Phương trình hoành độ giao điểm:
\(x^2+\left(m+1\right)x-m^2+1=0\)
\(\Delta=\left(m+1\right)^2+4\left(m^2-1\right)=5m^2+2m-3\)
a/ Để d tiếp xúc (P) thì pt có nghiệm kép
\(\Rightarrow\Delta=0\Rightarrow5m^2+2m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\frac{3}{5}\end{matrix}\right.\)
b/ Để pt có nghiệm \(\Rightarrow5m^2+2m-3\ge0\Rightarrow\left[{}\begin{matrix}m\le-1\\m\ge\frac{5}{3}\end{matrix}\right.\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-m-1\\x_1x_2=-m^2+1\end{matrix}\right.\)
\(x_1y_2+x_2y_1=1\)
\(\Leftrightarrow x_1\left(-x_2^2\right)+x_2\left(-x_1^2\right)=1\)
\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)=-1\)
\(\Leftrightarrow\left(-m-1\right)\left(-m^2+1\right)=-1\)
\(\Leftrightarrow m^3+m^2-m=0\)
\(\Leftrightarrow m\left(m^2+m-1\right)=0\Rightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=\frac{-1+\sqrt{5}}{2}\left(l\right)\\m=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\)