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Question-106977




Question Number 106977 by mr W last updated on 08/Aug/20
Answered by mr W last updated on 08/Aug/20
A(a,0) and B(b,0)  P lies on y=x^2   find the minimum of perimeter of  ΔPAB.
A(a,0)andB(b,0)Pliesony=x2findtheminimumofperimeterofΔPAB.
Commented by PRITHWISH SEN 2 last updated on 08/Aug/20
(b−a)+ (√(a^2 +b^2 ))
(ba)+a2+b2
Answered by mr W last updated on 08/Aug/20
METHOD I  say P(p,p^2 )  tangent at P is  y=p^2 +2p(x−p)  ⇒2px−y−p^2 =0  image of B in tangent is C(x_C ,y_C )  x_C =b−((4p(2pb−p^2 ))/(4p^2 +1))  y_C =((2(2pb−p^2 ))/(4p^2 +1))  A,P,C are collinear,  ((y_C −p^2 )/p^2 )=((x_C −p)/(p−a))  ((2(2pb−p^2 ))/(4p^2 +1))−p^2 =(p^2 /(p−a))[b−((4p(2pb−p^2 ))/(4p^2 +1))−p]  ⇒p^4 −(a+b)p^3 +(1/2)p^2 −(3/4)(a+b)p+ab=0  ⇒p=....
METHODIsayP(p,p2)tangentatPisy=p2+2p(xp)2pxyp2=0imageofBintangentisC(xC,yC)xC=b4p(2pbp2)4p2+1yC=2(2pbp2)4p2+1A,P,Carecollinear,yCp2p2=xCppa2(2pbp2)4p2+1p2=p2pa[b4p(2pbp2)4p2+1p]p4(a+b)p3+12p234(a+b)p+ab=0p=.
Commented by PRITHWISH SEN 2 last updated on 08/Aug/20
Sir, whether this image concept applicable in this  case ? Because the point A and B are fixed. What  do you think sir ?
Sir,whetherthisimageconceptapplicableinthiscase?BecausethepointAandBarefixed.Whatdoyouthinksir?
Commented by mr W last updated on 08/Aug/20
Commented by mr W last updated on 08/Aug/20
the shortest path A−P−B is that one  which a light ray from A to B follows.  that means the tangent at point P  should act like a mirror.
theshortestpathAPBisthatonewhichalightrayfromAtoBfollows.thatmeansthetangentatpointPshouldactlikeamirror.
Commented by PRITHWISH SEN 2 last updated on 08/Aug/20
yes sir but the oriantation of the tangent is   changing at every point . A−P−B^′  can′t be the  straight line always. Can PB′ (B′ the image of  B) be colinear with AB′ ?
yessirbuttheoriantationofthetangentischangingateverypoint.APBcantbethestraightlinealways.CanPB(BtheimageofB)becolinearwithAB?
Commented by mr W last updated on 08/Aug/20
only that point P is the right one,  when the image of B, i.e. point C is  collinear with A and P. there is only  one such a point P.
onlythatpointPistherightone,whentheimageofB,i.e.pointCiscollinearwithAandP.thereisonlyonesuchapointP.
Commented by PRITHWISH SEN 2 last updated on 08/Aug/20
OK sir now it is clear.
OKsirnowitisclear.
Commented by mr W last updated on 08/Aug/20
image the parabolla is curved mirror  and you stand at point A. now you   direct at a point at the mirror with  a laser pointer. the laser beam will be  reflected by the mirror. only when  the reflected laser beam meets the  point B, you have got the shortest  path.
imagetheparabollaiscurvedmirrorandyoustandatpointA.nowyoudirectatapointatthemirrorwithalaserpointer.thelaserbeamwillbereflectedbythemirror.onlywhenthereflectedlaserbeammeetsthepointB,youhavegottheshortestpath.
Commented by PRITHWISH SEN 2 last updated on 08/Aug/20
thank you sir
thankyousir
Answered by mr W last updated on 08/Aug/20
METHOD 2  say P(p,p^2 )  L=AP+BP=(√((p−a)^2 +p^4 ))+(√((b−p)^2 +p^4 ))  (dL/dp)=((2(p−a)+4p^3 )/(2(√((p−a)^2 +p^4 ))))+((2(p−b)+4p^3 )/(2(√((b−p)^2 +p^4 ))))=0  (((p−a)+2p^3 )/( (√((p−a)^2 +p^4 ))))=(((b−p)−2p^3 )/( (√((b−p)^2 +p^4 ))))  (((p−a)^2 +4(p−a)p^3 +4p^6 )/((p−a)^2 +p^4 ))=(((b−p)^2 −4(b−p)p^3 +4p^6 )/((b−p)^2 +p^4 ))  ⇒p^4 −(a+b)p^3 +(1/2)p^2 −(3/4)(a+b)p+ab=0  we get the same result as above.
METHOD2sayP(p,p2)L=AP+BP=(pa)2+p4+(bp)2+p4dLdp=2(pa)+4p32(pa)2+p4+2(pb)+4p32(bp)2+p4=0(pa)+2p3(pa)2+p4=(bp)2p3(bp)2+p4(pa)2+4(pa)p3+4p6(pa)2+p4=(bp)24(bp)p3+4p6(bp)2+p4p4(a+b)p3+12p234(a+b)p+ab=0wegetthesameresultasabove.

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