1.( 1) Let the slope of the straight line be k first, list the equation of the straight line y=k(x- 1), bring it into the elliptic equation, and find out two points (not written here). Note that there should be y coordinates at both points at this time. Further analysis, triangle abf2 should be cut into two sections by X axis, both of which can be F 66.

1.(2) In 1, you have discussed K and got the value, so the Y coordinates of A and B are all out. Let's look at the problem again. F 1F2, because it doesn't say who is A, it should be counted as two. At this time, F 1F2 is exactly the two triangles cut out by the X axis mentioned above, and you will know. . .

(1) First of all, the eccentricity is c/a, and it can be inferred from the problem that a=b (the condition of using eccentricity), then the slope of the straight line is k= 1, and then the equation of the straight line is y = x-a. By using the distance formula from point to straight line, the value of A satisfying the second condition can be obtained, and the problem is solved.

In 2 (2) 1, the equation of ellipse is solved, and the value of b is known. Note that the center of the circle is at the position of (0, -b), that is to say, the circle is symmetrical about the Y axis, and the ellipse is also symmetrical about the Y axis, so it is not difficult to know that the ef point is also symmetrical about the Y axis. Then, how can the intersection of a straight line and an ellipse be symmetrical about the Y axis, and the Y axis is obviously parallel to the X axis? . .