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[几何] 过椭圆外一点的割线截椭圆长度

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青青子衿 Posted 2019-2-26 14:31 |Read mode
Last edited by 青青子衿 2019-6-15 20:06考察“\(\text{X}\)射线”经过边界曲线形如标准椭圆\(\,C\,\)均匀介质\(\,D\,\)的长度.
\[C\colon\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\]
设“\(\text{X}\)射线”经过点\(\left(x_{\overset{\,}0},y_{\overset{\,}0}\right)\),方向角度为\(\,\varphi\),则射线的参数方程可写为(其中\(p=\cos\varphi\),  \(q=\sin\varphi\))
\begin{cases}
x=x_{\overset{\,}0}+p\,t\\
y=y_{\overset{\,}0}+q\,t
\end{cases}
当\(\Delta_1\ge0\)时,“\(\text{X}\)射线”经过均匀介质\(\,D\,\)(即与椭圆\(\,C\,\)相割),其中
\[\Delta_1=\dfrac{p^2}{a^2}+\dfrac{q^2}{b^2}-\dfrac{\left(x_{\overset{\,}0}q-y_{\overset{\,}0}p\right)^2}{a^2b^2}\]
“\(\text{X}\)射线”经过均匀介质\(\,D\,\)的长度为
\[l=\dfrac{2\sqrt{\Delta_1}}{\dfrac{p^2}{a^2}+\dfrac{q^2}{b^2}}\]

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色k Posted 2019-2-26 14:36
这……似曾相识,你以前是不是发过?在论坛还是群?

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kuing Posted 2019-2-26 15:08
原来是在群里……
QQ截图20190226150719.jpg

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 Author| 青青子衿 Posted 2019-6-15 19:34
\begin{gather*}
x_{\overset{\,}1}=\frac{\frac{q^2u}{b^2}-\frac{pqv}{b^2}+p\sqrt{\frac{p^2}{a^2}+\frac{q^2}{b^2}-\frac{\left(qu-pv\right)^2}{a^2b^2}}}{\frac{p^2}{a^2}+\frac{q^2}{b^2}}\\
y_{\overset{\,}1}=\frac{\frac{p^2v}{a^2}-\frac{pqu}{a^2}+q\sqrt{\frac{p^2}{a^2}+\frac{q^2}{b^2}-\frac{\left(qu-pv\right)^2}{a^2b^2}}}{\frac{p^2}{a^2}+\frac{q^2}{b^2}}\\
\\
\frac{{x_{\overset{\,}1}}^2}{a^2}+\frac{{y_{\overset{\,}1}}^2}{b^2}=1
\end{gather*}

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