Let x=0, then y= 1, then D(0, 1)

Let y=0, then x=-2, then a (-2,0).

And: A(-a, 0), D(0, b)

Then: a=2, b= 1,

Then: ellipse c: x 2/4+y 2 =1.

Then point b (2 2,0)

Since point S is the moving point above the X axis on ellipse C, let S(2cosα, sinα) (0 < 0.

SA linear equation: y=[(sinα)(x+2)]/(2cosα+2) (point inclination)

SB linear equation: y=[(sinα)(x-2)]/(2cosα-2) (point inclination)

Then the intersection point m (10/3,10 (2cosα+2)/(3sinα)-2 =10/3.

The intersection of SB and X is n (10/3,10 (2cosα-2)/(3sinα)+2) =10/3.

Then | Mn | =| ym-yn | =|10 (2cosα+2)/(3sinα)-2-[10 (2cosα-2)/(3sinα)+2] |

=|40/(3sinα)-4|

Because 0

So when sinα= 1, | Mn | min = [40/(3s in α)-4] min = 40/3-4 = 28/3.

At this time, s (0, 1)

Set the fixed point t (T(2cosβ, sinβ) ((0 < (0

Then |AS|= root number (12+2 2) = root number 5.

As a linear equation: x/2+y= 1 (intercept formula)

Then the distance d from the moving point T to the straight line is = | 2cosβ/2+sinβ-1|/(1/2 * radical number 5).

=2|cosβ+sinβ- 1|/ root number 5

Then the TSB area of the triangle = 1/2*|AS|*d= 1/2* radical number 5*2|cosβ+sinβ- 1|/ radical number 5= 1/5.

| cos β+sin β-1| =1/5, which can be divided into the following two situations:

(1) When cosβ+sinβ- 1=- 1/5, cosβ+sinβ= radical number 2*sin(β+π/4)=4/5, the solution is β+π/4 = kπ+Arcsin (2). Arcsin(2* radical 2/5)π/4; Because 0

So there is such a t-score, and ***8 points are in line with the meaning of the question.