Question

Tren toa do mat phang Oxy cho parabol (P): y=x² va duong thang (d): y=-2ax-4a

tim cac gia tri cau a de (d) cat (P) tai 2 diem phan biet co hoanh do x₁;x₂ thoa man |x₁| + |x₂| = 3

Answer 1

Phương trình hoành độ giao điểm: \(x^2+2ax+4a=0\)

\(\Delta'=a^2-4a>0\Rightarrow\left[{}\begin{matrix}a< 0\\a>4\end{matrix}\right.\)

Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2a\\x_1x_2=4a\end{matrix}\right.\)

\(\left|x_1\right|+\left|x_2\right|=3\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=9\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=9\)

\(\Leftrightarrow4a^2-8a+8\left|a\right|=9\)

- Với \(a>0\) \(\Rightarrow4a^2=9\Rightarrow a^2=\frac{9}{4}\Rightarrow a=\frac{3}{2}< 4\left(l\right)\)

- Với \(a< 0\Rightarrow4a^2-16a-9=0\Rightarrow\left[{}\begin{matrix}a=-\frac{1}{2}\\a=\frac{9}{2}>0\left(l\right)\end{matrix}\right.\)

Vậy \(a=-\frac{1}{2}\)