∴kpa 1=y/(x+a),kpa2=y/(x-a)

Judging from the meaning of the question, [y/(x+a)][y/(x-a)]=- 1/4,

That is x 2/a 2+y 2/(a/4) = 1,

∵c=√3,∴a^-(a^2/4)=3

∴ A 2 = 4, and the elliptic equation is x 2/4+y 2 =1.

2. Let the equation of the straight line L be y=kx+b, that is, kx-y+b=0.

∵A and B are on the circle X 2+Y 2 = 4, ∴OA=OB=2.

When ∠AOB = 90°, the area of △AOB is the largest, and the distance from point O to AB (that is, straight line L) is ∠ 2.

That is | b |/√ (k 2+ 1) = √ 2, ∴ b 2 = 2k 2+2 (1).

Substitute y=kx+b into the elliptic equation and get x 2/4+(kx+b) 2 = 1.

Permutation (4k 2+ 1) x 2+8kbx+4b 2-4 = 0.

∵△=(8kb)^2-4(4k^2+ 1)(4b^2-4)=0

∴b^2=4k^2+ 1 (2)

By (1) and (2), k 2 = 1/2, b 2 = 3.

∵k & lt； 0，b & gt0,

∴k=-√2/2,b=√3，

The equation of line l is y=-(√2/2)x+√3.