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a)Supposeourmirrorisshapedliketheparabolay=kx2wherekisanypositive

constant.Findthecoordinatesofitsfocusandtheequationofitsdirectrixinterms

ofk.

Assumingthefocusanddirectrixlieonafocalchordoflength4p,wherepisthe

distancefromthevertextoboththefocusanddirectrix,wecanutilizethefollowing

standardformofaparabolicequationwithvertex(h,k):

(x-h)2=4p(y-k)

Supposingourparabolaisdefinedbyequationy=kx2,thevertexliesattheorigin,at

point(0,0).Thestandardequationcanthenbealgebraicallymanipulatedasfollows:

x2=4py

y=

x2

4p

1

4p=k

1

4k=p

Thus,thefocusliesatpoint(0,

1

4k),andthefocalchordendsonthedirectrixat

point(0,

-1

4k).Theequationofthedirectrix,therefore,isy=

-1

4k.

b)Findtheequationofthelinetangenttotheparabolaatapoint(x0,y0)onthe

parabola.Then,findthey-interceptofthetangentline.

Giventheequationoftheparabolay=kx2,itfollows,bywayofthepowerand

productrules,thatderivativey'=2xk.

Bysubstitutingvariablex0forx,weseethattheslopeofthetangentlineatpoint(x0,

y0)is2x0k.

Usingthepoint-slopeformofaline,itfollowsthat

(y-y0)=2x0k(x-x0)

y=2kx0(x-x0)+y0

andthustheequationofthetangentat(x0,y0)is

y=2kx0x-2kx02+y0

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Substitutingkx02fory0,sincey=kx2,

y=2kx0x-2y0+y0

Theequationofthetangentatpoint(x0,y0)isthus

y=2kx0x-y0.

Substitutingvalue0forxinordertofindthey-intercept:

y=2kx0(0)-y0

y=-y0

Therefore,they-interceptofthetangentlineat(x0,y0)isatpoint(0,-y0).

c)Considerthetriangleformedbyapoint(x0,y0)ontheparabola,they-interceptof

thetangentlineatthispoint,andthefocus.Drawapictureofthistriangle.Provethe

triangleisisosceles.

Thisparticulartrianglewouldbefor

thespecificcaseofy=x2.The

triangleformedbythepoint(1,1)

ontheparabola,they-interceptof

thetangentlineatpoint(0,-1),and

thefocus(0,1/4)isshowninred.

Wewill,however,provethetriangle

isisoscelesforthegeneralcase.

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First,wewilldefineeachofthesepointsasaletter(A,B,orC)forthesakeof

convenience.

PointA=Focus=(0,

1

4k)

PointB=Pointonparabola=(x0,y0)

PointC=Y-interceptoftangent=(0,-y0)

Inorderforthetriangletobeisosceles,twoormoreofthelinesegmentsintriangle

ABCmustbeequal.Wewillexaminelinesegments

ABand

AC.

Inordertoprovethat

ABand

ACareofequallength,wemustusethedistance

formula:

d=

(x2 x1)2

+ (y2 y1)2

First,thedistancebetweenAandB:

d

AB = (x0 0)2+ (y0

1

4k)2

Substitutingx0for

y0

k(fromequationy=x2),

d

AB =y

0

k+ y0

2y

0

2k+

1

(4k)2

d

AB = y02+

y0

2k+

1

(4k)2

d

AB = y0+

1

4k

2

d

AB = y0+

1

4k

NowforthedistancebetweenAandC:

d

AC= (0 0)2

+ (y0

1

4k)2

d

AC= (y0 1

4k)2

d

AC= y0+

1

4kandtherefore,thetwodistancesareequal,andthetriangleis

isosceles!

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d)Supposeanincominglightraystrikesacurveatapoint(x0,y0).Ifthelightray

makesananglewithrespecttothetangentline,thenitisreflectedatanequal

angletothetangentline.Thisresultfromphysicsisknownbythephrase"theangle

ofincidenceequalstheangleofreflection."Usingthisfact,arguethatincominglight

raysparalleltotheaxisoftheparabolaareallreflectedtothefocus,independentof

thepointofincidence.Thus,aparabolicmirrorfocusesalllightraysparalleltotheaxistoapoint.

Werefertothefollowingdiagram,againofaspecificcasey=x^2,inorderto

illustratethisphenomenon.

Inorderforthelightray(depictedin

green)tobereflectedtothefocus,

angles1and2mustbeequaltoeach

other.

Weknowthatthelightrayisparallelto

theparabola'saxisofsymmetry,orthe

y-axis.

Uponinspection,wecanseethat

becauseangles1and3are

correspondinganglesonatransversal,

theymustbecongruent.

Inaddition,weknowthatbecausethe

triangle(inred)isanisoscelestriangle,

itsbaseangles,2and3,mustbe

congruent.

Therefore,bythetransitiveproperty,becauseangles1and3arecongruent,and

angles2and3arecongruent,angles1and2arecongruent.Thus,theangleof

incidenceandtheangleofreflectionareequal,andalllightraysparalleltothe

parabolicaxiswillbereflectedtothefocus.

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e)Thepathfollowedbyarayoflightfromthestartothefocusofthemirrorhas

anotherspecialproperty.Drawachordoftheparabolathatisabovethefocusand

paralleltothedirectrix.Considerarayoflightparalleltotheaxisasitcrossesthe

chord,hitstheparabolaandisreflectedtothefocus.Letd1bethedistancefromthe

chordtothepointofincidence(x0,y0)ontheparabolaandletd2bethedistance

from(x0,y0)tothefocus.Showthatthesumofthedistancesd1+d2isconstant,

independentofthisparticularpointofincidence.

Again,toillustratethis

phenomenon,welooktothis

diagramofspecificcasey=x^2.

Thechordisdrawninmaroon,

andthelightrayisagaingreen.

WehavealreadyproveninpartC

thatthedistancefromthepointofincidence(x0,y0)tothefocusis

equalto

y0+

1

4k.Fromthiswe

cansaythatd2is

y0+

1

4k.

Thisalsoappliestothepointat

whichthechord,whichwillbe

definedasy=cforanyintegerc

largerthan

1

4k

,intersectsthe

parabola.Inthiscase,the

distancewouldbedefinedbyc+

1

4k.

Fromthiswecanthengatherthatthedistancefromthechordtothepointof

incidenced1isequaltoc+

1

4k-(

y0+

1

4k),orc-y0.

Addingthesetwotogether(d1+d2),wegetthegeneralexpression:

c-y0+y0+

1

4k

Whichequals...

c+

1

4k,whichdoesnotinvolveeithercomponentof(x0,y0).Thus,thesumof

thedistancesremainsconstantregardlessofpointofincidence.

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