Đáp án:
\(\left( {C'} \right):\,\,y = - \frac{3}{2} - \frac{3}{2}\sin \left( {6 - 2x} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
Goi\,\,M\left( {x;y} \right) \in \left( C \right)\\
Goi\,\,M'\left( {x';y'} \right) = {V_{\left( {I;\frac{{ - 1}}{2}} \right)}}\left( M \right)\\
\Rightarrow \overrightarrow {IM'} = - \frac{1}{2}\overrightarrow {IM} \\
\Rightarrow \left\{ \begin{array}{l}
x' - 2 = - \frac{1}{2}\left( {x - 2} \right)\\
y' + 1 = - \frac{1}{2}\left( {y + 1} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x' = - \frac{1}{2}x + 3\\
y' = - \frac{1}{2}y - \frac{3}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2x' = - x + 6\\
2y' = - y - 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 6 - 2x'\\
y' = - 3 - 2y'
\end{array} \right. \Rightarrow M\left( {6 - 2x';\,\, - 3 - 2y'} \right)\\
M \in \left( C \right)\\
\Rightarrow - 3 - 2y' = 3\sin \left( {6 - 2x'} \right)\\
\Leftrightarrow 2y' = - 3 - 3\sin \left( {6 - 2x'} \right)\\
\Leftrightarrow y' = - \frac{3}{2} - \frac{3}{2}\sin \left( {6 - 2x'} \right)\\
Goi\,\,\left( {C'} \right) = {V_{\left( {I; - \frac{1}{2}} \right)}}\left( C \right)\\
\Rightarrow \left( {C'} \right):\,\,y = - \frac{3}{2} - \frac{3}{2}\sin \left( {6 - 2x} \right)
\end{array}\)
