a=2√3，

| f 1O | = | of 2 | = c = √( 12-3)= 3，

Perpendicular bisector whose y axis is |F 1F2|,

|F 1M|=|MF2|，

|F 1M|=|MF2|=|MP|，

Then F 1, F2 and m are on a circle with m as the center and |F 1P| as the diameter.

< f 1f2p = 90 degrees,

The triangle F 1F2P is a right triangle,

F 1P^2=F 1F2^2+PF2^2，

|F 1F2|=2c=6，

F 1P^2-PF^2=36，

(| pf 1 |+| PF2 |)(| pf 1 |-| PF2 |)= 36，( 1)

|PF 1|+|PF2|=2a=4√3，(2)

( 1)/(2),

(|PF 1|-|PF2|=3√3，(3)

(2)+(3),

2|PF 1|=7√3

|PF 1|=7√3/2，

|PF2|=√3/2，

|PF 1|/|PF2|=7，

∴│PF 1│ is 7 times that of │PF2.