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determiner-les-coordonnes-de-X-et-Y-r-1-




Question Number 213060 by a.lgnaoui last updated on 29/Oct/24
determiner les coordonnes de  X et  Y  r=1
$$\mathrm{determiner}\:\mathrm{les}\:\mathrm{coordonnes}\:\mathrm{de}\:\:\mathrm{X}\:\mathrm{et}\:\:\mathrm{Y} \\ $$$$\boldsymbol{\mathrm{r}}=\mathrm{1} \\ $$
Commented by a.lgnaoui last updated on 29/Oct/24
Commented by Frix last updated on 29/Oct/24
 { (((x^2 /(16))+(y^2 /4)=1)),(((x−2)^2 +(y−(√3))^2 =1)) :}   { ((x^2 =16−4y^2 )),((x^2 +y^2 −4x−2(√3)y+6=0)) :}  Inserting and solving for x  x=−((3y^2 +2(√3)y−22)/4)  Inserting and transforming  y^4 +((4(√3))/3)y^3 −((56)/9)y^2 −((88(√3))/9)y+((76)/3)=0  This has no useable exact solutions.  y≈1.36205409 ⇒ x≈2.92903306  y≈1.93380767 ⇒ x≈1.02056436  X= (((1.02)),((1.93)) )  Y= (((2.93)),((1.36)) )
$$\begin{cases}{\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}}\\{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} =\mathrm{16}−\mathrm{4}{y}^{\mathrm{2}} }\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}\sqrt{\mathrm{3}}{y}+\mathrm{6}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{Inserting}\:\mathrm{and}\:\mathrm{solving}\:\mathrm{for}\:{x} \\ $$$${x}=−\frac{\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{y}−\mathrm{22}}{\mathrm{4}} \\ $$$$\mathrm{Inserting}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${y}^{\mathrm{4}} +\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}{y}^{\mathrm{3}} −\frac{\mathrm{56}}{\mathrm{9}}{y}^{\mathrm{2}} −\frac{\mathrm{88}\sqrt{\mathrm{3}}}{\mathrm{9}}{y}+\frac{\mathrm{76}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions}. \\ $$$${y}\approx\mathrm{1}.\mathrm{36205409}\:\Rightarrow\:{x}\approx\mathrm{2}.\mathrm{92903306} \\ $$$${y}\approx\mathrm{1}.\mathrm{93380767}\:\Rightarrow\:{x}\approx\mathrm{1}.\mathrm{02056436} \\ $$$${X}=\begin{pmatrix}{\mathrm{1}.\mathrm{02}}\\{\mathrm{1}.\mathrm{93}}\end{pmatrix}\:\:{Y}=\begin{pmatrix}{\mathrm{2}.\mathrm{93}}\\{\mathrm{1}.\mathrm{36}}\end{pmatrix} \\ $$
Commented by a.lgnaoui last updated on 29/Oct/24
exact.thank you
$$\mathrm{exact}.\mathrm{thank}\:\mathrm{you}\: \\ $$

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