Known point p is on parabola y? =4x, the coordinates of the minimum time point P for the sum of the distance from point P to point M (7 7,8) and the distance from point P to Y axis.

Solution 1. Parabolic parameters: 2p=4, p=2, p/2= 1, so the focus f (1, 0); Alignment equation x =-1;

Connect FM, then the intersection of FM and parabola is the required point p.

The slope of the line where FM is located is k=4/3, and the equation of the line is y = (4/3) (x-1);

Substitute into the parabolic equation, (16/9)(x- 1)? =4x, i.e. 16(x? -2x+ 1)=36x，4x？ -8x+4 = 9x；

4x？ -17x+4 = (4x-1) (x-4) = 0, then x? = 1/4 (truncated), x? =4，y？ =4, that is, the coordinate of point P is (4,4);

The distance s from point p to m? =√[(7-4)? +(8-4)? ]=5; Distance s from point p to y axis? =4;

So the minimum value of the sum of two distances Smin=S? +S？ =9.

It is proved that the distance from P to Y axis is s because P is on the straight line of FM (that is, M, P and F are on the same straight line). =P to the alignment

Distance-1= ∣ pf ∣-1; Two distances and s? +S？ =∣mp∣+∣pf∣- 1=∣mf∣- 1。

Take another point p' different from p on the parabola. For the same reason, s? +S '？ =∣mp'∣+∣p'f∣- 1；

At this time, because M, P' and F are not in a straight line, there must be ∣ Member' ∣+∣ P' f ∣-1> ∣ MP ∣+∣ PF ∣-1.

In other words, there must be an s'? +S '？ & gts？ +S？ =9.