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. ?