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Question-142560




Question Number 142560 by ajfour last updated on 02/Jun/21
Answered by 1549442205PVT last updated on 02/Jun/21
put AC=x,CP=y⇒AO.OB=OP^2   ⇔(√(x^2 +c^2 ))(1−(√(c^2 +x^2 )))=c^2 +y^2 (1)  xy=c^2 (2)  (1)⇔(√(x^2 +c^2 ))=2c^2 +x^2 +y^2 =2xy+x^2 +y^2   ⇔(√(xy+x^2 ))=(x+y)^2 ⇔x(x+y)=(x+y)^4   ⇔x=(x+y)^3 (3)  From (2) we have x=(c^2 /y).Replace into  (3) we get (c^2 /y)=((c^2 /y)+y)^3 ⇔c^2 y^2 =(y^2 +c^2 )^3   ⇔t^3 −c^2 t+c^4 =0 (∗) (put y^2 +c^2 =t  ⇒c^2 y^2 =c^2 (y^2 +c^2 )−c^4 =c^2 t−c^4 )  t=−(m+n)⇔t+m+n=0  ⇔t^3 +m^3 +n^3 −3mnt=0.Hence,  (∗)⇒ { ((m^3 +n^3 =c^4 )),((mn=(c^2 /3))) :}⇔ { ((m^3 +n^3 =c^4 (4))),((m^3 n^3 =(c^6 /(27)))) :}  (m^3 −n^3 )^2 =(m^3 +n^3 )^2 −4m^3 n^3 =c^8 −((4c^6 )/(27))  ⇒m^3 −n^3 =c^3 (√((27c^2 −4)/(27)))(5).  (4) and (5)⇒m^3 =(1/2)(c^4 +c^3 (√((27c^2 −4)/(27))))  n^3 =(1/2)(c^4 −c^3 (√((27c^2 −4)/(27))))  t=−(m+n)=−[3(√((1/2)(c^4 +c^3 (√((27c^2 −4)/(27))))))+^3 (√((1/2)(c^4 −c^3 (√((27c^2 −4)/(27))))))]  PQ=2(√(c^2 +y^2 ))=2(√t)
$$\mathrm{put}\:\mathrm{AC}=\mathrm{x},\mathrm{CP}=\mathrm{y}\Rightarrow\mathrm{AO}.\mathrm{OB}=\mathrm{OP}^{\mathrm{2}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }\left(\mathrm{1}−\sqrt{\mathrm{c}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\right)=\mathrm{c}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$\mathrm{xy}=\mathrm{c}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }=\mathrm{2c}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{2xy}+\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{xy}+\mathrm{x}^{\mathrm{2}} }=\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \Leftrightarrow\mathrm{x}\left(\mathrm{x}+\mathrm{y}\right)=\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{x}=\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} \left(\mathrm{3}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}=\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{y}}.\mathrm{Replace}\:\mathrm{into} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{y}}=\left(\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{y}}+\mathrm{y}\right)^{\mathrm{3}} \Leftrightarrow\mathrm{c}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\left(\mathrm{y}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{t}^{\mathrm{3}} −\mathrm{c}^{\mathrm{2}} \mathrm{t}+\mathrm{c}^{\mathrm{4}} =\mathrm{0}\:\left(\ast\right)\:\left(\mathrm{put}\:\mathrm{y}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{t}\right. \\ $$$$\left.\Rightarrow\mathrm{c}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\mathrm{c}^{\mathrm{4}} =\mathrm{c}^{\mathrm{2}} \mathrm{t}−\mathrm{c}^{\mathrm{4}} \right) \\ $$$$\mathrm{t}=−\left(\mathrm{m}+\mathrm{n}\right)\Leftrightarrow\mathrm{t}+\mathrm{m}+\mathrm{n}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{t}^{\mathrm{3}} +\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} −\mathrm{3mnt}=\mathrm{0}.\mathrm{Hence}, \\ $$$$\left(\ast\right)\Rightarrow\begin{cases}{\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} =\mathrm{c}^{\mathrm{4}} }\\{\mathrm{mn}=\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{3}}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} =\mathrm{c}^{\mathrm{4}} \left(\mathrm{4}\right)}\\{\mathrm{m}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} =\frac{\mathrm{c}^{\mathrm{6}} }{\mathrm{27}}}\end{cases} \\ $$$$\left(\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} \right)^{\mathrm{2}} =\left(\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4m}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} =\mathrm{c}^{\mathrm{8}} −\frac{\mathrm{4c}^{\mathrm{6}} }{\mathrm{27}} \\ $$$$\Rightarrow\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =\mathrm{c}^{\mathrm{3}} \sqrt{\frac{\mathrm{27c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{27}}}\left(\mathrm{5}\right). \\ $$$$\left(\mathrm{4}\right)\:\mathrm{and}\:\left(\mathrm{5}\right)\Rightarrow\mathrm{m}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}^{\mathrm{4}} +\mathrm{c}^{\mathrm{3}} \sqrt{\frac{\mathrm{27c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{27}}}\right) \\ $$$$\mathrm{n}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}^{\mathrm{4}} −\mathrm{c}^{\mathrm{3}} \sqrt{\frac{\mathrm{27c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{27}}}\right) \\ $$$$\mathrm{t}=−\left(\mathrm{m}+\mathrm{n}\right)=−\left[\mathrm{3}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}^{\mathrm{4}} +\mathrm{c}^{\mathrm{3}} \sqrt{\frac{\mathrm{27c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{27}}}\right)}+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}^{\mathrm{4}} −\mathrm{c}^{\mathrm{3}} \sqrt{\frac{\mathrm{27c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{27}}}\right)}\right] \\ $$$$\mathrm{PQ}=\mathrm{2}\sqrt{\mathrm{c}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{t}} \\ $$
Commented by 1549442205PVT last updated on 02/Jun/21
Commented by 1549442205PVT last updated on 02/Jun/21
Excuse me,i mistaken and shall correct   Thank you very much
$$\mathrm{Excuse}\:\mathrm{me},\mathrm{i}\:\mathrm{mistaken}\:\mathrm{and}\:\mathrm{shall}\:\mathrm{correct}\: \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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