Forgot password
 Register account
View 1956|Reply 3

[几何] 过椭圆外一点的割线截椭圆长度

[Copy link]

461

Threads

957

Posts

4

Reputation

Show all posts

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
\[l=\dfrac{2\sqrt{\Delta_1}}{\dfrac{p^2}{a^2}+\dfrac{q^2}{b^2}}\]

13

Threads

898

Posts

8

Reputation

Show all posts

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?

673

Threads

110K

Posts

218

Reputation

Show all posts

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

461

Threads

957

Posts

4

Reputation

Show all posts

original poster 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*}

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-15 14:13 GMT+8

Powered by Discuz!

Processed in 0.015132 seconds, 26 queries