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A2-Level Maths: Core 4 for Edexcel C4.2 Coordinate geometry This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 29 © Boardworks Ltd 2006 Parametric equations of curves Contents Parametric equations of curves Converting from parametric to Cartesian form The parametric equations of some standard curves The area under a curve defined parametrically Examination-style question 2 of 29 © Boardworks Ltd 2006 Parametric equations of curves All of the curves we have looked at so far have been defined by a single equation in terms of x and y. Curves can also be defined by writing x and y in terms of a third variable or parameter. For example, using the parameter t a curve is defined by: x = t2 – 3 y = 2t We can plot this curve for –3 < t < 3 using a table of values: t x = t2 – 3 y = 2t –3 –2 –1 0 1 2 3 6 –6 1 –4 –2 –2 –3 0 –2 2 1 4 6 6 (–2, 2) (1, 4) (6, 6) (6, –6) (1, –4) (–2, –2) (–3, 0) 3 of 29 © Boardworks Ltd 2006 Parametric equations of curves Each value of t gives us a coordinate that we can then plot on a set of axes. In this example, drawing a smooth line through these points gives us a parabola. In most cases, a graphical calculator or a graph-plotting computer program can be used to produce curves that have been defined parametrically. 4 of 29 t=3 y t=2 t=1 t=0 t = –1 0 x t = –2 t = –3 © Boardworks Ltd 2006 Parametric equations of curves This diagram shows a sketch of the curve defined by: t3 x = +2 3 y y = t2 9 A Find the coordinates of the points A and B where the curve meets the x-axis. B x This point is called a cusp. The curve meets the x-axis when y = 0, that is when: t2 9 = 0 t2 = 9 t = 3 5 of 29 © Boardworks Ltd 2006 Parametric equations of curves When t = 3: 33 x = + 2 = 11 3 When t = –3: ( 3)3 x= + 2 = 7 3 So the coordinates of A are (–7, 0) and the coordinates of B are (11, 0). 6 of 29 © Boardworks Ltd 2006 Parametric equations of curves This diagram shows a sketch of the curve defined by: x = t2 – 1 y y = t3 – 4t x=1 A The curve meets the line x = 1 at points A and B. Find the exact length of the line segment AB. 0 x B When x = 1: t 2 1=1 t2 = 2 t= 2 7 of 29 © Boardworks Ltd 2006 Parametric equations of curves When t = 2 : y = ( 2)3 4 2 =2 2 4 2 = 2 2 When t = – 2 : y = ( 2)3 + 4 2 = 2 2 + 4 2 =2 2 So the coordinates of A are (1, 2 2 ) and the coordinates of B are (1, –2 2 ). The length of line segment AB = 2 2 2 2 =4 2 8 of 29 © Boardworks Ltd 2006 From parametric to Cartesian form Contents Parametric equations of curves Converting from parametric to Cartesian form The parametric equations of some standard curves The area under a curve defined parametrically Examination-style question 9 of 29 © Boardworks Ltd 2006 Converting from parametric to Cartesian form The Cartesian form of an equation only contains the two variables x and y. In many cases, a curve that has been defined parametrically can be expressed in Cartesian form by eliminating the parameter. For example, Find the Cartesian equation for the following pair of parametric equations: x = 3t + 1 y = 5 – 2t In examples of this type, we make t the subject of one of the equations and then substitute this expression into the other equation. 10 of 29 © Boardworks Ltd 2006 Converting from parametric to Cartesian form If x = 3t + 1 then x 1 t= 3 Substituting this into the second equation gives: x 1 y = 5 2 3 2 x + 2 y 5= 3 3 y 15 = 2x + 2 3 y =17 2x This Cartesian equation represents a straight line graph. 11 of 29 © Boardworks Ltd 2006 Converting from parametric to Cartesian form Find the Cartesian equation for the following pair of parametric equations: x = 5 – t2 y = 3t2 – 4 If x = 5 – t2 then: t2 = 5 x The second equation is written in terms of t2 so we can leave this as it is. Substituting this value of t2 into the second equation gives: y = 3(5 x) 4 y =15 3 x 4 y =11 3 x 12 of 29 © Boardworks Ltd 2006 Converting from parametric to Cartesian form Find the Cartesian equation for the following pair of parametric equations: x = 3 + 2 sin θ y = 1 + 2 cos θ We can eliminate the parameter θ using the identity sin2 θ + cos2 θ = 1. x – 3 = 2 sin θ y – 1 = 2 cos θ Squaring and adding these equations gives: ( x 3)2 + ( y 1)2 = 22 sin2 + 22 cos2 = 4(sin2 + cos2 ) =4 13 of 29 © Boardworks Ltd 2006 Converting from parametric to Cartesian form The Cartesian equation is therefore ( x 3)2 + ( y 1)2 = 4 This is the equation of a circle of radius 2 centred at the point (3,1). Find the Cartesian equation for the following pair of parametric equations: x = 2 cos θ y = cos 2θ Using the double angle formulae we can write: y = 2 cos2 θ – 1 x2 = 4 cos2 θ and so the Cartesian equation is: x2 y = 1 2 14 of 29 © Boardworks Ltd 2006 Contents Parametric equations of standard curves Parametric equations of curves Converting from parametric to Cartesian form The parametric equations of some standard curves The area under a curve defined parametrically Examination-style question 15 of 29 © Boardworks Ltd 2006 Parabolas 16 of 29 © Boardworks Ltd 2006 Rectangular hyperbolae 17 of 29 © Boardworks Ltd 2006 Circles centred at the origin 18 of 29 © Boardworks Ltd 2006 Circles centred at the point (a, b) 19 of 29 © Boardworks Ltd 2006 Ellipses 20 of 29 © Boardworks Ltd 2006 Contents The area under a curve defined parametrically Parametric equations of curves Converting from parametric to Cartesian form The parametric equations of some standard curves The area under a curve defined parametrically Examination-style question 21 of 29 © Boardworks Ltd 2006 The area under a curve defined parametrically We know that the area under the curve y = f(x) between the limits x = a and x = b is given by: b A = y dx a Suppose, however, that we wish to find the area under a curve that is defined in terms of a parameter t. We can write the area in terms of the parameter as: A= t2 t1 dx y dt dt where t1 and t2 are the limits x = a and x = b rewritten in terms of the parameter t. 22 of 29 © Boardworks Ltd 2006 The area under a curve defined parametrically For example, consider the curve defined by the parametric equations: x = 2t y = t2 + 3 Suppose we want to find the area under this curve between x = –2 and x = 4. y x Since t = 2 , these limits can be written in terms of t as: t = –1 and t = 2 dx Also, =2 dt 23 of 29 A –2 4 x © Boardworks Ltd 2006 The area under a curve defined parametrically The area, A, is given by: t2 dx A = y dt t1 dt dx = 2 and y = t2 + 3 gives: Substituting t1 = –1, t2 = 2, dt 2 A = 2(t 2 + 3) dt 1 2 = 2t 2 + 6 dt 1 = 32 t 3 + 6 x 2 1 2 = ( 16 3 +12) ( 3 6) = 24 So the required area is 24 units2. 24 of 29 © Boardworks Ltd 2006 Contents Examination-style question Parametric equations of curves Converting from parametric to Cartesian form The parametric equations of some standard curves The area under a curve defined parametrically Examination-style question 25 of 29 © Boardworks Ltd 2006 Examination-style question The diagram shows part of the curve C, defined by the parametric equations: 1 x = + 4t t 1 y = 4t t y y + 3x = 12 0 The line y + 3x = 12 cuts the curve C at points A and B. A C x B a) Find the coordinates of the points A and B. b) Show that the Cartesian equation of the curve C is x2 – y2 = 16. 26 of 29 © Boardworks Ltd 2006 Examination-style question 1 1 a) Substituting x = + 4t and y = 4t into y + 3x = 12 gives: t t 1 1 4t + 3 + 4t = 12 t t 1 3 4t + +12t = 12 t t 4 + 8t = 12 t 1+ 2t 2 = 3t 2t 2 3t +1= 0 (2t 1)(t 1) = 0 27 of 29 © Boardworks Ltd 2006 Examination-style question The line and the curve intersect when t = 21 and when t = 1. When t = 1 2 : x= 1 1 2 + 4( 21 ) = 4 y= 1 1 2 4( 21 ) = 0 A is the point (4, 0). When t = 1: 1 x = + 4(1) = 5 1 1 y = 4(1) = 3 1 B is the point (5, –3). 28 of 29 © Boardworks Ltd 2006 Examination-style question b) Squaring the parametric equations of the curve gives: 1 2 x = + 4t t 2 and 1 x = 2 + 8 +16t 2 t 2 1 2 y = 4t t 2 1 y = 2 8 +16t 2 t 2 Subtracting y2 from x2 gives: 1 1 2 x y = 2 + 8 +16t 2 + 8 16t 2 = 16 t t 2 2 x 2 y 2 = 16 29 of 29 © Boardworks Ltd 2006