<p> C4 Chapter 2</p><p>CO-ORDINATE GEOMETRY IN THE (x,y) PLANE</p><p>PARAMETRIC EQUATIONS</p><p>Coordinates x and y are expressed as functions of a parameter. x = f(t) and y = g(t)</p><p>Example</p><p>Draw the curve given by the parametric equations: 2 x = t + 3, y = t for – 3 £ t £ 3 </p><p>Draw up a table of values</p><p> t -3 -2 -1 0 1 2 3 x 0 1 2 3 4 5 6 y 9 4 1 0 1 4 9</p><p> y 10</p><p>8</p><p>6</p><p>4</p><p>2</p><p>-2 0 2 4 6 x</p><p>JMcC 1</p><p>2 x = t + 3, y = t for – 3 £ t £ 3 C4 Chapter 2</p><p>To find the Cartesian equation of the curve, you need to eliminate the parameter between the two equations</p><p>Example 1</p><p>Find the Cartesian equation of the curve given by the parametric equations: 2 x = t + 3, y = t </p><p>Make t the subject Þ t = x – 3 </p><p>2 Substitute into the y equation Þ y = (x – 3) </p><p>Check: this is the parabola y = x2 translated 3 units to the right (see previous example)</p><p>Example 2</p><p>Find the Cartesian equation of the curve given by the parametric equations: 1 2 x = , y = , t ¹ – 3, t ¹ 5 t + 3 t – 5</p><p>Make t the subject: 1 1 1 1 – 3x x = Þ t + 3 = Þ t = – 3 = t + 3 x x x Substitute into the y equation: </p><p>1 – 3x 1 – 8x t – 5 = – 5 = x x 1 – 8x 2x \ y = 2 ¸ Þ y = x 1 – 8x</p><p>Ex 2A p 10</p><p>JMcC 2</p><p>2 x = t + 3, y = t </p><p>2 SMuabkset itt utthee insutob jtehcet yÞ eqtua=tioxn–Þ3 y = (x – 3) </p><p>1 2 x = , y = , t ¹ – 3, t ¹ 5 t + 3 t – 5</p><p>MSuabkset itt1 utt1hee – isn13utobx– jte8hcext :y eq11ua–tio8nx2: x 1 1 – 3x xt\ –=y5= =2 ¸Þ t +–35Þ== y Þ= t = – 3 = t + 3 x x x x1 – 8xx x C4 Chapter 2</p><p>USING PARAMETRIC EQUATIONS TO SOLVE PROBLEMS</p><p>Example 1</p><p>Find the coordinates of the point where the curve with parametric equations: x = √t, y = 4t – 25 meets the x-axis</p><p>25 At the x – axis, y = 0 Þ 4t – 25 = 0 Þ t = 4</p><p>25 5 \x = t = ± = ± 4 2 </p><p>æ 5 ö æ 5 ö Points of intersection are ç , 0 ÷ and ç – , 0 ÷ è 2 ø è 2 ø</p><p>Example 2</p><p>Find the coordinates of the points of intersection of the curve with parametric equations: x = 2t, y = t2 and the straight line y = x + 3</p><p>Substitute the parameters into the equation of the line: </p><p>2 2 t = 2t + 3 Þ t – 2t – 3 = 0 \ (t – 3)(t + 1) = 0 Þ t = 3, – 1 </p><p>\ when t = 3 Þ x = 6 and y = 9 </p><p>\ when t = – 1 Þ x = – 2 and y = 1 </p><p>Points of intersection are (6, 9) and ( – 2, 1) </p><p>Ex 2B p 13</p><p>JMcC 3</p><p>25 5 25 \Atx th=e xt –=a±x is, y = =0 ± Þ 4t – 25 = 0 Þ t = æ 5 ö æ 45 ö Points of intersec4tion are2 ç , 0 ÷ and ç – , 0 ÷ è 2 ø è 2 ø</p><p>2 2 tSP\ou i(wb=ntsths2t–ei tnou3 f+t) e(int=3 the+3Þer–1s Þp)e1tac=rÞtxai–o0mn=2xe Þtat6=re –er t s3a – (=ni6n2d=t,3 o 9,0 ya)t h–n e= da1 en 9 qdy u a= (tio1– n 2o,f 1 t) he line: C4 Chapter 2</p><p>PARAMETRIC EQUATIONS INVOLVING TRIGONOMETRIC PARAMETERS</p><p>You will need to use some of the Trigonometric Identities from C3</p><p>Example 1</p><p>Find the Cartesian equation of the curve given by: x = cos t, y = sin 2t y = sin 2t = 2 sin t cos t = 2x sin t \ we need to find sin t in terms of x or y </p><p>2 2 2 2 sin t = 1 – cos t Þ sin t = 1 – cos t = 1 – x </p><p>2 \y = 2x 1 – x </p><p>Example 2</p><p>Find the Cartesian equation of the curve given by: x = sec t, y = 4 sin t</p><p>1 1 y = 4 sin t and x = sec t = Þ cos t = cos t x</p><p>2 2 2 2 x – 1 (x – 1) sin t = 1 – cos t = 1 – (1 ¸ x ) = 2 = x x </p><p>2 4 (x – 1) \ y = x</p><p>Ex 2C p 15</p><p>JMcC 4</p><p>2 22 2 2 s\yiny =wt=es =in2 e1x2etd–1 =tc–oo2 sxf sitn indÞ ts icnso itns itn =ter2m1xs –s oincf o txs otr= y 1 – x </p><p>2 1 2 1 2 y = 4 sin4 t ( x an–d2 1 )x = sec t = 2 Þ cxos– t1= (x – 1) s\in ty== 1 – cos t = 1 – (1 ¸ x ) = 2 = x cos t x x x C4 Chapter 2</p><p>AREA UNDER A CURVE GIVEN BY PARAMETRIC EQUATIONS</p><p> dx Since Area = y dx, Chain Rule Þ y dx = y dt ò ò ò dt</p><p>Example</p><p>The diagram shows the curve with parametric equations: 2 x = t , y = t(5 – t) The region R is bounded by the curve and the x-axis. Find the area of R.</p><p> y</p><p>6</p><p>4</p><p>R 2</p><p>0 5 10 15 20 25 x</p><p>Curve crosses x – axis at (25, 0) Þ t(5 – t) = 0 Þ t = 5 25 5 dx 5 Area = y dx = y dt = t(5 – t) ´ 2t dt ò0 ò0 dt ò0</p><p>5 3 4 5 2 3 é 10t 2t ù = (10t – 2t ) dt = ê – ú = 104·17 ò0 ë 3 4 û o</p><p>Ex 2D p 18 & Mixed Ex 2E p 19</p><p>JMcC 5</p><p> dx Since Area = 2 y dx, Chain Rule Þ y dx = y dt x =òt , y = t(5 – t) ò ò dt</p><p>Curve cros25ses x – axis5 at (d2x5, 0) Þ 5t(5 – t) = 0 Þ t = 5 Area5= y dx = y dt3 = 4 t(55 – t) ´ 2t dt 2 3 é ù ò0 ò0 dt10t ò20t = (10t – 2t ) dt = ê – ú = 104·17 ò0 ë 3 4 û o</p>