题目内容

(本题满分15分) 已知直线l1xmy与抛物线C:y2=4x交于O (坐标原点),A两点,直线l2xmym 与抛物线C交于BD两点.

(Ⅰ) 若 | BD | = 2 | OA |,求实数m的值;

(Ⅱ) 过ABD分别作y轴的垂线,垂足分别为A1B1D1.记S1S2分别为三角形OAA1和四边形BB1D1D的面积,求的取值范围.

 

【答案】

 

(Ⅰ) m

(Ⅱ) 的取值范围是(0,1)∪(1,+∞)

【解析】(Ⅰ) 解: 设B(x1y1), D(x2y2),

 得

Δ,得

y1y2=4m y1y2=-4m

又由 得

y2-4my=0,

所以y=0或4m.[来源:学&科&网]

A (4m2,4m).

由 | BD |=2 | OA |,得

(1+m2)(y1y2)2=4 (16m4+16m2),

而 (y1y2)2=16m2+16m

m.        ………………………… 6分

(Ⅱ) 解: 由(Ⅰ)得

x1x2m(y1y2)+2m=4m2+2m

 所以 

t

因为

所以-1<t<0或t>0.

所以 0<<1 或 >1,

即 0<<1 或 >1.

所以,的取值范围是(0,1)∪(1,+∞).  ………………………15分

 

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