The Witch of Maria Agnesi

The witch of Maria Agnesi shahbaz ahmed June 2024 Introduction Figure 1: geometry of the figure 1 Start with a circle of radius a, centered at the point (0,a),as shown in the figure. Choose a point A on the line y = 2a and connect it to the origin with a line segment.The segment crosses the circle at a point called B.Let P(x,y) be the point where the vertical line through A crosses the horizontal line through B.Let t be the radians measure of the angle that segment QA makes with the positives x-axis where 0 < t < π The witch is the curve traced by the point P as point A moves along the line y = 2a. The path of P(x,y) is a bell -shaped witch of Maria Agnesi as shown in figure below. Figure 2: The path of the point P(x,y) Reference to the figure 1. x-axis ,y = 2a and CD are parallel and OA is a transversal ∠t = ∠OAQ (alternates angles) Similarly ∠OAQ = ∠ABP 2 Hence ∠t = ∠OAQ = ∠ABP Also being angle in a semi-circle ∠OBQ is a right angle Similarly ∠ABQ = 90 being the supplementary angle of ∠OBQ Reference to the figure 1 In the right angled triangle OBQ ∠BOQ = 90 − t =⇒ ∠OQB = ∠t Also from figure 1 QA = CP = x OQ = diameter of the circle = 2a CQ = AP = AB sin ∠t y = OQ − CQ = 2a − AB sin ∠t Also in the right angled triangle ABQ cot ∠t = AB BQ AB = BQ cot ∠t ................. (1) In the right angled triangle OBQ QB OQ = QB 2a = cos ∠t QB = 2a cos ∠t ..................... (2) From (1) and (2) 2 ∠t) AB = 2a cos ∠t cot ∠B = 2a (cos sin ∠t Putting in the equation y = OQ − CQ = 2a − AB sin ∠t We have: 2 ∠t) y = OQ − CQ = 2a − 2a (cos sin ∠t sin ∠t y = 2a − 2a(cos ∠t)2 y = 2a[1 − (cos ∠t)2 ] 3 y = 2a sin2 ∠t ...................... (3) Now from the right angled triangle AOQ cot ∠t = QA OQ = x 2a x = 2a cot ∠t..................... (4) Equation in parametric form x = 2a cot ∠t, y = 2a sin2 ∠t From equation (3) y 2a = sin2 ∠t .................... (5) From equation (4) x 2a = cot ∠t Squaring x2 4a2 = cot2 ∠t x2 4a2 + 1 = cot2 ∠t + 1 x2 +4a2 4a2 = csc2 ∠t x2 +4a2 4a2 = 4a2 x2 +4a2 = sin2 ∠t ............... (6) 1 sin2 ∠t From equation (5) and equation (6) 4a2 x2 +4a2 = y 2a y(x2 + 4a2 ) = 8a3 The Cartesian equation y= 8a3 x2 +4a2 The path of the point P(x,y) for a circle of unit radius Putting a = 1 in the above equation y= 8 x2 +4 4 Figure 3: Witch of Agnesi 5