Question

已知点P(3,2),F为抛物线y2=4x的焦点,点M在该抛物线上移动,当|MP|+|MF|取最小值

已知点P(3,2),F为抛物线y2=4x的焦点,点M在该抛物线上移动,当|MP|+|MF|取最小值时,点M的坐标为?

Answer 1

y² = 4x = 2px

p = 2, F(1, 0)

令M(m²/4, m)

求导：2yy' = 4, y' = 2/y

抛物线在M处的切线为: y - m = (2/m)(x - m²/4)

y = 2x/m + m/2                                       (1)

斜率为2/m

其垂线的斜率为-m/2

上述切线过F的垂线方程为: y - 0 = (-m/2)(x - 1)            (2)

(1)(2)联立: 2x/m + m/2 = -mx/2 + m/2

(2/m + m/2)x = 0

2/m和m/2同号,其和不可能为0, 于是x = 0

二者的交点为A(0, 1)

令F关于切线的对称点为F'(a, b), 则A为FF'的中点:

0 = (a + 1)/2, a = -1

1 = (b + 0)/2, b = 2

F'(-1, 2)

F'P的方程为y = 2

M即为F'P与抛物线的交点, 坐标为(1, 2)

做法是取一点关于切线的对称点, 然后与另一点相连, 交点即为所求的点(不清楚可再问).