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.