过椭圆x^2/4+y^2/3=1的右焦点F2作一倾斜角为派/4的直线交与该椭圆与...
发布网友
发布时间:2024-10-16 16:19
我来回答
共1个回答
热心网友
时间:2024-11-10 17:39
解决方法:①题意,C
=√^
2-B
^
2
=
1,∴F2(1,0)的k
=tanπ/
4
=
1
∴直线方程为y-0
=
1(x-1的),它是为y
=
x-1的为y
=
x-1的生成到椭圆x
2/4
+
Y
2/3
=
1以简化的精加工7倍^
2-8倍速8
=
0
|
AB
|
=√1
+1
^
2
*√(8/7)^
2-4
*
-8
/
7
=
24/7
(
2)易问F1的直线距离D
=
|
-1-1
|
/√1
^
2
+(-1)^
2
=√2
∴S△AF1B
=
1
/
2
*
24/7
*√2
=
12√2/7
③7倍^
2-8倍速-8
=
0有X1
+
X2
=
8/7横坐标弦AB的中点(1次+×2)/
2
=
4/7
Y1
=的x1-1
Y2
=
x2的-1
Y1
+
Y2
=×1
+×2-2
=
10
-6
/
7
所以弦AB的中点的垂直坐标(Y1
+
Y2)/
2
=
-3
/
7
AB中点的坐标为(4/7,-3
/
7)