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\(\varphi x -\mu y=2\varphi \\x+(\frac{\mu}{\varphi})^2y=1+\frac{\mu^2}{\varphi^2}\)

 12.05.2019
 #1
avatar+22149 
+3

Löse die Parametergleichung ohne Fallunterscheidung nach x und y auf

\(\varphi x -\mu y=2\varphi \\ x+ \left(\dfrac{\mu}{\varphi} \right)^2y=1+\dfrac{\mu^2}{\varphi^2}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\varphi x -\mu y } &=& \mathbf{2\varphi} \quad | \quad :\varphi \\ x - \dfrac{\mu}{\varphi} y &=& 2 \\ \mathbf{\dfrac{\mu}{\varphi} y} &=& \mathbf{x-2} \qquad (1) \\ \hline x+ \left(\dfrac{\mu}{\varphi}\right)^2y &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu^2}{\varphi^2} y &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi } \left(\dfrac{\mu }{\varphi } y \right) &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi } (x-2) &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi }x -2\dfrac{\mu }{\varphi } &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi }x &=& 1+ 2\dfrac{\mu }{\varphi } + \dfrac{\mu^2}{\varphi^2} \\ x \left(1+ \dfrac{\mu }{\varphi } \right) &=& \left(1+ \dfrac{\mu }{\varphi } \right)^2 \\ \mathbf{ x } &=& \mathbf{1+ \dfrac{\mu }{\varphi }} \\ \hline \mathbf{\dfrac{\mu}{\varphi} y} &=& \mathbf{x-2} \qquad (1) \\ \dfrac{\mu}{\varphi} y &=& 1+ \dfrac{\mu }{\varphi }-2 \\ \dfrac{\mu}{\varphi} y &=& \dfrac{\mu }{\varphi }-1 \\ y &=& \dfrac{\mu }{\varphi }\dfrac{\varphi}{\mu} -\dfrac{\varphi}{\mu} \\ \mathbf{ y } &=& \mathbf{1 -\dfrac{\varphi}{\mu} } \\ \hline \end{array}\)

 

laugh

 
 13.05.2019
 #2
avatar+8220 
+2

Danke heureka!

 
 13.05.2019

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