If the slope of the straight line passing through point F 1 does not exist, that is, the linear equation is x= 1, at this time, A (1, 3/2), B (1, 3/2), AB ⊥ X axis, AB = 3, F655.

S △ PAB =1/2 * ab * f1p = 3/2.

So the slope of the straight line passing through point F 1 exists, not 0. Let the equation of straight line AB be y=k(x- 1)(k≠0), that is, kx-y-k=0, and let A (x 1, y 1), b (.

The distance of straight line AB at point P (2,0) is

d=|2k-0-k|/√(k？ + 1)=|k|/√(k？ + 1)

By y=kx-k, 3x? +4y？ = 12

get

(4k？ +3)x？ -8k？ x+4k？ - 12=0

x 1+x2=8k？ /(4k？ +3)

x 1x2=(4k？ - 12)/(4k？ +3)

(x 1+x2)？ -4x 1x2

=[64k^4-( 16k？ -48)(4k？ +3)]/(4k？ +3)?

=( 144k？ + 144)/(4k？ +3)?

AB=√( 1+k？ )*√[(x 1+x2)？ -4x 1x2]

= 12(k？ + 1)/(4k？ +3)

S△PAB= 1/2*AB*d

that is

6(k？ + 1)/(4k？ +3)*|k|/√(k？ + 1)=36/ 13

|k|*√(k？ + 1)/(4k？ +3)=6/ 13

k？ (k？ + 1)/(4k？ +3)? =36/ 169

Let x=k? (x≥0)，

x(x+ 1)/(4x+3)？ =36/ 169

36( 16x？ +24x+9)= 169x？ + 169

407x？ +864x+ 155=0

x= 1.93

So k=√x= 1.4.

The linear equation is y= 1.4x- 1.4.