回復 26# anyway13 的帖子
第6題
沿x軸推移y坐標的-1倍
\left\{ \begin{align}
& {{x}_{1}}=x-y \\
& {{y}_{1}}=y \\
\end{align} \right.\quad \Rightarrow \quad \left\{ \begin{align}
& x={{x}_{1}}+{{y}_{1}} \\
& y={{y}_{1}} \\
\end{align} \right.
代入{{x}^{2}}+{{y}^{2}}+2x=1,可得
{{\left( {{x}_{1}}+{{y}_{1}} \right)}^{2}}+{{y}_{1}}^{2}+2\left( {{x}_{1}}+{{y}_{1}} \right)=1
再對y=3x鏡射
鏡射矩陣=\left[ \begin{matrix}
\frac{1-{{3}^{2}}}{1+{{3}^{2}}} & \frac{2\times 3}{1+{{3}^{2}}} \\
\frac{2\times 3}{1+{{3}^{2}}} & -\frac{1-{{3}^{2}}}{1+{{3}^{2}}} \\
\end{matrix} \right]=\left[ \begin{matrix}
-\frac{4}{5} & \frac{3}{5} \\
\frac{3}{5} & \frac{4}{5} \\
\end{matrix} \right]
\Rightarrow \quad \left\{ \begin{align}
& x=-\frac{4}{5}{{x}_{1}}+\frac{3}{5}{{y}_{1}} \\
& y=\frac{3}{5}{{x}_{1}}+\frac{4}{5}{{y}_{1}} \\
\end{align} \right.\Rightarrow \quad \left\{ \begin{align}
& {{x}_{1}}=-\frac{4}{5}x+\frac{3}{5}y \\
& {{y}_{1}}=\frac{3}{5}x+\frac{4}{5}y \\
\end{align} \right.
代入{{\left( {{x}_{1}}+{{y}_{1}} \right)}^{2}}+{{y}_{1}}^{2}+2\left( {{x}_{1}}+{{y}_{1}} \right)=1,可得
2{{x}^{2}}+2xy+13{{y}^{2}}-2x+14y-5=0