If the point (e, √3/2) is on an ellipse, it is (e 2/a 2)+(3/b 2) = 1.
The equation of an elliptic circle is x 2/2+y 2 =1.
(2) solution: get f 1 (-10), F2( 1 0) from (10),
And ∵ the straight line AF 1 is parallel to the straight line BF2, ∴ let the equations of AF 1 and BF2 be x+ 1=my, x- 1=my, respectively.
Let A(X 1, Y 1), B(X2, Y2), y 1 > 0, y2 > 0,
∴ From x12/2+y12 =1,x 1+ 1=my 1, we can get (m 2+2) y/kloc.
∴y 1=m+√(2m^2+2)/(m^2+2)
∴|af 1|=[√2(m^2+ 1)+m√(m^2+ 1)]/m^2+2①
Similarly | bf2 | = [√ 2 (m2+1)-m √ (m2+1)]/m2+2 ②
(I)| af 1 |-| bf2 | =[2m √( m ^ 2+ 1)]/(m ^ 2+2)，∴[2m √( m ^ 2+ 1)]/(。
∫ Note that m > 0, ∴ m = √ 2.
∴ The slope of straight line AF 1 is 1/m=(√2)/2.
(2) It is proved that ∵ straight line AF 1 is parallel to ∴PB/PF 1=BF2/AF 1 straight line BF2, that is, PF1= AF1(AF/kloc-0
From point B on the ellipse, BF 1+BF2=2√2, ∴ PF1= AF1/(AF1+BF2) × (2 √ 2-BF2).
Similarly, pf2 = bf2/(af1+bf2) × (2 √ 2-af1).
∴pf 1+pf2=af 1/(af 1+bf2)×(2√2-bf2)+bf2/(af 1+bf2)×(2√2-af 1)=2√2-(2af×bf2)/(af 1+bf2)
From ① ②, AF1+BF2 = 2 √ 2 (m2+1)/(m2+2), AF1× BF2 = (m2+1)/(m2+2).
∴PF 1+PF2=(3√2)/2.
∴PF 1+PF2 is a fixed value.