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Question-82404

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Question Number 82404 by jagoll last updated on 21/Feb/20
Commented by jagoll last updated on 21/Feb/20
may be some one help me
$${may}\:{be}\:{some}\:{one}\:{help}\:{me} \\ $$
Commented by mr W last updated on 21/Feb/20
i think the minimum is when A is at the origin and D=(−1,1),B=(1,1). the minimum area is 2.
$${i}\:{think}\:{the}\:{minimum}\:{is}\:{when}\:{A}\:{is} \\ $$$${at}\:{the}\:{origin}\:{and}\:{D}=\left(−\mathrm{1},\mathrm{1}\right),{B}=\left(\mathrm{1},\mathrm{1}\right). \\ $$$${the}\:{minimum}\:{area}\:{is}\:\mathrm{2}. \\ $$
Commented by jagoll last updated on 21/Feb/20
point C sir ?
$${point}\:{C}\:{sir}\:? \\ $$
Commented by jagoll last updated on 21/Feb/20
does mean the point C must lie on y−axis
$${does}\:{mean}\:{the}\:{point}\:{C}\:{must} \\ $$$${lie}\:{on}\:{y}−{axis} \\ $$
Commented by mr W last updated on 21/Feb/20
my understanding is that three vertices should lie on the parabola and the fourth vextex not. whether the fourth vertex is outside or inside of the parabola, it doesn′t matter. but this is your question. what do you mean? should the fourth vertex only outside of the parabola or it doesn′t matter?
$${my}\:{understanding}\:{is}\:{that}\:{three}\:{vertices}\: \\ $$$${should}\:{lie}\:{on}\:{the}\:{parabola}\:{and}\:{the}\: \\ $$$${fourth}\:{vextex}\:{not}.\:{whether}\:{the}\:{fourth} \\ $$$${vertex}\:{is}\:{outside}\:{or}\:{inside}\:{of}\:{the} \\ $$$${parabola},\:{it}\:{doesn}'{t}\:{matter}.\:{but}\:{this} \\ $$$${is}\:{your}\:{question}.\:{what}\:{do}\:{you}\:{mean}? \\ $$$${should}\:{the}\:{fourth}\:{vertex}\:{only}\:{outside} \\ $$$${of}\:{the}\:{parabola}\:{or}\:{it}\:{doesn}'{t}\:{matter}? \\ $$
Commented by jagoll last updated on 21/Feb/20
if according to my understanding the fourth point is outside parabola. dish according to the picture shown in the problem. sir
$${if}\:{according}\:{to}\:{my}\:{understanding} \\ $$$${the}\:{fourth}\:{point}\:{is}\:{outside}\:{parabola}. \\ $$$${dish}\:{according}\:{to}\:{the}\:{picture} \\ $$$${shown}\:{in}\:{the}\:{problem}.\:{sir} \\ $$
Commented by mr W last updated on 21/Feb/20
solution for the case that the fourth vertex lies outside the parabola see below.
$${solution}\:{for}\:{the}\:{case}\:{that}\:{the}\:{fourth} \\ $$$${vertex}\:{lies}\:{outside}\:{the}\:{parabola} \\ $$$${see}\:{below}. \\ $$
Answered by mr W last updated on 21/Feb/20
A(−a,a^2 ) with a>0 D(d,d^2 ) with d>0 B(b,b^2 ) with b^2 <d^2 say eqn. of AD is y=a^2 +m(x+a) then eqn. of AD is y=a^2 −(1/m)(x+a) d^2 =a^2 +m(d+a) d^2 −md−a^2 −ma=0 ⇒d=a+m similarly ⇒b=a−(1/m) AD^2 =(d+a)^2 +(d^2 −a^2 )^2 AD^2 =(1+m^2 )(2a+m)^2 similarly AB^2 =(1+(1/m^2 ))(2a−(1/m))^2 AB=AD ⇒(1+(1/m^2 ))(2a−(1/m))^2 =(1+m^2 )(2a+m)^2 ⇒(2am−1)^2 =m^4 (2a+m)^2 side length of square l=(2a+m)(√(1+m^2 )) we get (through graphic method): at m≈0.3821and a≈2.2358 l_(min) ≈5.1961
$${A}\left(−{a},{a}^{\mathrm{2}} \right)\:{with}\:{a}>\mathrm{0} \\ $$$${D}\left({d},{d}^{\mathrm{2}} \right)\:{with}\:{d}>\mathrm{0} \\ $$$${B}\left({b},{b}^{\mathrm{2}} \right)\:{with}\:{b}^{\mathrm{2}} <{d}^{\mathrm{2}} \\ $$$${say}\:{eqn}.\:{of}\:{AD}\:{is}\:{y}={a}^{\mathrm{2}} +{m}\left({x}+{a}\right) \\ $$$${then}\:{eqn}.\:{of}\:{AD}\:{is}\:{y}={a}^{\mathrm{2}} −\frac{\mathrm{1}}{{m}}\left({x}+{a}\right) \\ $$$${d}^{\mathrm{2}} ={a}^{\mathrm{2}} +{m}\left({d}+{a}\right) \\ $$$${d}^{\mathrm{2}} −{md}−{a}^{\mathrm{2}} −{ma}=\mathrm{0} \\ $$$$\Rightarrow{d}={a}+{m} \\ $$$${similarly} \\ $$$$\Rightarrow{b}={a}−\frac{\mathrm{1}}{{m}} \\ $$$${AD}^{\mathrm{2}} =\left({d}+{a}\right)^{\mathrm{2}} +\left({d}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left(\mathrm{2}{a}+{m}\right)^{\mathrm{2}} \\ $$$${similarly} \\ $$$${AB}^{\mathrm{2}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right)\left(\mathrm{2}{a}−\frac{\mathrm{1}}{{m}}\right)^{\mathrm{2}} \\ $$$${AB}={AD} \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right)\left(\mathrm{2}{a}−\frac{\mathrm{1}}{{m}}\right)^{\mathrm{2}} =\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left(\mathrm{2}{a}+{m}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{am}−\mathrm{1}\right)^{\mathrm{2}} ={m}^{\mathrm{4}} \left(\mathrm{2}{a}+{m}\right)^{\mathrm{2}} \\ $$$${side}\:{length}\:{of}\:{square}\:{l}=\left(\mathrm{2}{a}+{m}\right)\sqrt{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$$ \\ $$$${we}\:{get}\:\left({through}\:{graphic}\:{method}\right): \\ $$$${at}\:{m}\approx\mathrm{0}.\mathrm{3821}{and}\:{a}\approx\mathrm{2}.\mathrm{2358} \\ $$$${l}_{{min}} \approx\mathrm{5}.\mathrm{1961} \\ $$
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 (exact) solution: from (2am−1)^2 =m^4 (2a+m)^2 ⇒m^2 (2a+m)=±(2am−1) ⇒2m(m±1)a=±1−m^3 ⇒a=((1−m^3 )/(2m(m+1))) (this is for the case that the fourth vertex lies inside) ⇒a=((−1−m^3 )/(2m(m−1))) (this is for our case) l=(2a+m)(√(1+m^2 )) ⇒l=(((1+m^2 )^(3/2) )/(m(1−m))) (dl/dm)=((3(1+m^2 )^(1/2) )/((1−m)))−(((1+m^2 )^(3/2) (1−2m))/(m^2 (1−m)^2 ))=0 (((1+m^2 )(1−2m))/(m^2 (1−m)))=3 ⇒m^3 −2m^2 −2m+1=0 ⇒(m+1)(m^2 −3m+1)=0 ⇒m=−1 ⇒m=((3+(√5))/2)≈0.3820 ⇒m=((3−(√5))/2) ⇒a=((1+m^3 )/(2m(1−m)))=2.2361 ⇒l_(min) =(((1+m^2 )^(3/2) )/(m(1−m)))=5.1962
$${analytical}\:\left({exact}\right)\:{solution}: \\ $$$${from}\:\left(\mathrm{2}{am}−\mathrm{1}\right)^{\mathrm{2}} ={m}^{\mathrm{4}} \left(\mathrm{2}{a}+{m}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{m}^{\mathrm{2}} \left(\mathrm{2}{a}+{m}\right)=\pm\left(\mathrm{2}{am}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{2}{m}\left({m}\pm\mathrm{1}\right){a}=\pm\mathrm{1}−{m}^{\mathrm{3}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}−{m}^{\mathrm{3}} }{\mathrm{2}{m}\left({m}+\mathrm{1}\right)}\:\:\left({this}\:{is}\:{for}\:{the}\:{case}\right. \\ $$$$\left.{that}\:{the}\:{fourth}\:{vertex}\:{lies}\:{inside}\right) \\ $$$$\Rightarrow{a}=\frac{−\mathrm{1}−{m}^{\mathrm{3}} }{\mathrm{2}{m}\left({m}−\mathrm{1}\right)}\:\:\left({this}\:{is}\:{for}\:{our}\:{case}\right) \\ $$$${l}=\left(\mathrm{2}{a}+{m}\right)\sqrt{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$$\Rightarrow{l}=\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{m}\left(\mathrm{1}−{m}\right)} \\ $$$$\frac{{dl}}{{dm}}=\frac{\mathrm{3}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}−{m}\right)}−\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{2}{m}\right)}{{m}^{\mathrm{2}} \left(\mathrm{1}−{m}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{m}\right)}{{m}^{\mathrm{2}} \left(\mathrm{1}−{m}\right)}=\mathrm{3} \\ $$$$\Rightarrow{m}^{\mathrm{3}} −\mathrm{2}{m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{m}=−\mathrm{1} \\ $$$$\Rightarrow{m}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{3820} \\ $$$$\Rightarrow{m}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}+{m}^{\mathrm{3}} }{\mathrm{2}{m}\left(\mathrm{1}−{m}\right)}=\mathrm{2}.\mathrm{2361} \\ $$$$\Rightarrow{l}_{{min}} =\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{m}\left(\mathrm{1}−{m}\right)}=\mathrm{5}.\mathrm{1962} \\ $$
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
yes sir thank you
$${yes}\:{sir}\:{thank}\:{you} \\ $$

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