certain domain of $\mathbb{R}^{3}$. If the family has an envelope $E$ in this domain, then the coordinates $x,\,y$ of an arbitrary point of $E$ and the value $c$ of the parameter determining the family member touching $E$ in $(x,\,y)$ satisfy the pair of equations

$\displaystyle\begin{cases}F(x,\,y,\,c)=0,\\ F^{{\prime}}_{c}(x,\,y,\,c)=0.\end{cases}$ | (1) |

I.e., one may in principle eliminate $c$ from such a pair of equations and obtain the equation of an envelope.

Example 1. Let us determine the envelope of the the family

$\displaystyle y=Cx+\frac{Ca}{\sqrt{1+C^{2}}}$ | (2) |

of lines, with $C$ the parameter ($a$ is a positive constant). Now the pair (1) for the envelope may be written

$\displaystyle F(x,\,y,\,C)\,:=\,Cx-y+\frac{Ca}{\sqrt{1+C^{2}}}=0,\quad F^{{% \prime}}_{C}(x,\,y,\,C)\equiv x+\frac{a}{(1+C^{2})\sqrt{1+C^{2}}}=0.$ | (3) |

It’s easier to first eliminate $x$ by taking its expression from the second equation and putting it to the first equation. It follows the expression $y=\frac{C^{3}a}{(1+C^{2})\sqrt{1+C^{2}}}$, and so we have the parametric presentation

$x=-\frac{a}{(1+C^{2})\sqrt{1+C^{2}}},\quad y=\frac{C^{3}a}{(1+C^{2})\sqrt{1+C^% {2}}}$ |

of the envelope. The parameter $C$ can be eliminated from these equations by squaring both equations, then taking cube roots and adding both equations. The result is symmetric equation

$\sqrt[3]{x^{2}}+\sqrt[3]{y^{2}}=\sqrt[3]{a^{2}},$ |

which represents an astroid. But the parametric form tells, that the envelope consists only of the left half of the astroid.

Example 2. What is the envelope of the family

$\displaystyle y-\frac{1}{2}a^{2}=-\frac{1}{4}(x-a)^{2},$ | (4) |

of parabolas, with $a$ the parameter?

With a fixed $a$, the equation presents a parabola which is congruent to the parabola $y=-\frac{1}{4}x^{2}$ and the apex of which is $(a,\,\frac{1}{2}a^{2})$. When $a$ is changed, the parabola is submitted to a translation such that the apex draws the parabola $y=\frac{1}{2}x^{2}.$

The pair (1) for the envelope of the parabolas (4) is simply

$y-\frac{1}{2}a^{2}+\frac{1}{4}(x-a)^{2}=0,\quad x=-a,$ |

which allows immediately eliminate $a$, giving

$\displaystyle y=-\frac{1}{2}x^{2}.$ | (5) |

Thus the envelope of the parabolas is a “narrower” parabola. One infers easily, that a parabola (4) touches the envelope (5) in the point $(-a,\,-\frac{1}{2}a^{2})$ which is symmetric with the apex of (4) with respect to the origin.

The converse of the above theorem is not true. In fact, we have the

Proposition. The curve

$\displaystyle x=x(c),\quad y=y(c),$ | (6) |

given in this parametric form and satisfying the condition (1), is not necessarily the envelope of the family $F(x,\,y,\,c)=0$ of curves, but may as well be the locus of the special points of these curves, namely in the case that the functions (6) satisfy except (1) also both of the equations

$F^{{\prime}}_{x}(x,\,y,\,c)=0,\quad F^{{\prime}}_{y}(x,\,y,\,c)=0.$ |

Examples. Let’s look some simple cases illustrating the above proposition.

a) The family $(x-c)^{2}-y=0$ consists of congruent parabolas having their vertices on the $x$-axis. Differentiating the equation with respect to $c$ gives $x-c=0$, and thus the corresponding pair (1) yields the result $x=c,\;y=0$, i.e. the $x$-axis, which also is the envelope.

b) In the case of the family $(x-c)^{2}-y^{3}=0$ (or $y=\sqrt[3]{(x-c)^{2}}$) the pair (1) defines again the $x$-axis, which now isn’t the envelope but the locus of the special points (sharp vertices) of the curves.

c) The third family $(x-c)^{3}-y^{2}=0$ of the semicubical parabolas also gives from (1) the $x$-axis, which this time is simultaneously the envelope of the curves and the locus of the special points.

## Mathematics Subject Classification

53A04*no label found*51N20

*no label found*26B05

*no label found*26A24

*no label found*

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