Sllus Ojective: 0. 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 M,. E: Find the midpoint of the line segment joining Let, 7,nd,, nd,, 7, nd,. is. Sustitute the vlues into the formul nd simplif. 7 9 M,,,3 Conic Sections: curves tht re formed the intersection of plne nd doulenpped cone Prol: the conic formed connecting ll points equidistnt from point (the focus) k 4 p h or nd line (the directri) with the eqution of the form h 4 p k Verte: the point, hk,, tht lies on the is of smmetr hlfw etween the focus nd directri Ais of Smmetr: the line of smmetr of prol tht psses through the verte. For k 4 p h, the is of smmetr is horizontl, nd hs the eqution h. For h 4 p k, the is of smmetr is verticl, nd hs the eqution k. p : the distnce the focus nd directri re from the verte Focus: point on the is of smmetr of prol equidistnt from the verte s the directri Directri: line perpendiculr to the is of smmetr equidistnt from the verte s the focus Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Prol 4 p, p 0 Prol 4 p, p 0 Focus 0, p Directri p Verte 0,0 Directri p Verte 0,0 Focus 0, p Prol 4 p, p 0 Prol 4 p, p 0 Directri p Directri p Verte 0,0 Focus p,0 Focus p,0 Verte 0,0 E: Sketch the grph of the prol 3. Verte: 0,0 Becuse is squred, the is of smmetr is horizontl: 0 Find p: 3 4p 3 3 p 4 Becuse p is negtive, the prol will open to the left. Directri: 3 Focus: 4 3,0 4 - - Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Stndrd Form of the Eqution of Prol: Horizontl Ais (opens left or right): k 4 p h Verte: hk, Ais of Smmetr: k Verticl Ais (opens up or down): h 4 p k Verte: hk, Ais of Smmetr: h Focus: h p, k Focus: hk, p E: Sketch the grph of the prol 7 8 3 directri. Directri: h p Directri: k p. Identif the verte, focus, nd Becuse it is in the form h 4 p k verticl is., the prol hs 0 Find p: 4 p 8 p Verte: 7,3 p is positive, so the prol opens up. -0 - -0-0 Focus: p units up from the verte 7,3 7, - Directri: horizontl line p units down from the verte 3 E: Write n eqution of the prol whose verte is t 3, nd whose focus is t 4,. -0 Begin with sketch. The prol opens towrd the focus, so it opens right. Find p: The distnce from the focus to the verte is. The prol opens right, so p =. Becuse the prol hs horizontl is, we will use the eqution k 4 p h The verte 3, hk,, nd p =. 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 identif the focus. Pge 3 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Becuse the prol hs verticl is, we will use the eqution h 4 p k 0,0 hk,, so we now hve 4 p. The prol psses through the point 8,, p p p p 4 8 4 648 8. Use this to find p.. The verte is The eqution is 48 3. The focus is p units up from the verte: 0,8 You Tr: Grph the eqution 6 4 8. Identif the verte, focus, nd directri. QOD: Prols cn e found mn 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? Smple CCSD Common Em Prctice Question(s): Which eqution represents the grph elow? A. B. C. D. Pge 4 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Sllus Ojective: 0. 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 nd, is d,. Use the Pthgoren Theorem to show this on the coordinte plne forming right tringle with the points,,,, nd,. E: Find the distnce etween the points, 4 Let,, nd,, 4 nd,.. Sustitute the vlues into the formul nd simplif. d 4 7 3 499 8 Circle: the conic formed connecting ll points equidistnt from point with the eqution of h k r the form Center of Circle: the point, hk,, tht is equidistnt from ll points on the circle Rdius: the distnce r etween the center nd n point on the circle E: Show tht the eqution of circle with center hk, nd rdius r is h k r. Let one point on the circle e, nd the center e hk,. The rdius, r, is the distnce etween the center nd point on the circle. r h h d r h k. Squring oth sides, we hve E: Drw the circle given the eqution 4. Write in stndrd form. 4 Identif the center nd the rdius. Center: 0,0 Rdius: r 4 r - Sketch the grph plotting the center then plotting points on the circle units ove, elow, nd to the left nd right of the center. - Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
E: Write n eqution of the circle with center, 4 nd rdius 3. Center:, 4 hk, Rdius: r 3 Eqution: h k r 4 3 4 9 E: Write n eqution of the circle shown. The center of the circle is 0,0 hk, Eqution: r. To find r, we will use the point given on the circle,. r r r 6 - - The eqution of the circle is 6. E: A street light cn e seen on the ground within 30 d of its center. You re driving nd re 0 d est nd d south of the light. Write n inequlit to descrie the region on the ground tht is lit the light. Is the street light visile? Write the eqution of the circle (use the center 0,0 ). 30 900 The region lit the light is the region inside the circle, so we wnt to include ll distnces less thn or equl to the rdius. 900 To check if the street light is visile, sustitute the point 0, into the inequlit. 0 900 00 6 900 true YES, the street light is visile. You Tr: Grph the circle. 4 36 QOD: Descrie how to grph circle on the grphing clcultor. Pge 6 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Smple CCSD Common Em Prctice Question(s): Wht grph represents? A. B. C. D. Pge 7 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Sllus Ojective: 0. 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 k Center: : Horizontl Mjor Ais; hk, Vertices: h, k Co-Vertices: hk, Foci: h c, k h k Center: : Verticl Mjor Ais; 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 8 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
E: Drw the ellipse given 4 00. Identif the foci. Write the eqution in stndrd form. (Must e set equl to.) 4 00 00 00 00 4 Center: 0,0 4 Length of Mjor Ais (horizontl) = = 0 Length of Minor Ais (verticl) = = 4 - c 4 c Foci:,0,,0 - E: Write n eqution of the ellipse with center t, 0. The verte is t 0, 3 0,0, verte t 0, 3, nd co-verte t h k, so the ellipse hs verticl mjor is. We will use. The center is 0,0 hk,. The distnce etween the center nd the verte is 3, so 3. The distnce etween the center nd the co-verte is, so. 3 9 6 E: Grph the ellipse foci. Center: 6, Length of Mjor Ais (verticl) = = 0 Length of Minor Ais (horizontl) = = 0. Identif the center, vertices, co-vertices, nd 00 Pge 9 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Vertices: 6, 06, 8 nd 6, Co-Vertices: 6,, nd, c 00 7 c 7 3 Foci: 6, 36, 6.66 nd 6,0.66 0-0 - 0 - -0 E: The plnet Jupiter rnges from 460. million miles w to 07.0 million miles w 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 07.0 967. 483.6 c 07.0 483.6 c 07.0 c 3.4 c 3.4 483.6 333.4 33868.96 333.4 You Tr: Write n eqution of the ellipse with the center t 0,0, verte t 4,0, nd focus t QOD: How cn ou tell from the eqution of n ellipse whether the mjor is is horizontl or verticl?,0. Pge 0 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Smple CCSD Common Em Prctice Question(s): Which is the eqution for the grph elow? E. 3 9 6 F. 3 3 4 G. 3 9 6 H. 3 3 4 Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Sllus Ojective: 0. The student will sketch the grph of conic section with centers either t or not t the origin. (HYPERBOLAS) Hperol: 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 hperol Trnsverse Ais: the line segment joining the vertices Center: the midpoint of the trnsverse is Eqution of Hperol h k Center: : Horizontl Trnsverse Ais hk, Vertices: h, k k h Center: : Verticl Trnsverse Ais hk, Vertices: hk, Note: The foci of the hperol lie on the trnsverse is, c units from the center where c. Hperols hve slnt smptotes. Drw the rectngle formed the vertices nd the points hk, for horizontl nd h, k for verticl. The lines tht pss through the corners of this rectngle re the slnt smptotes of the hperol. E: Drw the hperol given the eqution smptotes. Write the eqution in stndrd form (set equl to ). 9 6 44. Find the vertices, foci, nd 9 6 44 44 44 44 6 9 Center: 0,0 Vertices: 4,0 nd 4,0 6 4 Foci: c c 6 9 c,0 nd,0 Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Asmptotes: 3 3 nd 4 4 0 Note: The hperol itself is onl the curve. (Dsh the smptotes.) -0-0 - -0 E: Write n eqution of the hperol with foci t 0, nd 0,. nd 0, nd vertices t 0, The foci nd vertices lie on the -is, so the trnsverse is is verticl. We will use the eqution k h. Center hk, 0,0. This is the midpoint of the vertices. The foci re units from the center, so c. The vertices re unit from the center, so. c 3 3 3 E: Grph the hperol. 6 hk, 4 Center,, Vertices:, 4, nd6, Note: The hperol itself is onl the curve. (Dsh the smptotes.) 0-0 - 0 - Pge 3 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections -0
E: The digrm shows the hperolic 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 k. Sustitute in point on the hperol 4,6, nd solve for. 4 6 36 36 4 3 4 3 36 4 You Tr: Grph the hperol 6 6. Identif the vertices nd foci. QOD: Wht re the smptotes of the hperol?? Wht re the smptotes of the hperol Pge 4 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Smple CCSD Common Em Prctice Question(s):. Which grph est represents A. 4 4? B. C. D. Pge of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
Sllus Ojective: 0. The student will clssif conic section given its eqution with its center either t or not t the origin. Generl Form of Second-Degree Eqution: A B C D E F 0 Discriminnt: B 4AC Clssifing Conic from Its Eqution A B C D E 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: Clssif the conic given 8 6 0. Then write in stndrd form nd grph. B 4AC 0 4 8 Becuse A C, this is n ellipse. Complete the squre: 8 6 6 0 6 4 0 Rewrite in stndrd form (set equl to ): 4 0 0 0 0 4 0 Center: 0, 4.36 0 3.6 Vertices: 0, 4 3.60, 0.838, 0, 7.6 0-0 - 0 - -0 Co-Vertices: 0.36, 4.36, 4,.36, 4 Pge 6 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
E: Clssif the conic given grph. 4 6 4 4 0. Then write in stndrd form nd B 4AC 0 4 4 6 0 This is hperol. Complete the squre: 4 4 4 4 4 40 6 4 4 6 Rewrite in stndrd form (set equl to ): 4 6 6 6 6 4 6 Center:, 4, Trnsverse Ais is horizontl. Vertices:, 0,, 4, 0-0 - 0 - -0 You Tr: Clssif the conic given 0 94 0. Then write in stndrd form nd grph. QOD: Descrie how to determine which conic section n eqution represents in generl form.. Wht tpe of conic section hs the eqution 4 7 A. circle B. ellipse C. hperol D. prol? 3 8 Pge 7 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections
. Which is the eqution of n ellipse? A. B. C. 4 00 6 36 9 6 0 D. 0 0 Pge 8 of 8 McDougl Littell: 0. 0.6 Alg II Notes Unit 0: Conic Sections