and expansion and simplification, which is apparently the equation for a parabola with axis along a=0, or y=x. Example 3 Let I ⊂ R be an open interval and let γ : I → R2 be a smooth plane curve parametrised by arc length. Consider the one-parameter family of normal lines to γ(I). A line is normal to γ at γ(t) if it passes through γ(t) and is perpendicular to the tangent vector to γ at γ(t). Let T denote the unit tangent vector to γ and let N denote the unit normal vector. Using a dot to denote the dot product, the generating family for the one-parameter family of normal lines is given by where Clearly (x − γ)·T = 0 if and only if x − γ is perpendicular to T, or equivalently, if and only if x − γ is parallel to N, or equivalently, if and only if x = γ + λN for some λ ∈ R. It follows that is exactly the normal line to γ at γ(t0). To find the discriminant of F we need to compute its partial derivative with respect to t: where κ is the plane curve curvature of γ. It has been seen that F = 0 if and only if x - γ = λN for some λ ∈ R. Assuming that F = 0 gives Assuming that κ ≠ 0 it follows that λ = 1/κ and so This is exactly the evolute of the curve γ. Example 4 The following example shows that in some cases the envelope of a family of curves may be seen as the topologic boundary of a union of sets, whose boundaries are the curves of the envelope. For and consider the (open) right triangle in a Cartesian plane with vertices , and Fix an exponent , and consider the union of all the triangles subjected to the constraint , that is the open set To write a Cartesian representation for , start with any , satisfying and any . The Hölder inequality in with respect to the conjugated exponents and gives: , with equality if and only if . In terms of a union of sets the latter inequality reads: the point belongs to the set , that is, it belongs to some with , if and only if it satisfies Moreover, the boundary in of the set is the