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Question Number 144534 by imjagoll last updated on 26/Jun/21
Findtheshortestdistancefromtheorigintothehyperbolax2+8xy+7y2=225,z=0
Answered by liberty last updated on 26/Jun/21
wemustfindtheminimumvalueofx2+y2(thesquareofthedistancefromtheorigintoanypointinthexyplane)subjecttoconstraintx2+8xy+7y2=225byLangrangemultiplier∅=x2+8xy+7y2+λ(x2+y2)∅x=2x+8y+2λx=0or(λ+1)x+4y=0ϕy=8x+14y+2λy=0or4x+(λ+7)y=0|λ+144λ+7|=0→{λ=−9λ=1whenλ=−9wegety=2xandsubstitutioninx2+8xy+7y2=225yieldsx2=5,y2=20sotheshortestdistanceis5+20=5whenλ=1givex=−2yandsubstitutioninx2+8xy+7y2=225yields−5y2=225forwhichnorealsolutionexists.
Answered by mr W last updated on 26/Jun/21
r=distancefrom(0,0)to(x,y)x=rcosθy=rsinθr2cos2θ+8r2cosθsinθ+7r2sin2θ=225r2(4+4sin2θ−3cos2θ)=225r2=2254+5sin(2θ−tan−134)⩾2254+5⇒rmin=2259=5
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