Question

椭圆x2/a2+y2/b2=1(a>b>0)的一个焦点F与抛物线y2=4x的焦点重合，且截抛物线的准线

倾斜角为45度地直线l过点F
（1）求该椭圆的方程（2）设椭圆的另一个焦点为F1，问抛物线y2=4x上是否存在一点M，是的M与F1关于直线l对称，若存在，求出点M的坐标，若不存在，说明理由

Answer 1

(1)
y² = 4x = 2*2x = 2px, p = 2
抛物线焦点F(p/2, 0), 即(1, 0); 准线x = -p/2 = -1
F与抛物线y2=4x的焦点重合, c = 1, a² = b² + c² = b² + 1 (1)
直线l倾斜角为45°, 斜率为k = tan45° = 1; 且过点(1, 0), 其方程为：y - 0 = k(x - 1), y = x - 1
直线l与抛物线的准线总相交，所以“截抛物线的准线倾斜角为45度地直线l过点F”不清楚。 清楚后应当可以列另一个方程，与(1)联立即可解出a, b, 从而得出椭圆的方程。

(2) F1(-1, 0)
设M(r, s)存在，MF1与 y = x - 1垂直，MF1的斜率为-1，其方程为 y - 0 = -(x + 1), y = -x - 1
y = x - 1与y = -x - 1的交点N(0, -1)为MF1的中点:
0 = (-1 + r)/2, r = 1
-1 = (0 + s)/2, s = -2
M(1, -2)
代入y² = 4x, 的确满足，所以M(1, -2)为所求的点。

Answer 2

y^2=4x
2p=4,p/2=1焦点(1,0) 准线x=a^2/c
直线l: y=x-1
(x-1)^2=4x
x^2-2x+1=4x
x^2-6x+9=8
(x-3)^2=8
x=3+2√2=a^2/c c=1
a^2=3+2√2,
b^2=2+2√2
椭圆方程：x^2/(3+2√2)+y^2/(2+2√2)=1

2
过F1(-1,0)斜率-1直线l'：y=-(x+1)
l各l'相交N(0,-1) x-1=-x-1,x=0 ,y=-1
Nx=[Mx+(-1)]/2
Mx=1
Ny=(My)/2
My=-2
M(1,-2)在抛物线上

Answer 3

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