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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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