第15章 15.25

$\frac{x^2}{\left(\dfrac{1}{\sqrt3}\right)^2}+\frac{y^2}{\left(\dfrac{1}{3}\right)^2}=1$

より $\left(0,\pm\dfrac13\right)$、$\left(\pm\dfrac{1}{\sqrt3},0\right)$、$\left(\dfrac13,-\dfrac{\sqrt2}{3\sqrt3}\right)$

$\begin{pmatrix}1 & -\alpha \\ \beta & \sqrt3\gamma\end{pmatrix}\begin{pmatrix}0 \\ \pm\dfrac13\end{pmatrix}=\begin{pmatrix}\mp\dfrac{\alpha}{3} \\ \pm\dfrac{\gamma}{\sqrt3}\end{pmatrix}$

$\begin{pmatrix}1 & -\alpha \\ \beta & \sqrt3\gamma\end{pmatrix}\begin{pmatrix}\pm\dfrac{1}{\sqrt3} \\ 0\end{pmatrix}=\begin{pmatrix}\pm\dfrac{1}{\sqrt3} \\ \pm\dfrac{1}{\sqrt3}\beta\end{pmatrix}$

$\begin{pmatrix}1 & -\alpha \\ \beta & \sqrt3\gamma\end{pmatrix}\begin{pmatrix}\dfrac13 \\ -\dfrac{\sqrt2}{3\sqrt3}\end{pmatrix}=\begin{pmatrix}\dfrac13+\dfrac{\sqrt2}{3\sqrt3}\alpha \\ \dfrac13\beta-\dfrac{\sqrt2}{3}\gamma\end{pmatrix}$

これら点が単位円 $x^2+y^2=1$ にあればよいので

$\frac{\alpha^2}{9}+\frac{\gamma^2}{3}=1 \qquad \alpha=3\sqrt{1-\frac{\gamma^2}{3}} \qquad ①$

$\frac13+\frac13\beta^2=1 \qquad \beta^2=2 \qquad \beta=\sqrt2 \qquad ②$

$\left(\frac13+\frac{\sqrt2}{3\sqrt3}\alpha\right)^2+\left(\frac13\beta-\frac{\sqrt2}{3}\gamma\right)^2=1 \qquad \left(\frac13+\frac{\sqrt2}{3\sqrt3}\alpha\right)^2+\left(\frac{\sqrt2}{3}-\frac{\sqrt2}{3}\gamma\right)^2=1 \qquad ③$

①を③に代入すると

$\left(\frac13+\frac{\sqrt2}{\sqrt3}\sqrt{1-\frac{\gamma^2}{3}}\right)^2+\left(\frac{\sqrt2}{3}-\frac{\sqrt2}{3}\gamma\right)^2=1 \qquad \frac19+\frac{2\sqrt2}{3\sqrt3}\sqrt{1-\frac{\gamma^2}{3}}+\frac23\left(1-\frac{\gamma^2}{3}\right)+\frac29-\frac49\gamma+\frac29\gamma^2=1$

$1+2\sqrt6\sqrt{1-\frac{\gamma^2}{3}}+6-2\gamma^2+2-4\gamma+2\gamma^2=1 \qquad 8-4\gamma=-2\sqrt6\sqrt{1-\frac{\gamma^2}{3}}$

$64-64\gamma+16\gamma^2=24\left(1-\frac{\gamma^2}{3}\right) \qquad 8-8\gamma+2\gamma^2=3-\gamma^2 \qquad 3\gamma^2-8\gamma+5=0 \qquad (3\gamma-5)(\gamma-1)=0$

$\gamma=\dfrac53,1$ これを①に代入すると$\gamma=\dfrac53$では不適なので $\gamma=1$、$\alpha=\sqrt6$ となる

よって $\alpha=\sqrt6$、$\beta=\sqrt2$、$\gamma=1$