证明曲线 $x^2-3xy-40=0$ 与直线 $3x+y+k=0$ 对于常数 $k$ 的所有取值都有交点。

用直线方程消去 $y$，把交点问题转成二次方程有实根的问题。判别式中的平方项无论 $k$ 取何值都非负。

The line gives $y=-3x-k$. Substituting into the curve equation gives

The discriminant is $(3k)^2-4(10)(-40)=9k^2+1600>0$ for every real $k$. Thus there are two distinct real values of $x$ for every $k$, each giving a point on both the line and the curve.