this case by an appropriate substitution. For example, if the family is given by Cθ with an equation of the form u(x, y)cos θ+v(x, y)sin θ=w(x, y), then putting t=eiθ, cos θ=(t+1/t)/2, sin θ=(t-1/t)/2i changes the equation of the curve to or The equation of the envelope is then given by setting the discriminant to 0: or Alternative definitions The envelope E1 is the limit of intersections of nearby curves Ct. The envelope E2 is a curve tangent to all of the Ct. The envelope E3 is the boundary of the region filled by the curves Ct. Then , and , where is the set of points defined at the beginning of this subsection's parent section. Examples Example 1 These definitions E1, E2, and E3 of the envelope may be different sets. Consider for instance the curve parametrised by where . The one-parameter family of curves will be given by the tangent lines to γ. First we calculate the discriminant . The generating function is Calculating the partial derivative . It follows that either or . First assume that . Substituting into F: and so, assuming that t ≠ 0, it follows that if and only if . Next, assuming that and substituting into F gives . So, assuming , it follows that if and only if . Thus the discriminant is the original curve and its tangent line at γ(0): Next we calculate E1. One curve is given by and a nearby curve is given by where ε is some very small number. The intersection point comes from looking at the limit of as ε tends to zero. Notice that if and only if If then L has only a single factor of ε. Assuming that then the intersection is given by Since it follows that . The y value is calculated by knowing that this point must lie on a tangent line to the original curve γ: that . Substituting and solving gives y = t3. When , L is divisible by ε2. Assuming that then the intersection is given by It follows that , and knowing that gives . It follows that Next we calculate E2. The curve itself is the curve that is tangent to all of its own tangent lines. It follows that Finally we calculate E3. Every point in the plane has at least one tangent line to γ passing through it, and so