|
kuing
发表于 2014-3-17 14:40
\begin{align*}
& ax^2-xy+b=0 \\
\longrightarrow{} & a(x\cos\alpha+y\sin\alpha)^2-(x\cos\alpha+y\sin\alpha)(y\cos\alpha -x\sin\alpha)+b=0 \\
\iff{} & (a\cos^2\alpha+\sin\alpha\cos\alpha)x^2+(a\sin^2\alpha-\sin\alpha \cos\alpha)y^2+(2a\sin\alpha\cos\alpha+\sin^2\alpha-\cos^2\alpha)xy+b=0,
\end{align*}
当 $2a\sin\alpha\cos\alpha+\sin^2\alpha-\cos^2\alpha=0$ 时
\begin{align*}
&(a\cos^2\alpha+\sin\alpha\cos\alpha)(a\sin^2\alpha-\sin\alpha\cos\alpha) \\
={}&\frac14(2a\sin\alpha\cos\alpha+2\sin^2\alpha)(2a\sin\alpha\cos\alpha -2\cos^2\alpha) \\
={}&\frac14(\cos^2\alpha-\sin^2\alpha+2\sin^2\alpha)(\cos^2\alpha -\sin^2\alpha-2\cos^2\alpha) \\
={}&{-}\frac14,
\end{align*}
因 $2a\sin\alpha\cos\alpha+\sin^2\alpha-\cos^2\alpha=0$ 必有解,故若 $b\ne0$,则此时必为双曲线。 |
|