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发表于 2023-5-30 02:23
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本帖最后由 dodonaomikiki 于 2023-5-30 02:29 编辑
\begin{align*}
'PROOF\\
Set \qquad \ell: x+2=t(y+1)\\
\begin{cases} x+2=t(y+1) \\ x^2+4y^2=4 \end{cases}\\
\Longrightarrow (t^2+4)y^2+(2t^2-4t)y+(t^2-4t)&=0\\
Set:A(x1, y1), B(x2, y2)\\
\Longrightarrow y1+y2&=\frac{ 4t -2t^2 }{ t^2+4 }, y1y2=\frac{ t^2-4t }{ t^2+4 }\\
And \qquad set:\\
C( x_C,-1 ), D( x_D, -1 )\\
Cauz \qquad (PAC) \qquad are \qquad co-llinear\\
\Longrightarrow \frac{ y1}{ x1+2} &= \frac{ -1}{ x_C+2 } \\
\Longrightarrow x_C&= \frac{-x1-xy1-2 }{y1} \\
&= \frac{( -ty1-t+2 )-2y1-2 }{y1} \\
&= \frac{( -t-2 )y1-t }{y1} \\
\end{align*}
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