The math problem of finding the maximum value in senior two.

1、po^2=x^2+y^2=x^2+( 1-x^2/4)=3/4x^2+ 1-2 & lt； = X & lt=2

The maximum value of PO^2 is 3/4 * 2 2+1= 3+1= 4. The maximum absolute value of po is 2.

The minimum value of PO^2 is 3/4 * 0 2+1= 0+1=1. The minimum value of the absolute value of po is 1.

pf 1^2=(x+3^ 1/2)^2+y^2=(x+3^ 1/2)^2+( 1- 1/4x^2)=(3^ 1/2/2x+2)^2

-2 & lt； = X & lt= 2 -3^ 1/2<； =3^ 1/2/2x<； =3^ 1/2

2-3^ 1/2<； =3^ 1/2/2x+2<； =2+3^ 1/2

The absolute value of PF 1 ranges from 2-3 1/2 to 2+3 1/2.

PF 1 absolute value +PF2 absolute value =4 PF 1 absolute value multiplied by PF2 absolute value.