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源自知乎提问
题：已知直线 y=(4a-4)x+4a^2-24a+16，则与之始终相切的抛物线的解析式为_____.
包络线
设 $f(x,y,a)=(4a-4)x-y+4a^2-24a+16$ ，则有 \[\left\{\begin{aligned}f(x,y,a)&=0,\\[1ex]
\frac{\partial f(x,y,a)}{\partial a}&=0,\end{aligned}\right.\] 即 \[\left\{\begin{aligned}(4a-4)x-y+4a^2-24a+16&=0,\\[1ex]
4x+8a-24&=0,\end{aligned}\right.\] 两式消 a 便得 \[y=-x^2+8x-20,\] 这即为所求抛物线的方程. |
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