题目内容

8.若{(x,y)|$\left\{\begin{array}{l}{x+y=1}\\{x-y-3=0}\end{array}\right.$}⊆{(x,y)|y=ax2+1},则a=-$\frac{1}{2}$.

分析 首先解方程组,从而代入可得-1=4a+1,从而解得.

解答 解:解方程组$\left\{\begin{array}{l}{x+y=1}\\{x-y-3=0}\end{array}\right.$得,
$\left\{\begin{array}{l}{y=-1}\\{x=2}\end{array}\right.$;
∵{(x,y)|$\left\{\begin{array}{l}{x+y=1}\\{x-y-3=0}\end{array}\right.$}⊆{(x,y)|y=ax2+1},
∴-1=4a+1,
∴a=-$\frac{1}{2}$;
故答案为:-$\frac{1}{2}$.

点评 本题考查了方程组的求解及集合间关系的应用.

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