Obviously, △ABO is a right triangle with AB as the hypotenuse, m is the midpoint of AB, | vector om | = √ 5/2,

∴|OM|=√5/2,∴|AB|=√5。

According to Pythagorean theorem, there are | OA | 2 +| OB | 2 =| AB | 2, ∴ A 2+B 2 = 5. ①

∵e=c/a=√3/2,∴2c=√3a,∴4c^2=3a^2,∴4(a^2-b^2)=3a^2,∴a^2=4b^2。 ②

Substituting ② into ① gives 4b 2+b 2 = 5, ∴ b 2 = 1, ∴ a 2 = 4.

∴ The elliptic equation satisfying the condition is: x 2/4+y 2 =1.

The second question:

1. When the straight line is PQ⊥x axis, the equation of PQ is obviously: X =- 1.

Let x =- 1 in x 2/4+y 2 =1,and we get:1/4+y 2 =1,∴ y 2 = 3/4, ∴ y.

Without loss of generality, let p be above the x axis and then q be below the x axis.

∴ The coordinates of points P and Q are (-1, √3/2) and (-1, -√ 3/2) respectively.

If the point (-1, 0) is C, then: | OC | = 1, the distances from points P and Q to OC are both √3/2.

∴△POC area =△ QOC area = (1/2) | OC |× (√ 3/2) = √ 3/4.

∴△POQ area =△ POC area +△ QOC area = =2△POC area = √ 3/2.

Second, when PQ has a slope, the slope is not 0, otherwise the straight lines of P, O and Q*** cannot form a triangle.

When PQ has a non-zero slope, let the slope be k, then the equation of: l is: y = k (x+1) = kx+k.

∵ both p and q are on a straight line Y = KX+K, ∴ let the coordinates of p and q be (m, km+k) and (n, KN+k) respectively.

Simultaneous: y = kx+k, x 2/4+y 2 =1,if y is eliminated, x 2/4+(kx+k) 2 =1,

∴x^2+4k^2x^2+8k^2x+4k^2- 1=0,∴( 1+4k^2)x^2+8k^2x+4k^2- 1=0。

Obviously, m and n are the roots of the equation (1+4k2) x2+8k2x+4k2-1= 0. According to Vieta's theorem, there are:

m+n=-8k^2/( 1+4k^2)、mn=(4k^2- 1)/( 1+4k^2)。

According to the distance formula between two points, there are:

|pq|=√[(m-n)^2+(km+k-kn-k)^2]=√[( 1+k^2)(m-n)^2]。

Kx-y+k = 0 is obtained from the equation y = kx+k of PQ.

The formula of the distance from point to straight line is: the distance from point O to PQ is d = | k |/√ (k 2+ 1).

So:

△ area of △POQ

=( 1/2)|pq|d=( 1/2)[|k|/√(k^2+ 1)]√[( 1+k^2)(m-n)^2]

=( 1/2)|k|√(m-n)^2=( 1/2)|k|√[(m+n)^2-4mn]

=( 1/2)|k|√{[-8k^2/( 1+4k^2)]^2-4(4k^2- 1)/( 1+4k^2)}

=|k|√{[4k^2/( 1+4k^2)]^2-(4k^2- 1)/( 1+4k^2)}

=|k|√[ 16k^4-( 16k^4- 1)]/( 1+4k^2)=|k|/( 1+4k^2)

= 1/(4|k|+ 1/|k|).

∵4|k|| 1/|k|≧2√[4|k|( 1/|k|)]=4,∴ 1/(4|k|+ 1/|k|)≦ 1/4。

∴ When 4 | k | =1| k |, the maximum area of △POQ is 1/4.

From 4 | k | =1| k |, we get | k | = 1/4, ∴ k = 1/2, or k =- 1/2.

∴ When the area of △POQ has a maximum, the equation of straight line L is:

Y = (1/2) (X+ 1), or Y =-( 1/2) (X+ 1),

Namely: x-2y+ 1 = 0, or x+2y+ 1 = 0.