Given that a is a real number, the intersection of the line 2x-y + 5 = 0 and the line X-Y + A + 4 = 0 is not on the ellipse x2 + 2Y2 = 11, the value range of a is obtained

Given that a is a real number, the intersection of the line 2x-y + 5 = 0 and the line X-Y + A + 4 = 0 is not on the ellipse x2 + 2Y2 = 11, the value range of a is obtained

The intersection point of two straight lines is the solution of the system of equations 2x − y + 5 = 0x − y + A + 4 = 0, where (x, y) = (A-1, 2a-3). The point is not on the ellipse x2 + 2Y2 = 11 if and only if (A-1) 2 + 2 (2a-3) 2 = 11, the solution is a = - 2, or a = − 49, ｜ a ≠ - 2 and a ≠ - 49

It is known that the parabola y & # 178; = 4x, the ellipse X & # 178 / / M + Y & # 178 / / 8 = 1, they have the common focus F2, and compared with P and Q,
F1 is the other intersection of the ellipse
Find the value of (1) M
(2) The coordinates of P and Q
(3) Area of △ pf1f2

A:
1）
Parabola y & # 178; = 4x = 2px, P = 2
Focus F2 (1,0), guide line x = - 1
Ellipse X & # / M + Y & # / 8 = 1
Focus F2 (1,0), C = 1
So: M-8 = C & # 178; = 1
The solution is m = 9
2）
Y & # 178; = 4x is substituted into ellipse X & # 178 / / 9 + Y & # 178 / / 8 = 1 to get:
x²/9+4x/8=1
2x²+9x-18=0
(2x-3)(x+6)=0
x1=3/2,x2=-6
Because: x > = 0
So: x = 3 / 2 is substituted into Y & # 178; = 4x to get y & # 178; = 6, y = √ 6 or y = - √ 6
So: points P and Q are (3 / 2, √ 6), (3 / 2, √ 6)
3）
Focus F1 (- C, 0) = (- 1,0)
So: F1F2 = 2
So: S = F1F2 * distance from point P to X axis / 2 = 2 * √ 6 / 2 = √ 6
So: the area of triangle pf1f2 is √ 6