a: Để (d) vuông góc (d') thì \(\left(a-3\right)\cdot2=-1\)
=>\(a-3=-\dfrac{1}{2}\)
=>\(a=-\dfrac{1}{2}+3=\dfrac{5}{2}\)
b: Phương trình hoành độ giao điểm là:
\(x^2=2x+m^2-4m+9\)
=>\(x^2-2x-\left(m^2-4m+9\right)=0\)
\(\Delta=\left(-2\right)^2-4\cdot1\cdot\left(-m^2+4m-9\right)\)
\(=4+4m^2-16m+36\)
\(=4m^2-16m+40\)
\(=4m^2-16m+16+24\)
\(=\left(2m-4\right)^2+24>=24>0\forall m\)
=>(P) luôn cắt (d) tại hai điểm phân biệt có hoành độ là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2-\sqrt{4m^2-16m+40}}{2}=1-\sqrt{m^2-4m+10}\\x_2=\dfrac{2+\sqrt{4m^2-16m+40}}{2}=1+\sqrt{m^2-4m+10}\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=-\left(m^2-4m+9\right)\end{matrix}\right.\)
\(\left|x_1-2023\right|-\left|x_2+2023\right|=y_1+y_2-48\)
=>\(\left|1-\sqrt{m^2-4m+10}-2023\right|-\left|1+\sqrt{m^2-4m+10}+2023\right|=y_1+y_2-48\)
=>\(\sqrt{m^2-4m+10}+2022-\left|\sqrt{m^2-4m+10}+2024\right|=x_1^2+x_2^2-48\)
=>\(x_1^2+x_2^2-48=\sqrt{m^2-4m+10}+2022-\sqrt{m^2-4m+10}-2024=-2\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-48+2=0\)
=>\(2^2-2\left(-m^2+4m-9\right)-46=0\)
=>\(4+2m^2-8m+18-46=0\)
=>\(2m^2-8m-24=0\)
=>\(m^2-4m-12=0\)
=>(m-6)(m+2)=0
=>\(\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\)
