The elliptic equation is

x？ /2+y？ = 1

(2) Let x= 1 and get Y = √ 2/2, so there is B( 1, √ 2/2). P ( 1，m)，0

So the equation of straight line AC is

y=k(x- 1)+m，0 & ltm & lt√2/2

Substitute into elliptic equation and simplify.

x？ +2[k？ (x？ -2x+ 1)+2km(x- 1)+m？ ]-2=0

(2k？ + 1)x？ -(4k？ -2km)x+(2k？ -2km+m？ -2)=0 ①

From the meaning of the question, we can see that the two roots x 1 and x2 of the unary quadratic equation ① about x should satisfy:

x 1+x2=(4k？ -2km)/(2k？ + 1)=2

Get km=- 1

So k=- 1/m

And the discriminant δ = [-(4k? -2km)]？ -4(2k？ + 1)(2k？ -2km+m？ -2)>0

Replace it with km=- 1

(2k？ + 1)? -(2k？ + 1)(2k？ +2+m？ -2)>0

(2k？ + 1)[(2k？ + 1)-(2k？ +m？ )]>0

(2k？ + 1)( 1-m？ )>0

Obviously, it is established.

Therefore, there is k =-1/m.

(3)ABCF2 is a parallelogram. Since AP=PC and BP=PF2 are required, there are m=√2/4 and k=-2√2.

Because (Vieta's theorem):

x 1+x2=2

x 1x2=(2k？ -2km+m？ -2)/(2k？ + 1)= 129/ 136

So |AC|=√( 1+k? )*|x2-x 1|=√{( 1+k？ )[(x 1+x2)？ -4x 1x2]}

=√[( 1+8)(2? -4* 129/ 136)]=3√(7/34)

The inclination angle (angle with the positive direction of the X axis) α of the straight line AC satisfies tanα=-2√2.

So sin < APB = sin (α-90) =-cos α =-{-1√ [(1+(-2 √ 2))? ]}= 1/3

Then the area of the parallelogram

s = 4× 1/2×3 √( 7/34)/2×√2/4× 1/3 =√ 1 19/68