GENERAL ¨ ARTICLE Trigonometric Characterization of Some Plane Curves B Barua and J Das There is a way to describe a family of plane curves different from that using Cartesian or po- B Barua and J Das (née lar co-ordinates. This is a trigonometric equation Chaudhuri) are at Indian involving two angles. In this article, we highlight Society of Nonlinear Analysis (INSA), Kolkata. the fact that trigonometric equations are conve- nient to describe certain one-parameter families of plane curves. In some cases, the trigonomet- ric form makes it easier to characterize the family than if one were to use the equivalent Cartesian or polar representations. 1. Introduction The study of one-parameter families of plane curves is a well-known, interesting and important topic in plane analytic geometry. These curves can be represented by an equation of the type f(x, y, a)=0org(r, θ, b)= 0, where a, b are real parameters. Here, (x, y)and(r, θ) denote respectively the Cartesian and the plane polar coordinates [1, 2]. There is another way of representing a family of plane curves. It involves two angles related to a point on a curve: θ, the vectorial angle of the point with reference to a pole and a polar axis, and ψ, the angle made by the tangent to the curve at that point with the polar axis (Figure 1). A relation between tan θ and tan ψ gives the trigonometric equation of the family of curves. In this article, trigonometric equations of some known plane curves are deduced and it is shown that these equations reveal some geometric characteristics of the families of the curves under consideration. In Section 2, Keywords families of conics are discussed, and in Section 3, some Plane curves, trigonometric equations. RESONANCE ¨ March 2015 235 GENERAL ¨ ARTICLE Figure 1. P O θ ψ other plane curves. In Section 4, a few trigonometric equations are considered which lead to known families of plane curves. 2. Conics Consider the equation x2 + λy2 =1 (λ ∈ R : parameter). (2.1) It is well known that this represents an ellipse if λ>0 (the case λ = 1 represents a circle, which is a special case of an ellipse) and a hyperbola if λ<0. For λ =0, it represents a pair of parallel lines, x = ± 1, which is a degenerate conic. Let λ = 0. To deduce the trigonometric equation, we differentiate (2.1) with respect to x,toget yy/x − 1 , = λ where y =dy/dx. This relation can be written as The product of the tangents of the two θ ψ − 1 . tan tan = λ (2.2) angles is a negative constant for an Thus, the product of the tangents of the two angles is ellipse (– 1 for a a negative constant for an ellipse (Figure 2) (−1fora circle), and a circle), and a positive constant for a hyperbola (Figure positive constant for 3). a hyperbola. 236 RESONANCE ¨March 2015 GENERAL ¨ ARTICLE Conversely, we can start with the trigonometric equa- Figure 2. (left) tion, and get back the equation in terms of Cartesian Figure 3. (right) coordinates. Given a real number k,lettheequation tan θ tan ψ = −k (2.3) hold. Taking tan θ = y/x and tan ψ = y, (2.3) can be written as yy = −kx. On integration, this gives: 2 2 2 kx + y = C (C ∈ R). (2.4) Equation (2.4) represents a family of ellipses if k>0, and a family of hyperbolas if k<0. The Cartesian equation of a parabola is y2 =4ax, (a ∈ R : parameter). (2.5) Differentiating with respect to x,oneobtains2yy =4a and hence, y2 =2yyx or y(y − 2xy)=0. (2.6) If y = 0, then from (2.5), x = 0 and hence (0, 0) is the only solution, which is a degenerate case. If y = 0, then from (2.6) we get 2tanψ =tanθ. (2.7) RESONANCE ¨ March 2015 237 GENERAL ¨ ARTICLE Figure 4. Figure 5. P Conversely, if (2.7) is true for any curve, then taking tan θ = y/x and tan ψ = y we get the equation 2y = y/x, which on integration gives a family of parabolas, given by y2 = Cx,C∈ R (Figure 4). θ 2.1 Applications T O ψ N Example 1. Let us find the locus of points such that the distance between the foot of the perpendicular on the x-axis from a point on the required locus and the point of intersection of the tangent to the locus at that point with the x-axis is bisected by the pole. (See Figure 5). PN θ ψ PN PN From the figure, ON =tan and tan = TN = 2TO = PN 1 θ ψ θ 2ON = 2 tan . Therefore, 2tan =tan .So,by(2.7), Figure 6. the required locus is a family of parabolas with the pole as vertex. Example 2. Let P be a point on an ellipse, N be the foot P of the perpendicular from P on its major axis, and T be the point of intersection of the tangent at P with its major axis. If O is the centre of the ellipse, let us prove that ON · OT = constant (see Figure 6). θ ψ Taking the centre O of the ellipse as the pole and its O N major axis as the polar axis, we have PN PN tan θ = , tan ψ = − . ON NT So 2 PN tan θ tan ψ = − < 0. ON(OT − ON) 238 RESONANCE ¨March 2015 GENERAL ¨ ARTICLE We know from Section 2 that, for an ellipse, tan θ tan ψ =–a2 for some real number a. Hence, B 2 2 PN = a2(ON · OT − ON ), 2 2 2 2 i.e., a ON · OT = a ON +PN . For an ellipse, the right-hand side of this equation is a constant. Hence, ON · OT = constant. A' O A 3. Other Algebraic Curves In this section, various types of families of algebraic curves are considered. Equations of these curves are written either in Cartesian coordinates (x, y)orinterms B' of plane polar coordinates (r, θ). In some cases, para- metric equations are also considered. Figure 7. 3.1 Astroid In Cartesian coordinates, the equation of an astroid (Fig- ure 7) is x2/3 + y2/3 = a2/3 , (a ∈ R). (3.1) Differentiating both sides of (3.1) with respect to x,we get y 1/3 y − . = x (3.2) Writing tan θ = y/x and tan ψ = y,wegetfrom(3.2) tan ψ = − tan1/3 θ, or, tan3 ψ +tanθ =0. (3.3) Conversely, from (3.3) we can deduce (3.2). Integrating both sides, we get back the equation 2/3 2/3 x + y = C (C ∈ R). Thus, (3.3) is the trigonometric representation of a fam- ily of astroids. RESONANCE ¨ March 2015 239 GENERAL ¨ ARTICLE 3.2 Quadrifolium B Arose-petalcurve(Figure 8) with four petals is called a quadrifolium. In plane polar coordinates (r, θ), the equation of the curve is r = a cos 2θ. (3.4) Differentiating both sides with respect to θ and writing dr/dθ =˙r,weget A' O A r˙ = −2a sin 2θ. (3.5) Differentiating the relations r2 = x2 + y2and tan θ = y/x x B' with respect to ,weget r x yy r r + . ˙ = θ = xy − y Figure 8. ≡ d Note that, ( dx). Substituting this value ofr ˙ in (3.5) we get 1+tanθ tan ψ 2tan2θ = . (3.6) tan θ − tan ψ Again, from the relations, x = r cos θ, y = r sin θ,weget y˙ r˙ tan θ + r y = = . x˙ r˙ − r tan θ Substituting this expression for tan ψ (= y) in (3.6), we get r θ − ˙ . 2tan2 = r On integration, this gives r = C cos 2θ. This represents a family of quadrifoliums. Hence (3.6) is the trigonometric representation of this family. 240 RESONANCE ¨March 2015 GENERAL ¨ ARTICLE 3.3 Nodal Cubic Curve The equation of the curve in Cartesian coordinates is 2 3 2 y = x + x . (3.7) Parametrically, it can be written as x = t2 − 1, (3.8) 3 y = t − t. x y dx dy Writing ˙ and ˙ for dt and dt respectively, we get 2 dy y˙ 3t − 1 y = = = =tanψ. (3.9) dx x˙ 2t Also, y t(t2 − 1) = = t =tanθ. (3.10) x t2 − 1 Hence, 3tan2 θ − 1 tan ψ = 2tanθ or, 2 2tanθ tan ψ =3tan θ − 1. (3.11) Conversely, let (3.11) be true for a given plane curve. We get: 2y =3y/x − x/y. Solving this equation, we Figure 9. get, 2 3 2 y = Cx + x , (C ∈ R : parameter). (3.12) This represents a family of nodal cubic curves. Hence, (3.11) is the trigonometric equation of such a family (Figure 9). 3.4 Applications B Example 3. Let us prove that the envelope of a family of A ellipses for which the sum of the major and minor axes is a constant, is an astroid. A= (0 , 0) for t = ± 1, B = (-1 , 0) for t = If a, b be the major and minor semi-axes of an ellipse, 0. then, according to the hypothesis, a + b = k,where RESONANCE ¨ March 2015 241 GENERAL ¨ ARTICLE k ∈ R is the given constant. Hence, the equation of the ellipse can be written as x2 y2 F (a) ≡ + − 1=0. (3.13) a2 (k − a)2 To find the envelope, we have ∂aF = 0, from which we get y2 k − a 3 2 θ. x2 = a =tan (3.14) Also, for the ellipse, the trigonometric equation is k − a 2 θ ψ − tan tan = a or 1/2 k − a 1/3 3 ψ − − θ tan \tan T tan = a = tan is the trigonometric or equation of an 3 tan ψ +tanθ =0.