圆锥曲线练习
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)