1, an ellipse left alignment line l, intersects with vertical lines AA 1⊥l, BB 1⊥l, and the vertical feet are A 1 and B 1, BH ⊥ aa/kloc-0, respectively.

Accord to that second definition of ellipse,

| af 1 |/| aa 1 | = | BF 1 |/| bb 1 | = e，

| af 1 |/| BF 1 | = | aa 1 |/| bb 1 | = 2，

∴|AA 1|=2|BB 1|，

∵|A 1H|=|BB 1|，

∴|AA 1|=2|A 1H|，

∴H is the midpoint of AA 1,

|AH|=|BB 1|，

HAB=〈AF 1O=60，

∴|AH|/|AB|=cos60 = 1/2，

|BF 1|=|AB|/3，

| bb 1 |/(3bf 1 | = 1/2，

1/(3 | BF 1 | |/bb 1 |)= 1/2，

1/3e= 1/2，

∴e=2/3.

2. OM⊥PQ, vertical ruler m,

According to the formula of focal chord length,

| pq | = (2b2/a)/[1-(ecos θ) 2], (easily proved by the second definition),

Where θ is the angle between the focus chord and the X axis,

The elliptic equation is: x 2/2+y 2 = 1,

a=√2，b= 1，c= 1，

e=c/a=√2/2，

∴|pq|=(2* 1/√2)/[ 1-(cosθ)^2/2]

=2√2/[2-(cosθ)^2]，

|OM|=|OF 1|*sinθ=sinθ，

∴s△pqo=|om|*|pq|/2=√2sinθ/[ 1+ 1-(cosθ)^2]

=√2sinθ/[ 1+(sinθ)^2]，

Let sinθ=t,-1≤t≤ 1,

S=√2t/( 1+t^2)，

St^2-√2t+S=0，

When the discriminant △ = 2-4 s 2 ≥ 0,

S^2≤ 1/2，

-√2/2≤S≤√2/2，

S(max)=√2/2，

At this moment,

√2/2=√2t/( 1+t^2)，

(t- 1)^2=0，

t= 1，

sinθ=π/2，

That is, when PQ is perpendicular to the X axis, the area of triangle POQ is the largest, but at this time |PQ| is the smallest. ?