# Envelope of a Family of Curves

Consider a one-parameter family of plane curves defined by the equation

$f\left( {x,y,C} \right) = 0,$

where C is a parameter.

The envelope of this family of curves is a curve such that at each point it touches tangentially one of the curves of the family (Figure 1).

The parametric equations of the envelope are defined by the system of equations

$\left\{ \begin{array}{l} f\left( {x,y,C} \right) = 0\\ {f'_C}\left( {x,y,C} \right) = 0 \end{array} \right.,$

that is, by the original equation of the family of curves and the equation obtained by differentiating the original equation with respect to the parameter $$C.$$ Eliminating the parameter $$C$$ from these equations, we can get the equation of the envelope in explicit or implicit form.

The above system of equations is a necessary condition for the existence of an envelope. Besides the envelope curve, the solution of this system may comprise, for example, singular points of the curves of the family that do not belong to the envelope. The set of all solutions of the system is called the discriminant curve. Thus, in general, the envelope is a part of the discriminant curve.

To find the equation of the envelope uniquely, the sufficient conditions are used. They assume that the following inequalities are satisfied (in addition to the above system of equations):

$\left| {\begin{array}{*{20}{c}} {\frac{{\partial f}}{{\partial x}}} & {\frac{{\partial f}}{{\partial y}}}\\ {\frac{{\partial {f'_C}}}{{\partial x}}} & {\frac{{\partial {f'_C}}}{{\partial y}}} \end{array}} \right| \ne 0,\;\;\;\frac{{{\partial ^2}f}}{{\partial {C^2}}} \ne 0.$

Note that not any one-parameter family of curves has an envelope. A classic counter-example is the family of concentric circles (Figure $$2$$), which is described by the equation

${x^2} + {y^2} = {C^2}.$

There is no envelope for the given set of curves.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the envelope of the family of circles given by the equation $\left( {x - C} \right)^2 + \left( {y - C} \right)^2 = 1.$

### Example 2

Find the envelope of the family of ellipses defined by the equation $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ provided that $${a^2} + {b^2} = 1.$$

### Example 1.

Find the envelope of the family of circles given by the equation $\left( {x - C} \right)^2 + \left( {y - C} \right)^2 = 1.$

Solution.

We write the system of equations:

$\left\{ \begin{array}{l} f\left( {x,y,C} \right) = 0\\ {f'_C}\left( {x,y,C} \right) = 0 \end{array} \right..$

The first equation describes the family of curves and is given in the problem definition. Differentiating it with respect to the parameter $$C,$$ we get

$2\left( {x - C} \right) \cdot \left( { - 1} \right) + 2\left( {y - C} \right) \cdot \left( { - 1} \right) = 0,\;\;\Rightarrow x - C + y - C = 0,\;\;\Rightarrow x + y - 2C = 0.$

Thus, the system of equations can be written as

$\left\{ \begin{array}{l} {\left( {x - C} \right)^2} + {\left( {y - C} \right)^2} = 1\\ x + y - 2C = 0 \end{array} \right..$

We express $$C$$ from the second equation and substitute it in the first equation:

$C = \frac{{x + y}}{2},\;\;\Rightarrow {\left( {x - \frac{{x + y}}{2}} \right)^2} + {\left( {y - \frac{{x + y}}{2}} \right)^2} = 1, \;\;\Rightarrow {\left( {x - \frac{x}{2} - \frac{y}{2}} \right)^2} + {\left( {y - \frac{x}{2} - \frac{y}{2}} \right)^2} = 1, \;\;\Rightarrow \frac{{2{{\left( {y - x} \right)}^2}}}{4} = 1, \;\;\Rightarrow {\left( {y - x} \right)^2} = 2, \;\;\Rightarrow y - x = \pm \sqrt 2 ,\;\;\Rightarrow y = x \pm \sqrt 2 .$

Note that the family of circles does not contain singular points, so the resulting solution is the equation of the envelope. It consists of two straight lines:

$y = x - \sqrt 2 \;\;\text{and}\;\;y = x + \sqrt 2 .$

Schematically, the family of circles and two envelope lines are shown in Figure $$3.$$

### Example 2.

Find the envelope of the family of ellipses defined by the equation $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ provided that $${a^2} + {b^2} = 1.$$

Solution.

The equation for the given family of curves can be written as

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{1 - {a^2}}} = 1,$

where the semi-axis $$a$$ is a parameter and $$0 \lt a \lt 1.$$ Differentiating this equation with respect to the parameter $$a,$$ we obtain the second equation:

$- \frac{{2{x^2}}}{{{a^3}}} - \frac{{{y^2} \cdot \left( { - 2a} \right)}}{{{{\left( {1 - {a^2}} \right)}^2}}} = 0, \;\;\Rightarrow - \frac{{2{x^2}}}{{{a^3}}} + \frac{{2{y^2}a}}{{{{\left( {1 - {a^2}} \right)}^2}}} = 0,\;\;\Rightarrow\frac{{{y^2}a}}{{{{\left( {1 - {a^2}} \right)}^2}}} = \frac{{{x^2}}}{{{a^3}}},\;\;\Rightarrow {y^2}{a^4} = {x^2}{\left( {1 - {a^2}} \right)^2}.$

Take the square root of both sides of then equation:

${y^2}{a^4} = {x^2}{\left( {1 - {a^2}} \right)^2}, \;\;\Rightarrow \left| y \right|{a^2} = \left| x \right|\left( {1 - {a^2}} \right).$

Express $${{a^2}}$$ from the last equation:

$\left| y \right|{a^2} = \left| x \right| - \left| x \right|{a^2}, \;\;\Rightarrow \left( {\left| x \right| + \left| y \right|} \right){a^2} = \left| x \right|, \;\;\Rightarrow {a^2} = \frac{{\left| x \right|}}{{\left| x \right| + \left| y \right|}}.$

Substituting this into the first equation and assuming that the family of curves has no singular points, we find the envelope:

$\frac{{{x^2}}}{{\frac{{\left| x \right|}}{{\left| x \right| + \left| y \right|}}}} + \frac{{{y^2}}}{{1 - \frac{{\left| x \right|}}{{\left| x \right| + \left| y \right|}}}} = 1,\;\;\Rightarrow \frac{{{x^2}\left( {\left| x \right| + \left| y \right|} \right)}}{{\left| x \right|}} + \frac{{{y^2}\left( {\left| x \right| + \left| y \right|} \right)}}{{\left| x \right| + \left| y \right| - \left| x \right|}} = 1,\;\;\Rightarrow {\left| x \right|^2} + \left| x \right|\left| y \right| + \left| x \right|\left| y \right| + {\left| y \right|^2} = 1,\;\;\Rightarrow {\left( {\left| x \right| + \left| y \right|} \right)^2} = 1,\;\;\Rightarrow \left| x \right| + \left| y \right| = \pm 1.$

Obviously, the negative root here does not make sense. Therefore, the final equation of the envelope is given by

$\left| x \right| + \left| y \right| = 1.$

This equation describes a square (Figure $$4$$).

See more problems on Page 2.