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Question Number 82404 by jagoll last updated on 21/Feb/20

Commented by jagoll last updated on 21/Feb/20

$$maybesomeonehelpme$$

Commented by mr W last updated on 21/Feb/20

$$ithinktheminimumiswhenAis$$$$attheoriginandD=(-1,1),B=(1,1).$$$$theminimumareais2.$$

Commented by jagoll last updated on 21/Feb/20

$$pointCsir?$$

Commented by jagoll last updated on 21/Feb/20

$$doesmeanthepointCmust$$$$lieony-axis$$

Commented by mr W last updated on 21/Feb/20

$$myunderstandingisthatthreevertices$$$$shouldlieontheparabolaandthe$$$$fourthvextexnot.whetherthefourth$$$$vertexisoutsideorinsideofthe$$$$parabola,it{doesn}^{\prime }tmatter.butthis$$$$isyourquestion.whatdoyoumean?$$$$shouldthefourthvertexonlyoutside$$$$oftheparabolaorit{doesn}^{\prime }tmatter?$$

Commented by jagoll last updated on 21/Feb/20

$$ifaccordingtomyunderstanding$$$$thefourthpointisoutsideparabola.$$$$dishaccordingtothepicture$$$$shownintheproblem.sir$$

Commented by mr W last updated on 21/Feb/20

$$solutionforthecasethatthefourth$$$$vertexliesoutsidetheparabola$$$$seebelow.$$

Answered by mr W last updated on 21/Feb/20

$$A(-a,a^2)witha>0$$$$D(d,d^2)withd>0$$$$B(b,b^2)with{b}^2<d^2$$$$sayeqn.ofADisy=a^2+m(x+a)$$$$theneqn.ofADisy=a^2-\frac{1}{m}(x+a)$$$$d^2=a^2+m(d+a)$$$$d^2-md-a^2-ma=0$$$$\Rightarrow d=a+m$$$$similarly$$$$\Rightarrow b=a-\frac{1}{m}$$$${AD}^2={(d+a)}^2+{(d^2-a^2)}^2$$$${AD}^2=(1+m^2){(2a+m)}^2$$$$similarly$$$${AB}^2=(1+\frac{1}{m^2}){(2a-\frac{1}{m})}^2$$$$AB=AD$$$$\Rightarrow (1+\frac{1}{m^2}){(2a-\frac{1}{m})}^2=(1+m^2){(2a+m)}^2$$$$\Rightarrow {(2am-1)}^2=m^4{(2a+m)}^2$$$$sidelengthofsquarel=(2a+m)\sqrt{1+m^2}$$$$$$$$weget\left(throughgraphicmethod\right):$$$$atm\approx 0.3821anda\approx 2.2358$$$$l_{min}\approx 5.1961$$

Commented by mr W last updated on 21/Feb/20

Commented by mr W last updated on 21/Feb/20

Commented by mr W last updated on 21/Feb/20

$$analytical\left(exact\right)solution:$$$$from{(2am-1)}^2=m^4{(2a+m)}^2$$$$\Rightarrow m^2(2a+m)=\pm (2am-1)$$$$\Rightarrow 2m(m\pm 1)a=\pm 1-m^3$$$$\Rightarrow a=\frac{1-m^3}{2m(m+1)}(thisisforthecase$$$$thatthefourthvertexliesinside)$$$$\Rightarrow a=\frac{-1-m^3}{2m(m-1)}\left(thisisforourcase\right)$$$$l=(2a+m)\sqrt{1+m^2}$$$$\Rightarrow l=\frac{{(1+m^2)}^{\frac{3}{2}}}{m(1-m)}$$$$\frac{dl}{dm}=\frac{3{(1+m^2)}^{\frac{1}{2}}}{(1-m)}-\frac{{(1+m^2)}^{\frac{3}{2}}(1-2m)}{m^2{(1-m)}^2}=0$$$$\frac{(1+m^2)(1-2m)}{m^2(1-m)}=3$$$$\Rightarrow m^3-2m^2-2m+1=0$$$$\Rightarrow (m+1)(m^2-3m+1)=0$$$$\Rightarrow m=-1$$$$\Rightarrow m=\frac{3+\sqrt{5}}{2}\approx 0.3820$$$$\Rightarrow m=\frac{3-\sqrt{5}}{2}$$$$\Rightarrow a=\frac{1+m^3}{2m(1-m)}=2.2361$$$$\Rightarrow l_{min}=\frac{{(1+m^2)}^{\frac{3}{2}}}{m(1-m)}=5.1962$$

Commented by mr W last updated on 21/Feb/20

Commented by mr W last updated on 21/Feb/20

Commented by mr W last updated on 21/Feb/20

Commented by jagoll last updated on 21/Feb/20

$$yessirthankyou$$

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