Zum Inhalt springen

Wir ziehen die Verdoppelungsformeln aus Aufgabe heran. Es ist

{\displaystyle {}{\begin{aligned}\alpha &={\frac {3w^{2}-2(\lambda _{1}+\lambda _{2}+\lambda _{3})w+\lambda _{1}\lambda _{2}+\lambda _{1}\lambda _{3}+\lambda _{2}\lambda _{3}}{2z}}\\&={\frac {3(\mu _{1}\mu _{2}+\mu _{1}\mu _{3}+\mu _{2}\mu _{3})^{2}+2(\mu _{1}^{2}+\mu _{2}^{2}+\mu _{3}^{2})(\mu _{1}\mu _{2}+\mu _{1}\mu _{3}+\mu _{2}\mu _{3})+\mu _{1}^{2}\mu _{2}^{2}+\mu _{1}^{2}\mu _{3}^{2}+\mu _{2}^{2}\mu _{3}^{2}}{-2(\mu _{1}+\mu _{2}+\mu _{3})(\mu _{1}\mu _{2}+\mu _{1}\mu _{3}+\mu _{2}\mu _{3})+2\mu _{1}\mu _{2}\mu _{3}}}\\&={\frac {4\mu _{1}^{2}\mu _{2}^{2}+4\mu _{1}^{2}\mu _{3}^{2}+4\mu _{2}^{2}\mu _{3}^{2}+8\mu _{1}^{2}\mu _{2}\mu _{3}+8\mu _{1}\mu _{2}^{2}\mu _{3}+8\mu _{1}\mu _{2}\mu _{3}^{2}+2(\mu _{1}\mu _{2}^{3}+\mu _{1}\mu _{3}^{3}+\mu _{2}\mu _{3}^{3}+\mu _{1}^{3}\mu _{3}+\mu _{1}^{3}\mu _{2}+\mu _{2}^{3}\mu _{3})}{-2\mu _{1}^{2}\mu _{2}-2\mu _{1}\mu _{2}^{2}-2\mu _{1}^{2}\mu _{3}-2\mu _{1}\mu _{3}^{2}-2\mu _{2}^{2}\mu _{3}-2\mu _{2}\mu _{3}^{2}-4\mu _{1}\mu _{2}\mu _{3}}}\\&=-\mu _{1}-\mu _{2}-\mu _{3}\end{aligned}}}

und

${\displaystyle {}2w-\lambda _{1}-\lambda _{2}-\lambda _{3}=2\mu _{1}\mu _{2}+2\mu _{1}\mu _{3}+2\mu _{2}\mu _{3}+\mu _{1}^{2}+\mu _{2}^{2}+\mu _{3}^{2}=(\mu _{1}+\mu _{2}+\mu _{3})^{2}\,.}$

Somit ist die erste Koordinate von ${\displaystyle {}2(w,z)}$ gleich ${\displaystyle {}0}$. Die Geradengleichung der Tangente an ${\displaystyle {}(w,z)}$ ist

${\displaystyle {}Y=\alpha X+\beta \,,}$

wobei wir ${\displaystyle {}\beta }$ über

{\displaystyle {}{\begin{aligned}\beta &=z-\alpha w\\&=-(\mu _{1}+\mu _{2}+\mu _{3})(\mu _{1}\mu _{2}+\mu _{1}\mu _{3}+\mu _{2}\mu _{3})+\mu _{1}\mu _{2}\mu _{3}+(\mu _{1}+\mu _{2}+\mu _{3})(\mu _{1}\mu _{2}+\mu _{1}\mu _{3}+\mu _{2}\mu _{3})\\&=\mu _{1}\mu _{2}\mu _{3}\end{aligned}}}
berechnen.