一道英文数学题Let (s,t) be a point in the first quadrant (not on a

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一道英文数学题
Let (s,t) be a point in the first quadrant (not on a coordinate axis) that is on the graph of y = 9−x2
and let L be a line tangent to y = 9−x2 at (s,t).Then L will cut off a triangle in the first quadrant.
Find the (s,t) that corresponds to the triangle of that type that has the minimum area.What is
the value of

L是y=9-x²在(s,t)处的切线,导数y‘=-2x,所以L斜率为-2s,L方程为y-t=-2s(x-s),即y=-2sx+2s²+t
且(s,t)在抛物线上,有t=9-s²,所以L方程化简为：y=-2sx+s²+9
L与坐标轴的交点为((s²+9)/2s,0)(0,s²+9),
由于L与坐标轴围成的三角形在第一象限
则三角形面积S=1/2 *(s²+9)/2s * (s²+9)=(s²+9)²/4s=(s³+18s+81/s)/4
S'=(3s²+18-81/s²)/4
当S'=0时,即3s²+18-81/s²=0,即(s+9/s)(3s-9/s)=0,解得s=±√3,又因为(s,t)在第一象限,所以s>0,所以s=√3,可以验证下,当0

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