Syllus Ojective: 10.1 The student will sketch the grph of conic section with centers either t or not t the origin. (PARABOLAS) Review: The Midpoint Formul The midpoint M of the line segment connecting the points (, y ) nd (, ) M + y + y. 1 1 =, 1 1 E: Find the midpoint of the line segment joining ( 7,1) nd (,5) Let (, y ) = ( 7,1) nd (, ) (,5) 1 1. y is y =. Sustitute the vlues into the formul nd simplify. 1+ y1+ y 7+ 1+ 5 9 M =, =, =,3 Conic Sections: curves tht re formed y the intersection of plne nd doulenpped cone Prol: the conic formed y connecting ll points equidistnt from point (the focus) nd line (the directri) with the eqution of the form ( h) = 4 p( y k) Verte ( hk, ) ( y k) = 4 p( h) : the point,, tht lies on the is of symmetry hlfwy etween the focus nd directri Ais of Symmetry: the line of symmetry of prol tht psses through the verte. ( y k) = 4 p( h) ( h) 4 p( y k) For, the is of symmetry is horizontl, nd hs the eqution = h. For =, the is of symmetry is verticl, nd hs the eqution y = k. p : the distnce the focus nd directri re from the verte Focus: point on the is of symmetry of prol equidistnt from the verte s the directri Directri: line perpendiculr to the is of symmetry equidistnt from the verte s the focus or Pge 1 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Prol = 4 py, p > 0 Prol = 4 py, p < 0 Focus ( 0, p) Directri y = p Verte ( 0,0 ) Directri y = p Verte ( 0,0 ) Focus ( 0, p ) Prol y = 4 p, p > 0 Prol y = 4 p, p > 0 Directri = p Verte ( 0,0 ) Focus ( p,0) Focus ( p,0) Verte ( 0,0 ) Directri = p E: Sketch the grph of the prol 1 3 = y. Verte: ( 0,0 ) Becuse y is squred, the is of symmetry is horizontl: y = 0 4p = 3 Find p: y = 3 3 p = 4 the prol will open to the left. Becuse p is negtive, Directri: 3 = Focus: 4 3,0 4 Pge of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Stndrd Form of the Eqution of Prol: Horizontl Ais (opens left or right): ( y k) = 4 p( h) Verte: ( hk, ) Ais of Symmetry: y = k Focus: ( h pk, ) Verticl Ais (opens up or down): ( h) = 4 p( y k) Verte: ( hk, ) Ais of Symmetry: = h Focus: ( hk, p) E: Sketch the grph of the prol ( 7) 8( y 3) directri. Becuse it is in the form ( h) 4 p( y k) verticl is. =, the prol hs + Directri: = h p + Directri: y = k p + =. Identify the verte, focus, nd Find p: 4 p = 8 p = p is positive, so the prol opens up. Verte: ( 7,3) Focus: p units up from the verte ( 7, 3 + ) = ( 7, 5) Directri: horizontl line p units down from the verte y = 3 y = 1 E: Write n eqution of the prol whose verte is t ( 3, ) nd whose focus is t ( 4, ). Begin with sketch. The prol opens towrd the focus, so it opens right. Find p: The distnce from the focus to the verte is 1. The prol opens right, so p = 1. Becuse the prol hs horizontl is, we will use the eqution ( y k) 4 p( h) The verte ( 3, ) = ( hk, ), nd p = 1. ( y ) = 4( 3) =. E: A store uses prolic mirror to see ll of the isles in the store. A cross section of the mirror is shown. Write n eqution for the cross section of the mirror nd identify the focus. Pge 3 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Becuse the prol hs verticl is, we will use the eqution ( h) 4 p( y k) ( 0,0 ) ( hk, ) =, so we now hve = 4 py. The prol psses through the point ( 8, ) ( y, ) ( ) ( ) py p p p = 4 8 = 4 64 = 8 = 8 =. Use this to find p. =. The verte is The eqution is = 4( 8) y = 3 y. The focus is p units up from the verte: ( 0,8 ) You Try: Grph the eqution ( y 6) = 4( 8). Identify the verte, focus, nd directri. QOD: Prols cn e found mny plces in rel life. Find t lest three rel-life emples of prols. Wht is the significnce of the focus in these rel-life emples? Pge 4 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Syllus Ojective: 10.1 The student will sketch the grph of conic section with centers either t or not t the origin. (CIRCLES) Review: The Distnce Formul The distnce d etween the points (, y ) nd (, y ) is d ( ) ( y y ) 1 1 = +. 1 1 Use the Pythgoren Theorem to show this on the coordinte plne y forming right tringle with the,,, y. points ( y ), ( y ), nd ( ) 1 1 1 E: Find the distnce etween the points (, 4) nd ( 5, 1) Let (, y ) = ( 5, 1) nd (, ) (, 4) 1 1. y =. Sustitute the vlues into the formul nd simplify. ( ) ( ) ( ) ( ) ( ) ( ) ( ) d = 1 + y y1 = 5 + 1 4 = 7 + 3 = 49 + 9 = 58 Circle: the conic formed y connecting ll points equidistnt from point with the eqution of the form ( ) ( ) h + y k = r Center of Circle ( hk, ) : the point,, tht is equidistnt from ll points on the circle Rdius: the distnce r etween the center nd ny point on the circle E: Show tht the eqution of circle with center ( ) ( ) h y k r + =. ( hk, ) nd rdius r is Let one point on the circle e ( y, ) nd the center e (, ) center nd point on the circle. = ( ) + ( ) r = ( h) + ( y h) d y y 1 1 hk. The rdius, r, is the distnce etween the r = h + y k. Squring oth sides, we hve ( ) ( ) E: Drw the circle given y the eqution y = 4. Write in stndrd form. + y = 4 Identify the center nd the rdius. Center: ( 0,0 ) Rdius: r = 4 r = Sketch the grph y plotting the center then plotting points on the circle units Pge 5 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
ove, elow, nd to the left nd right of the center. E: Write n eqution of the circle with center ( 1, 4) nd rdius 3. Center: ( 1, 4 ) ( hk, ) = Rdius: r = 3 Eqution: ( ) ( ) ( ) ( ) ( ) ( ) ( ) h + y k = r 1 + y 4 = 3 + 1 + y 4 = 9 E: Write n eqution of the circle shown. The center of the circle is ( 0,0 ) ( hk, ) =. Eqution: + y = r To find r, we will use the point given on the circle ( 5,1 ). ( ) ( ) + y = r 5 + 1 = r r = 6 The eqution of the circle is + y = 6. E: A street light cn e seen on the ground within 30 yd of its center. You re driving nd re 10 yd est nd 5 yd south of the light. Write n inequlity to descrie the region on the ground tht is lit y the light. Is the street light visile? Write the eqution of the circle (use the center ( 0,0 ) ). + y = 30 + y = 900 The region lit y the light is the region inside the circle, so we wnt to include ll distnces less thn or equl to the rdius. + y 900 To check if the street light is visile, sustitute the point ( 10,5 ) into the inequlity. ( ) ( ) 10 + 5 900 100 + 65 900 true YES, the street light is visile. You Try : Grph the circle. ( ) ( y ) + 4 + 5 = 36 QOD: Descrie how to grph circle on the grphing clcultor. Pge 6 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Syllus Ojective: 10.1 The student will sketch the grph of conic section with centers either t or not t the origin. (ELLIPSES) Ellipse: the set of ll points P such tht the sum of the distnces etween P nd two distinct fied points is constnt. Foci: the two fied points tht crete n ellipse Vertices: the two points t which the line through the foci intersect the ellipse Mjor Ais: the line segment joining the two vertices Center of the Ellipse: the midpoint of the mjor is Co-Vertices: the two points t which the line perpendiculr to the mjor is t the center intersects the ellipse Minor Ais: the line segment joining the co-vertices Eqution of n Ellipse: ( h) ( y k) + = 1 : Horizontl Mjor Ais Center: ( hk, ) Vertices: ( h± k, ) Co-Vertices: ( hk, ± ) Foci: ( h± ck, ) ( h) ( y k) + = : Verticl Mjor Ais 1 Center: ( hk, ) Vertices: ( hk, ± ) Co-Vertices: ( h± k, ) Foci: ( hk, ± c) Note: The foci of the ellipse lie on the mjor is, c units from the center where c =. Ellipse centered t the origin with horizontl mjor is: Pge 7 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
E: Drw the ellipse given y 4 5y 100 + =. Identify the foci. Write the eqution in stndrd form. (Must e set equl to 1.) 4 5y 100 + = 100 100 100 Center: ( 0,0 ) y + = 1 5 4 Length of Mjor Ais (horizontl) = = 10 Length of Minor Ais (verticl) = = 4 c = 5 4 = 1 c = 1 Foci: ( 1,0 ),( 1,0) E: Write n eqution of the ellipse with center t ( 0,0 ), verte t ( 0, 3) ( 1, 0)., nd co-verte t 1 The verte is t ( 0, 3) ( h ) ( y k ), so the ellipse hs verticl mjor is. We will use + =. The center is ( 0,0 ) ( hk, ) =. y + = 1 The distnce etween the center nd the verte is 3, so = 3. The distnce etween the center nd the co-verte is 1, so = 1. y y + = 1 + = 1 1 3 9 ( 6 ) ( y ) E: Grph the ellipse foci. Center: ( 6, ) Length of Mjor Ais (verticl) = = 0 Length of Minor Ais (horizontl) = = 10 + = 1. Identify the center, vertices, co-vertices, nd 5 100 Vertices: ( 6, ± 10) = ( 6, 8) nd ( 6,1) Co-Vertices: ( 6 ± 5,) = ( 1, ) nd ( 11, ) Pge 8 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
c = = 100 5 = 75 c = 75 = 5 3 Foci: ( 6, ± 5 3) ( 6, 6.66) nd ( 6,10.66) E: The plnet Jupiter rnges from 460. million miles wy to 507.0 million miles wy from the sun. The center of the sun is focus of the orit. If Jupiter s orit is ellipticl, write n eqution for its orit in millions of miles. c = 460. + c = 507.0 = 967. = 483.6 + c = 507.0 483.6 + c = 507.0 c = 3.4 c = ( 3.4) ( 483.6) = = 3331.4 y + = 1 33868.96 3331.4 You Try : Write n eqution of the ellipse with the center t ( 0,0), verte t ( 4,0 ), nd focus t (,0 ). QOD: How cn you tell from the eqution of n ellipse whether the mjor is is horizontl or verticl? Pge 9 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Syllus Ojective: 10.1 The student will sketch the grph of conic section with centers either t or not t the origin. (HYPERBOLAS) Hyperol: the set of ll points P such tht the difference of the distnces from P to two fied points, clled the foci is constnt Vertices: the two points t which the line through the foci intersects the hyperol Trnsverse Ais: the line segment joining the vertices Center: the midpoint of the trnsverse is Eqution of Hyperol ( h) ( y k) = 1 : Horizontl Trnsverse Ais Center: ( hk, ) Vertices: ( h± k, ) ( y k) ( h) = : Verticl Trnsverse Ais 1 Center: ( hk, ) Vertices: ( hk, ± ) Note: The foci of the hyperol lie on the trnsverse is, c units from the center where c = +. Hyperols hve slnt symptotes. Drw the rectngle formed y the vertices nd the points ( hk, ± ) for horizontl nd ( h k, ) slnt symptotes of the hyperol. ± for verticl. The lines tht pss through the corners of this rectngle re the E: Drw the hyperol given y the eqution symptotes. Write the eqution in stndrd form (set equl to 1). 9 16y 144 =. Find the vertices, foci, nd 9 16y 144 = 144 144 144 y = 1 16 9 Center: ( 0,0 ) Vertices: ( 4,0 ) nd ( 4,0 ) = 16 = 4 Foci: c = + c = 16 + 9 = 5 ( ) ( ) c = 5 5,0 nd 5,0 Pge 10 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Asymptotes: 3 3 y = nd y = 4 4 Note: The hyperol itself is only the curve. (Dsh the symptotes.) E: Write n eqution of the hyperol with foci t ( 0, ) nd ( 0, ) nd vertices t ( 0, 1) nd ( 0,1 ). The foci nd vertices lie on the y-is, so the trnsverse is is verticl. We will use the eqution ( y k) ( h) = 1. Center ( hk, ) = ( 0,0). This is the midpoint of the vertices. y = 1 The foci re units from the center, so c =. The vertices re 1 unit from the center, so = 1. c = + = 1+ = 3 y = 1 y = 1 1 3 3 E: Grph the hyperol ( ) hk = = 1, = 4 Center (, ) ( 5, ) Vertices: ( 5± 1, ) = ( 4, ) n ( d6, ) Note: The hyperol itself is only the curve. (Dsh the symptotes.) ( y + ) 5 = 1. 16 Pge 11 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
E: The digrm shows the hyperolic cross section of lrge hourglss. Write n eqution tht models the curved sides. Center ( hk, ) = ( 0,0) = Vertices: (,0) nd (,0) The trnsverse is is horizontl, so we will use the eqution ( h) ( y k) = 1. y 1 = Sustitute in point on the hyperol ( 4,6 ) ( y, ) = nd solve for. 4 6 36 36 = 1 4 = 1 = 3 4 3 = 36 = 1 y = 1 4 1 : Grph the hyperol + 1 16y = 16. Identify the vertices nd foci. You Try ( ) y QOD: Wht re the symptotes of the hyperol = 1? Wht re the symptotes of the hyperol y =? 1 Pge 1 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
Syllus Ojective: 10. The student will clssify conic section given its eqution with its center either t or not t the origin. Generl Form of Second-Degree Eqution: A By Cy D Ey F + + + + + = 0 Discriminnt: B 4AC Clssifying Conic from Its Eqution A By Cy D Ey F + + + + + = 0 If If If If B B B B 4AC < 0 nd B = 0 nd A= C: CIRCLE 4AC < 0 nd B 0 or A C: ELLIPSE 4AC = 0: PARABOLA 4AC > 0: HYPERBOLA Grphing from the Generl Form To grph from generl form, complete the squre for oth vriles to write in stndrd form. E: Clssify the conic given y + y + 8y+ 6= 0. Then write in stndrd form nd grph. B ( )( ) 4AC = 0 4 1 = 8 Becuse A C, this is n ellipse. Complete the squre: Rewrite in stndrd form (set equl to 1): Center: ( 0, 4) ( y + 4) 10 + = 10 10 10 ( y + 4) + = 1 5 10 = 5.36 = 10 3.16 Vertices: ( 0, 4 ± 3.16) = ( 0, 0.838 ),( 0, 7.16) Co-Vertices: ( 0 ±.36, 4) = (.36, 4 ),(.36, 4) ( ) + y + 8y+ 16 + 6 = 0 + 16 ( y ) + + 4 = 10 Pge 13 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections
E: Clssify the conic given y grph. 4 y 16 4 y 4 0 =. Then write in stndrd form nd B ( )( ) 4AC = 0 4 4 1 = 16 > 0 This is hyperol. Complete the squre: Rewrite in stndrd form (set equl to 1): Center: (, ) ( ) ( y+ ) 4 16 = 16 16 16 ( ) ( y+ ) = 1 4 16 = 4, = Trnsverse Ais is horizontl. Vertices: ( ±, ) = ( 0, ),( 4, ) ( ) ( y y ) 4 4 + 4 + 4 + 4 4 = 0 + 16 4 ( ) ( y ) 4 + = 16 You Try: Clssify the conic given y y 0y+ 94 = 0. Then write in stndrd form nd grph. QOD: Descrie how to determine which conic section n eqution represents in generl form. Pge 14 of 14 McDougl Littell: 10.1 10.6 Alg II Notes Unit 10: Conic Sections