大學(xué)課件：parabolas講義整理

ColleenBeaudoinForFCIT大學(xué)課件：parabolasGeometricdefinition:aconehasaplane intersectingit,crossingits verticalaxisAlgebraicdefinition:Allpointsthatareequidistantfromagivenline(thedirectrix)andafixedpointnotonthedirectrix(thefocus)Geometricdefinition:aconeyAnypointontheparabolaisequidistanttothefocusandthedirectrix.Example:PointA:d1=d1PointB:d2=d2AByAnypointontheparabolaisFocusVertexDirectrixAxisofSymmetryxyFocusxyOnevariableissquaredandoneisnot.(Howdoesthisdifferfromlinearequations?)Therearemanywaystheequationofaparabolacanbewritten.Wewillgetthequadraticpart(variablethatissquared)ontheleftoftheequalsignandthelinearpart(variableistothefirstpower)ontherightoftheequalsign.Equation: (x-h)2=c(y–k)

OR

(y-k)2=c(x–h)Onevariableissquaredandon(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:1.Putinstandardform(above)–squaredtermonleft2.Decidewhichwaytheparabolaopens. Lookattherightside.Ify:+c→opensup Ify:-c→opensdown Ifx:+c→opensright Ifx:-c→opensleft(x-h)2=c(y–k)OR(y-k)(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:3.Plotthevertex(h,k)Notewhathappenstothesigns.4.Plotthefocus:move│?c│fromthevertexinthedirectionthattheparabolaopens.Markwithanf.5.Drawthedirectrix:│?c│fromthevertexintheoppositedirectionofthefocus(Rememberthatthedirectrixisaline.)(x-h)2=c(y–k)OR(y-k)(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:6.Plottheendpointsofthelatusrectum/focalchord(widthatthefocus).Thewidthisthe│c│atthefocus.7.Sketchtheparabolabygoingthroughthevertexandtheendpointsofthelatusrectum.(Besuretoextendthecurveandputarrows.)8.Identifytheaxisofsymmetry.(Thelinethatgoesthroughthevertexdividingtheparabolainhalf.)(x-h)2=c(y–k)OR(y-k)Exp.1:Graph(x-5)2=12(y–6)Tograph:1.Putinstandardform–squaredtermonleft Done2.Decidewhichwaytheparabolaopens. Lookattherightside.Ify:+c→opensup Ify:-c→opensdown Ifx:+c→opensright Ifx:-c→opensleft Upbecauseyisontherightand12ispositiveExp.1:Graph(x-5)2=12(yExp.1:Graph(x-5)2=12(y–6)Tograph:3.Plotthevertex(h,k)Notewhathappenstothesigns. (5,6)4.Plotthefocus:move│?c│fromthevertexinthedirectionthattheparabolaopens.Markwithanf.

(5,9):foundbymovingup3fromthevertex5.Drawthedirectrix:│?c│fromthevertexintheoppositedirectionofthefocus(Rememberthatthedirectrixisaline.) y=3:foundbymovingdown3fromthevertexExp.1:Graph(x-5)2=12(yExp.1:Graph(x-5)2=12(y–6)Tograph:6.Plottheendpointsofthelatusrectum/focalchord(widthatthefocus).Thewidthisthe│c│atthefocus. L.R.=12withendpointsat(-1,9)&(11,9)7.Sketchtheparabolabygoingthroughthevertexandtheendpointsofthelatusrectum.(Besuretoextendthecurveandputarrows.)8.Identifytheaxisofsymmetry.(Thelinethatgoesthroughthevertexdividingtheparabolainhalf.)

x=5Exp.1:Graph(x-5)2=12(yVertex:(5,6)Focus:(5,9)Directrix:y=3L.R.:12Axis:x=5fVertex:(5,6)fExp.2:Graph(y+3)2=-4(x–2)Tograph:1.Putinstandardform Done2.Decidewhichwaytheparabolaopens.

Leftbecausexisontherightand4isnegative3.Plotthevertex(h,k) (2,-3)4.Plotthefocus: (1,-3):foundbymovingleft1fromthevertex5.Drawthedirectrix: x=3:foundbymovingright1fromthevertexExp.2:Graph(y+3)2=-4(xExp.2:Graph(y+3)2=-4(x–2)Tograph:6.Plottheendpointsofthelatusrectum L.R.=4withendpointsat(1,-1)&(1,-5)7.Sketchtheparabola8.Identifytheaxisofsymmetry. y=-3

Exp.2:Graph(y+3)2=-4(xVertex:(2,-3)Focus:(1,-3)Directrix:x=3L.R.:4Axis:y=-3fVertex:(2,-3)fWhat’sthefirststep?

Putinstandardform. y2-4y+1=x y2-4y+4=x-1+4Completethesquare.

(y–2)2=x+3

(y–2)2=1(x+3)Nowyoutrygraphingtheparabolaandlabelingalltheparts.What’sthefirststep?Giventhefollowinginformation,writetheequationoftheparabola.Vertexis(0,0)andFocusisat(0,2)GiventhefollowinginformatioHowcanyoutellthegraphofanequationwillbeaparabola?What’sthestandardformofaparabola?Whatarethestepsforgraphingaparabola?Whatarecommonerrorspeoplemakewhengraphingparabolas?HowcanyoutellthegraphofConicSectionStandardFormofEquationParabolaCircleEllipseHyperbolaConicSectionStandardFormof19ColleenBeaudoinForFCIT大學(xué)課件：parabolasGeometricdefinition:aconehasaplane intersectingit,crossingits verticalaxisAlgebraicdefinition:Allpointsthatareequidistantfromagivenline(thedirectrix)andafixedpointnotonthedirectrix(thefocus)Geometricdefinition:aconeyAnypointontheparabolaisequidistanttothefocusandthedirectrix.Example:PointA:d1=d1PointB:d2=d2AByAnypointontheparabolaisFocusVertexDirectrixAxisofSymmetryxyFocusxyOnevariableissquaredandoneisnot.(Howdoesthisdifferfromlinearequations?)Therearemanywaystheequationofaparabolacanbewritten.Wewillgetthequadraticpart(variablethatissquared)ontheleftoftheequalsignandthelinearpart(variableistothefirstpower)ontherightoftheequalsign.Equation: (x-h)2=c(y–k)

OR

(y-k)2=c(x–h)Onevariableissquaredandon(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:1.Putinstandardform(above)–squaredtermonleft2.Decidewhichwaytheparabolaopens. Lookattherightside.Ify:+c→opensup Ify:-c→opensdown Ifx:+c→opensright Ifx:-c→opensleft(x-h)2=c(y–k)OR(y-k)(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:3.Plotthevertex(h,k)Notewhathappenstothesigns.4.Plotthefocus:move│?c│fromthevertexinthedirectionthattheparabolaopens.Markwithanf.5.Drawthedirectrix:│?c│fromthevertexintheoppositedirectionofthefocus(Rememberthatthedirectrixisaline.)(x-h)2=c(y–k)OR(y-k)(x-h)2=c(y–k)OR(y-k)2=c(x–h)

wherethevertexisat(h,k)and|c|isthewidthatthefocusTograph:6.Plottheendpointsofthelatusrectum/focalchord(widthatthefocus).Thewidthisthe│c│atthefocus.7.Sketchtheparabolabygoingthroughthevertexandtheendpointsofthelatusrectum.(Besuretoextendthecurveandputarrows.)8.Identifytheaxisofsymmetry.(Thelinethatgoesthroughthevertexdividingtheparabolainhalf.)(x-h)2=c(y–k)OR(y-k)Exp.1:Graph(x-5)2=12(y–6)Tograph:1.Putinstandardform–squaredtermonleft Done2.Decidewhichwaytheparabolaopens. Lookattherightside.Ify:+c→opensup Ify:-c→opensdown Ifx:+c→opensright Ifx:-c→opensleft Upbecauseyisontherightand12ispositiveExp.1:Graph(x-5)2=12(yExp.1:Graph(x-5)2=12(y–6)Tograph:3.Plotthevertex(h,k)Notewhathappenstothesigns. (5,6)4.Plotthefocus:move│?c│fromthevertexinthedirectionthattheparabolaopens.Markwithanf.

(5,9):foundbymovingup3fromthevertex5.Drawthedirectrix:│?c│fromthevertexintheoppositedirectionofthefocus(Rememberthatthedirectrixisaline.) y=3:foundbymovingdown3fromthevertexExp.1:Graph(x-5)2=12(yExp.1:Graph(x-5)2=12(y–6)Tograph:6.Plottheendpointsofthelatusrectum/focalchord(widthatthefocus).Thewidthisthe│c│atthefocus. L.R.=12withendpointsat(-1,9)&(11,9)7.Sketchtheparabolabygoingthroughthevertexandtheendpointsofthelatusrectum.(Besuretoextendthecurveandputarrows.)8.Identifytheaxisofsymmetry.(Thelinethatgoesthroughthevertexdividingtheparabolainhalf.)

x=5Exp.1:Graph(x-5)2=12(yVertex:(5,6)Focus:(5,9)Directrix:y=3L.R.:12Axis:x=5fVertex:(5,6)fExp.2:Graph(y+3)2=-4(x–2)Tograph:1.Putinstandardform Done2.Decidewhichwaytheparabolaopens.

Leftbecausexisontherightand4isnegative3.Plotthevertex(h,k) (2,-3)4.Plotthefocus: (1,-3):foundbymovingleft1fromthevertex5.Drawthedirectrix: x=3:foundbymovingright1fromthevertexExp.2:Graph(y+3)2=-4(xExp.2:Graph(y+3)2=