# determining envelope

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 (http://planetmath.org/Congruence) 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.

Title determining envelope DeterminingEnvelope 2013-03-22 17:10:48 2013-03-22 17:10:48 pahio (2872) pahio (2872) 10 pahio (2872) Topic msc 53A04 msc 51N20 msc 26B05 msc 26A24 SingularSolution