题目内容
已知抛物线y2=-x与直线y=k(x+1)相交于A、B两点.
(1)求证:OA⊥OB;
(2)当△OAB的面积等于
时,求k的值.
(1)求证:OA⊥OB;
(2)当△OAB的面积等于
| 10 |
(1)由方程y2=-x,y=k(x+1)
消去x后,整理得
ky2+y-k=0.
设A(x1,y1)、B(x2,y2),由韦达定理y1•y2=-1.
∵A、B在抛物线y2=-x上,
∴y12=-x1,y22=-x2,y12•y22=x1x2.
∵kOA•kOB=
•
=
=
=-1,
∴OA⊥OB.
(2)设直线与x轴交于N,又显然k≠0,
∴令y=0,则x=-1,即N(-1,0).
∵S△OAB=S△OAN+S△OBN
=
|ON||y1|+
|ON||y2|
=
|ON|•|y1-y2|,
∴S△OAB=
•1•
=
.
∵S△OAB=
,
∴
=
.解得k=±
.
消去x后,整理得
ky2+y-k=0.
设A(x1,y1)、B(x2,y2),由韦达定理y1•y2=-1.
∵A、B在抛物线y2=-x上,
∴y12=-x1,y22=-x2,y12•y22=x1x2.
∵kOA•kOB=
| y1 |
| x1 |
| y2 |
| x2 |
| y1y2 |
| x1x2 |
| 1 |
| y1y2 |
∴OA⊥OB.
(2)设直线与x轴交于N,又显然k≠0,
∴令y=0,则x=-1,即N(-1,0).
∵S△OAB=S△OAN+S△OBN
=
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 2 |
∴S△OAB=
| 1 |
| 2 |
| (y1+y2)2-4y1y2 |
=
| 1 |
| 2 |
(
|
∵S△OAB=
| 10 |
∴
| 10 |
| 1 |
| 2 |
|
| 1 |
| 6 |
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