A,B thuộc (P) sao cho A,B đối xứng qua M(-1;5)
=>\(\left\{{}\begin{matrix}x_A+x_B=2\cdot\left(-1\right)=-2\\y_A+y_B=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1+x_2=-2\\\dfrac{1}{2}\left(x_1^2+x_2^2\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=-2\\x_1^2+x_2^2=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1+x_2=-2\\\left(x_1+x_2\right)^2-2x_1x_2=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=\dfrac{\left(x_1+x_2\right)^2-20}{2}=\dfrac{4-20}{2}=-8\end{matrix}\right.\)
=>x1,x2 là các nghiệm của phương trình:
\(A^2+2A-8=0\)
=>(A+4)(A-2)=0
=>\(\left[{}\begin{matrix}A=-4\\A=2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=-4\\x_2=2\end{matrix}\right.\\\left\{{}\begin{matrix}x_1=2\\x_2=-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y_1=\dfrac{1}{2}\cdot\left(-4\right)^2=8\\y_2=\dfrac{1}{2}\cdot2^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=8\end{matrix}\right.\end{matrix}\right.\)
vậy: \(\left[{}\begin{matrix}A\left(-4;8\right);B\left(2;2\right)\\A\left(2;2\right);B\left(-4;8\right)\end{matrix}\right.\)
