IAS 3.1 Conic Sections

Transcription

1 Year 13 Mathematics IAS 3.1 Conic Sections Robert Lakeland & Carl Nugent Contents Achievement Standard The Straight Line The Parametric Form of a Straight Line The Circle The Parametric Form of a Circle The Ellipse The Parabola The Hperbola The Equations of Tangents and Normals The Intersection Between a Line and a Conic Section Directri and Eccentricit Practice Internal Assessment Answers Innovative Publisher of Mathematics Tets

2 IAS 3.1 Conic Sections NCEA 3 Internal Achievement Standard 3.1 Conic Sections This achievement standard involves appling the geometr of conic sections in solving problems. Achievement Achievement with Merit Achievement with Ecellence Appl the geometr of conic sections in solving problems. Appl the geometr of conic sections, using relational thinking, in solving problems. Appl the geometr of conic sections using etended abstract thinking, in solving problems. This achievement standard is derived from Level 8 of The New Zealand Curriculum and is related to the achievement objectives appl the geometr of conic sections in the Mathematics strand of the Mathematics and Statistics Learning Area. Appl the geometr of conic sections in solving problems involves: selecting and using methods demonstrating knowledge of concepts and terms communicating using appropriate representations. Relational thinking involves one or more of: selecting and carring out a logical sequence of steps connecting different concepts or representations demonstrating understanding of concepts forming and using a model; and relating findings to a contet, or communicating thinking using appropriate mathematical statements. Etended abstract thinking involves one or more of: devising a strateg to investigate or solve a problem identifing relevant concepts in contet developing a chain of logical reasoning, or proof forming a generalisation; and using correct mathematical statements, or communicating mathematical insight. Problems are situations that provide opportunities to appl knowledge or understanding of mathematical concepts and methods. Situations will be set in real-life or mathematical contets. Methods include a selection from those related to: graphs and equations of the circle, ellipse, parabola and hperbola Cartesian and parametric forms properties of conic sections tangents and normals. IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

3 IAS 3.1 Conic Sections 9 The Circle A circle is formed when we cut straight across the A circle is the path or locus of points which are an the cone. equal distance from a given point, the centre. When the centre of the circle is the origin then the Both the Casio 9750GII and equation of the circle radius r is given b the TI-8 Plus can graph conics. Detailed below are the + = r instructions related to circles, but ou can also use our graphics When the centre is (a, b) then the equation becomes calculator, in a similar wa for ( a) + ( b) = r the other conics studied in this chapter, i.e. ellipse, parabola and hperbola. Completing the Square On the Casio 9750GII from the main MENU If the equation has been epanded select CONICS MENU 7. You are e.g = 0 presented with a number of options. then we need to factorise it (complete the square) in Scroll down to order to find the values of a, b and r. (X H) + (Y K) = R or We do this b first grouping all the variables on the AX + AY + BX + CY + D = 0 left while leaving the constants on the right. An alternative to completing the square is to graph the equation given in epanded form + + ulake + = 0 e.g Ltd = = You select We now make each of the two variables into a AX + AY + BX + CY + D = 0 form perfect square b using the pattern and enter A = 1, B =, C = and D =. ( ± a) = ± a + a 1 EXE (- ) EXE EXE We compare DRAW G-Solv CNTR RADS +? = ± a + a EXE F SHIFT F5 F1 OR F Ltd ulake to identif the value for a (in this case 3 see eample net page) and replace? with a (in this case ( 3) = 9) to both sides of the equation to complete the pattern. We compare The Equation of a Circle + +? = ± a + a to identif the value for a (in this case see eample net page) and replace? with a (in this case () = ) to both sides of the equation to complete the pattern. Therefore = ( 3) + ( + ) = 9 ( 3) + ( + ) = 3 We can now read off the centre and radius of the circle, i.e. centre (3, ), radius 3. Select Graph Solve to find the radius (3) and centre (3, ). You are then able to use these values to write the equation in the form ( 3) + ( + ) = 3 On our TI-8 Plus ou access the conics menu via the application button and scroll down until ou get to the conics sub-menu. CIRCLE APPS ENTER ENTER After selecting circle choose the form of the equation ou require. 1: (X H) + (Y K) = R : AX + AY + BX + CY + D = 0 In this case. Enter 1 ENTER (- ) ENTER ENTER ENTER GRAPH Circles, Ellipses, Parabolas etc. are called conic sections as the are formed when we cut through a cone. IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

4 IAS 3.1 Conic Sections 17 Find the coordinates of the centre, the length of the major and minor aes and the coordinates of the foci. Sketch the ellipse. ( + ) 1 ( ) + = 1 9 Show that The centre of this ellipse is (, and a = and b = 3 Length of major ais is ( a) = 8 Length of minor ais is ( b) = To find the foci we use c = a b, so c =. Foci are (., ) and ( +., i.e. (., ) and (0., (-, 7) (, (-, ) ). - (-, 1) = 0 is the equation of an ellipse and find its centre and the length of each ais = 0 Collect like terms together and factorise each term 5( 8) + ( + ) + 31 = 0 Complete the square for each variable. The term being subtracted is to balance what has been added. 5( ) + ( + + ) + 31 = 0 5(( ) 1) + (( + ) ) + 31 = 0 5( ) 00 + ( + ) = 0 5( ) + ( + ) = 100 Divide through b 100 to get it into the form of an ellipse. 5( ) - - In down a pair of parametric equations for the Ltd ulake Write down the Cartesian equation of the ellipse with parametric equations = Write cost and = sin t Centre is (5, ) with a = 1 and b = 100 ( ) + ( + ) 100 ( + ) + 5 = 1 = 1 Ellipse, centre (, ) with major ais ( 5) = 10 and minor ais ( ) =. ellipse ( ) 9 ( + 3) + = 1 ( 5) 1 + ( ) ( ) = 1 1( 5) ( ) + 1 = 1 Centre is (, 3) with a = 3 and b = = + 3 cos t and = 3 + sin t IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

5 IAS 3.1 Conic Sections 3 Give the equation of the parabola with verte (3, ), ais of smmetr = and a point A parabola has a verte at (3, 0) and (5, ) which is on the curve. Locate its focus. focus at (3, 0.5), find its equation and Using intercept. focus (3, 0.5) ( 1 ) = a( 1 ) 0 Verte of the parabola is (3, ) therefore ( ) = a( 3) Substituting (5, ) to find a As the parabola is arranged verticall its equation must be in the form ( ) = a(5 3) ( 1 ) = a( 1 ) 1 = 8a verte = (3, 0) a = a = 0.5 So equation of the parabola is ( 3) = 0.5( 0) ( ) = 8( 3) ( 3) = Using a = to find the focus intercept ( = 0) Focus = (3 + a, ) (0 3) = = (5, ) =.5 Show that = 0 A parabola has parametric equations = 1 + t and = t is the equation of a parabola and sketch the graph. Locate its verte, focus and the directri. Find the Cartesian equation of the parabola and In its intercepts. Sketch the graph. Ltd ulake Rearranging and completing the square = 0 + = 5 ( + 1) 1 = 5 Making t the subject. t = 1 Substituting = 1 ( + 1) = ( + 1) = ( 1) Verte (1, 1) and focus a = 1 focus = (, 1) Equation of the directri is = 0 1( + ) = ( 1) ( 1) = 1( + ) which is the form of the parabola = a with verte translated to (, 1). intercepts = 0 =. and =. (-, 1) (, 1) 0 0 (1, -1) (, -3) IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

6 8 IAS 3.1 Conic Sections Merit Use our knowledge of conics to solve the following problems. 81. The wire supporting a suspension bridge is best modelled b a parabola. It is connected to the bridge b chain lengths. The first chain is 8 m and the shortest length is 3 m. If the are placed at 5 m intervals and the bridge is 50 m long find the length of chain used. Hint: Draw some aes on the diagram that will minimise our work. 8. A student constructing the parabolic reflector of a satellite dish is working to the following measurements. Find the position of the focus. Longest chain is 8 m Shortest chain is 3 m 50 m long with supports at 5 m intervals 1 m across.5 cm deep 83. The span on the underside of a bridge can be Bridge height is 3. m modelled b a parabola with a maimum height of 3. m and width of.5 m. Investigate if a barge with dimensions above the waterline of. m. m across and 1.8 m high (includes a gap for Barge 1.8 m clearance) can fit under the bridge. The bridge width is.5 m Ltd ulake 8. A car headlight is in the shape of a parabolic dish. The bulb is placed at the focus (f). The headlight is 0 mm across the front and 80 mm at its greatest depth. a) Find the distance from the verte of the parabolic dish to the bulb. b) Find the pair of parametric equations for the parabola. bulb 80 mm f 0 mm IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

7 3 IAS 3.1 Conic Sections The Equations of Tangents and Normals Tangents and Normals using Implicit Differentiation Implicit Differentiation Implicit differentiation is eplained in A tangent is a line that touches a curve at a point the NuLake EAS 3. Booklet or our EAS Calculus without crossing it. Workbook. You will need to understand this A normal is a line that is perpendicular to a tangent if ou want to reach Merit level, although it is at a point on a curve. possible to differentiate using the parametric form To find the equation of a tangent or normal it is (see eamples below and later in this section). usuall necessar to differentiate the function concerned first, to find the slope (gradient) at the designated point. If the equation includes a component, this will require differentiating implicitl. Earlier we used the formula 1 = m( 1 ) to find the equation of the desired straight line with tangent m. We also used the relationship between the gradients of perpendicular lines m 1 m = 1 1 OR m 1 = m which will be needed here. Find the equation of the tangent to the circle Find the equation of the normal to the parabola + = 9 at the point P (5, ). = 8( ) at (, ) and find where this normal intersects the curve again. Differentiating the circle implicitl Differentiating implicitl Ltd ulake + = 9 = 8( ) + d d = 0 d d = d d = At the point (5, ) the gradient of the tangent is 5 Tangent is = ( 5) = = 0 5. = 8 1 d d = 8 d d = At the point (, ) the gradient of the tangent is 1. Gradient of the normal is 1. Normal is + = 1( ) + = = 8 Substitute for ( 8) = 8( ) 1 + = 8 1 gives + 80 = 0 ( )( 0) = 0 =, 0 The point of intersection is (0, 1) IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent

8 58 IAS 3.1 Conic Sections Page 1 cont... Page 15 cont... Page 18 cont a) = + 3 cos t 39. a) = cos t 3. Centre (, 3) = sin t = + 5 sin t Major ais = 10 Minor ais = b) b) Foci (, 3) and (, 3) End points ( 7, 3) to (3, 3) End points (, 0) to (, ). Centre (, 1) Major ais = Minor ais = Foci ( 0.83, 1) and (.83, 1) - - End points ( 1, 1) to (5, 1) End points (, 0) to (, - ) 5. Centre (0, 0) 3. = Major ais = cos t 3 Minor ais = 0. = cos t = 3 + sin t Foci ( 1, 0) and (1, 0) = + 3 sin t End points (± 5,0) End points (0, ±). Centre (, 5) Major ais = - - Minor ais = 1-3 Foci (.87, 5) and ( 1.13, 5) - End points ( 3, 5) to ( 1, 5) End points (,.5) to (, 5.5) - 7. Centre (, 0) -8 ulake Major ais Ltd = 8 Minor ais = 37. a) = cos t Foci (1.35, 0) and (.5, 0) 1. a) = cos t = sin t End points (0, 0) to (8, 0) = + 10 sin t End points (, 3) to (, 3) b) b) ( + ) ( 1) 8. + = Centre (, 1) - Foci (, 0.73) and (,.73) Ltd ulake Page a) ( ) + ( 5) = 10 b) Page 18. Centre (0, 0) Major ais = 1 Minor ais = 10 Foci ( 3.3, 0) and (3.3, 0) End points (±, 0) and (0, ±5) ( + 1.5) ( 1) + = = cos t = 1 + sin t 5 Foci (.85, 1) and (1.85, 1) IAS 3.1 Year 13 Mathematics and Statistics Published b N New Zealand Robert Lakeland & Carl Nugent