两边对x求导得：2(y-k)y'=4c
y'=2c/(y-k) 切线斜率为2c/(y0-k)
(y0 − k)² = 4c(x0 − h),∴(y0 − k) = 4c(x0 − h)/(y0 − k),
点斜式方程：
y=y0+2c/(y0-k) * (x-x0)
=k+(y0-k) + 2c[(x-h)-(x0-h)]/(y0-k)
=k+ 4c(x0-h)/(y0-k)
+ 2c(x-h)/(y0-k) - 2c(x0-h)/(y0-k)
=k+ 2c(x-h)/(y0-k) +2c(x0-h)/(y0-k)
=k+ 4c[(x-h)+(x0-h)]/[2(y0-k)]
∴(y-k)(y0-k)=4c[(x-h)+(x0-h)]/2
图上的答案好像少了一个加号,你可以取特殊值x=x0验证