题目内容
设F为椭圆
+
=1的右焦点,过椭圆中心作一直线与椭圆交于P,Q两点,当三角形PFQ的面积最大时,
•
的值为______.
| x2 |
| 4 |
| y2 |
| 3 |
| PF |
| QF |
椭圆
+
=1的右焦点F(1,0)
①当直线PQ的斜率存在时,设直线PQ的方程为y=kx(k≠0)
代入椭圆方程可得,x2=
PQ=
=
原点到AB的距离d=|
|
S=
d×PQ=|
×
×
|=|
|=
<
②当直线的斜率不存在时,P(0,
),Q(0,-
),S=
×2
×1=
Smax=
,此时
=(1,-
),
=(1,
)
∴
•
=1×1-
×
=-2
故答案为:-2
| x2 |
| 4 |
| y2 |
| 3 |
①当直线PQ的斜率存在时,设直线PQ的方程为y=kx(k≠0)
代入椭圆方程可得,x2=
| 12 |
| 3+4k2 |
PQ=
| (1+k2)[(x1+x2)2-4x1x2] |
|
原点到AB的距离d=|
| k | ||
|
S=
| 1 |
| 2 |
| 1 |
| 2 |
| k | ||
|
|
2
| ||
|
2
| ||||
|
| 3 |
②当直线的斜率不存在时,P(0,
| 3 |
| 3 |
| 1 |
| 2 |
| 3 |
| 3 |
Smax=
| 3 |
| PF |
| 3 |
| QF |
| 3 |
∴
| PF |
| QF |
| 3 |
| 3 |
故答案为:-2
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