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Question-111584




Question Number 111584 by ajfour last updated on 04/Sep/20
Answered by ajfour last updated on 04/Sep/20
C(0,y)   ; A(x,0)  ;  B(z,z^2 )  (√((x−z)^2 +z^4 ))+(√(z^2 +(y−z^2 )^2 )) = a  △=z^3 +(1/2)z(y−z^2 )+(1/2)(x−z)z^2   let  x−z=z^2 tan θ  &    y−z^2 =ztan φ  ⇒  z^2 sec θ+zsec φ=a  &    △=z^3 +(z^2 /2)tan φ+(z^4 /2)tan θ   someone please help..
$${C}\left(\mathrm{0},{y}\right)\:\:\:;\:{A}\left({x},\mathrm{0}\right)\:\:;\:\:{B}\left({z},{z}^{\mathrm{2}} \right) \\ $$$$\sqrt{\left({x}−{z}\right)^{\mathrm{2}} +{z}^{\mathrm{4}} }+\sqrt{{z}^{\mathrm{2}} +\left({y}−{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:{a} \\ $$$$\bigtriangleup={z}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{z}\left({y}−{z}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{z}\right){z}^{\mathrm{2}} \\ $$$${let}\:\:{x}−{z}={z}^{\mathrm{2}} \mathrm{tan}\:\theta \\ $$$$\&\:\:\:\:{y}−{z}^{\mathrm{2}} ={z}\mathrm{tan}\:\phi \\ $$$$\Rightarrow\:\:{z}^{\mathrm{2}} \mathrm{sec}\:\theta+{z}\mathrm{sec}\:\phi={a} \\ $$$$\&\:\:\:\:\bigtriangleup={z}^{\mathrm{3}} +\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}\:\phi+\frac{{z}^{\mathrm{4}} }{\mathrm{2}}\mathrm{tan}\:\theta \\ $$$$\:{someone}\:{please}\:{help}.. \\ $$
Commented by mr W last updated on 05/Sep/20
P=z^3 +(z^2 /2)tan φ+(z^4 /2)tan θ+λ(z^2 sec θ+zsecφ−a)  (∂P/∂z)=0, (∂P/∂θ)=0, (∂P/∂φ)=0, (∂P/∂λ)=0  .... hard to solve ...
$${P}={z}^{\mathrm{3}} +\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}\:\phi+\frac{{z}^{\mathrm{4}} }{\mathrm{2}}\mathrm{tan}\:\theta+\lambda\left({z}^{\mathrm{2}} \mathrm{sec}\:\theta+{z}\mathrm{sec}\phi−{a}\right) \\ $$$$\frac{\partial{P}}{\partial{z}}=\mathrm{0},\:\frac{\partial{P}}{\partial\theta}=\mathrm{0},\:\frac{\partial{P}}{\partial\phi}=\mathrm{0},\:\frac{\partial{P}}{\partial\lambda}=\mathrm{0} \\ $$$$….\:{hard}\:{to}\:{solve}\:… \\ $$
Answered by ajfour last updated on 06/Sep/20

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