设
A
(
x
1
,
y
1
)
,
B
(
x
2
,
y
2
)
A\left( x_{1}, y_{1} \right), B\left( x_{2}, y_{2} \right)
A(x1,y1),B(x2,y2)
l
:
y
=
k
(
x
+
2
)
l: y = k\left( x+2 \right)
l:y=k(x+2)
显然
y
=
0
y=0
y=0符合题意
当
k
≠
0
k\neq 0
k=0
联立
l
l
l和
C
C
C
(
k
2
+
1
2
)
x
2
+
4
k
2
x
+
4
k
2
−
1
=
0
\left(k^2 + \frac{1}{2}\right)x^2 + 4k^2x + 4k^2 - 1=0
(k2+21)x2+4k2x+4k2−1=0
Δ
>
0
⇒
−
2
2
<
k
<
2
2
\Delta > 0 \Rightarrow - \frac{\sqrt{ 2 }}{2} < k < \frac{\sqrt{ 2 }}{2}
Δ>0⇒−22<k<22
由韦达定理
x
1
+
x
2
=
−
4
k
2
1
2
+
k
2
x_{1}+x_{2} =- \frac{4k^2}{\frac{1}{2} + k^2}
x1+x2=−21+k24k2
A
A
A和
B
B
B的中点为
D
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
D \left( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right)
D(2x1+x2,2y1+y2)
由
G
D
GD
GD和
l
l
l垂直得
y
1
+
y
2
2
+
1
2
x
1
+
x
2
2
−
0
⋅
k
=
−
1
k
x
1
+
x
2
2
+
2
k
+
1
2
x
1
+
x
2
2
−
0
⋅
k
=
−
1
k
2
+
4
k
2
x
1
+
x
2
+
k
x
1
+
x
2
=
−
1
k
2
−
1
2
−
k
2
−
1
2
+
k
2
4
k
=
−
1
k
=
1
±
2
2
\begin{aligned} \frac{\frac{y_{1}+y_{2}}{2} + \frac{1}{2}}{\frac{x_{1}+x_{2}}{2}-0} \cdot k &= -1\\ \frac{k\frac{x_{1}+x_{2}}{2} + 2k + \frac{1}{2}}{\frac{x_{1}+x_{2}}{2}-0} \cdot k &= -1\\ k^2 + \frac{4 k^2}{x_{1} + x_{2}} + \frac{k}{x_{1}+x_{2}} &=-1 \\ k^2 - \frac{1}{2} - k^2 - \frac{\frac{1}{2} + k^2}{4k} &=-1 \\ k &= 1 \pm \frac{\sqrt{ 2 }}{2} \end{aligned}
2x1+x2−02y1+y2+21⋅k2x1+x2−0k2x1+x2+2k+21⋅kk2+x1+x24k2+x1+x2kk2−21−k2−4k21+k2k=−1=−1=−1=−1=1±22
因此 l : y = 0 l: y=0 l:y=0或 y = ( 1 − 2 2 ) ( x + 2 ) y=\left(1-\frac{\sqrt{ 2 }}{2}\right)\left( x+2 \right) y=(1−22)(x+2)
