(2) Let the midpoint be (x, y), a (x 1, y 1) and b (x2, y2), and the linear equation can be divided into two cases: the dip angle α=π/2, α belongs to [0, π/2] and (π/2, π).
The linear equation of the latter is y=(x- 1)tanα, the linear equation and the elliptic equation are simultaneous equations, and the unary quadratic equation about x after eliminating y,
From the relationship between root and coefficient and the formula of midpoint coordinate, we can get the midpoint X coordinate, and substitute it into a linear equation to get Y. The two coordinate values are a parametric equation about tanα, the value of X is [0, 1], and the elimination parameter is an elliptic equation. Please verify the special midpoint of α=π/2.
(3) Let AB's linear equation y=(x- 1)tanα and the midpoint M(x0, y0), then the midline equation of the line segment AB is written obliquely from this point, and the linear equation AB is connected with the elliptic equation. The coordinate of the midpoint x0 can be obtained from the relationship between roots and coefficients and the coordinate formula of the midpoint, and y0 can be obtained by substituting it into the linear equation. These two coordinate values are parametric equations about tanα, and then brought into AB.