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 $F_x^{\prime }$, $F_y^{\prime }$ and $F_c^{\prime }$ in a certain domain of ${ℝ}^3$.  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

$\{\begin{array}{cc}F(x,y,c)=0,\hfill & \\ F_c^{\prime }(x,y,c)=0.\hfill & \end{array}$(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

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

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

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

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

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

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

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

which allows immediately eliminate $a$, giving

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

$x=x(c),y=y(c),$(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

$$F_x^{\prime }(x,y,c)=0,F_y^{\prime }(x,y,c)=0.$$

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.