Solution: (1) According to the topic, the F coordinate is (1, 0), and AC is perpendicular to the X axis, so let point A (1, y), then

K=[y-(-2)]/( 1-0)=7/2, and the solution is: y=3/2.

That is, point A is (1, 3/2), and it is obtained by substituting it into the elliptic equation.

1? /a？ + (3/2)? /b？ = 1, and a? -B? = c？ = 1 。

The two stand together and the solution is: a? = 4，b？ = 3 。 So, the elliptic equation is:

x？ /4 + y？ /3 = 1 .....................................................................①。

(2) According to the topic, the coordinates of point A and point B can be set to (2cosα, √3sinα) and (2cosβ, √3sinβ) respectively, so the coordinates of point C are (2cosα, -√3sinα). Then the equation of straight line AB is:

(y-√3 sinα)/(x-2 cosα)=(√3 sinβ-√3 sinα)]/(2 cosβ-2 cosα)

Let y = 0 of the above formula, simplify the arrangement, and get

X = 2 cos [(α-β)/2]/cos [(α+β)/2], which is the abscissa of point P.

In addition, the equation of the straight line BC is:

[y-(-√3 sinα)]/(x-2 cosα)=[√3 sinβ-(-√3 sinα)]/(2 cosβ-2 cosα)

Let y = 0 of the above formula, simplify the arrangement, and get

X = 2 cos [(α+β)/2]/cos [(α-β)/2], which is the abscissa of point Q.

Note at the same time that the abscissa of point P has the same sign as that of point Q, so

| Op ||| OQ |

= | the abscissa of point p | times the abscissa of point q |

= { 2 cos[(α-β)/2]/cos[(α+β)/2]} { 2 cos[(α+β)/2]/cos[(α-β)/2]}

= 4 (fixed value).

Note that the product of |OP| and |OQ| has nothing to do with the point (0). -2).