a: Phương trình hoành độ giao điểm là:
\(-x^2=2x+m-1\)
=>\(x^2+2x+m-1=0\)
\(\text{Δ}=2^2-4\cdot1\left(m-1\right)=4-4m+4=-4m+8\)
Để (d) cắt (P) tại hai điểm phân biệt thì Δ>0
=>-4m+8>0
=>-4m>-8
=>m<2
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)
\(\left(y_1-y_2\right)^2=16\)
=>\(\left[\left(-x_1^2\right)-\left(-x_2^2\right)\right]^2=16\)
=>\(\left(-x_1^2+x_2^2\right)^2=16\)
=>\(\left(x_1^2-x_2^2\right)^2=16\)
=>\(\left(x_1-x_2\right)^2\cdot\left(x_1+x_2\right)^2=16\)
=>\(\left(x_1-x_2\right)^2\cdot\left(-2\right)^2=16\)
=>\(\left(x_1-x_2\right)^2=4\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=4\)
=>\(\left(-2\right)^2-4\left(m-1\right)=4\)
=>4(m-1)=0
=>m-1=0
=>m=1(nhận)