(x-3)²+(y-4)²=9, 这是圆,可用参数法
令x=3+3cost, y=4+3sint
3x+4y=3(3+3cost)+4(4+3sint)
=9cost+12sint+25
=15sin(t+p)+25, 这里tanp=9/12=3/4
因此最大值为15+25=40, 最小值为-15+25=10
k=y/x=(4+3sint)/(3+3cost)
3k+3kcost=4+3sint
3sint-3kcost=3k-4
3√(1+k²)sin(t-u)=3k-4, 这里tanu=k
sin(t-u)=(3k-4)/[3√(1+k²)]
因为|sin(t-u)|<=1
得:|3k-4|/[3√(1+k²)]<=1
平方:9k²-24k+16<=9+9k²
得:k>=7/24