求双曲抛物面上的直母线

青青子衿

求在（等轴）双曲抛物面\(\,x^2-y^2=2z\,\)上，且过点\(P(3,1,4)\)的两条直线（其被称为直母线）

1. Show[ParametricPlot3D[{1/2 (u + v), 1/2 (u - v), 1/2 u v}, {u, -5, 5}, {v, -5, 5}],
2. ParametricPlot3D[{t + 3, t + 1, 2 t + 4}, {t, -5, 5}],
3. ParametricPlot3D[{t + 3, -t + 1, 4 t + 4}, {t, -5, 5}]]

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kuing

要求就求个一般情况吧，空间解析几何我不熟悉，也不知道有没有更快的方法，只想到了以下的方法。

设双曲抛物面为
\[z=\frac{x^2}{a^2}-\frac{y^2}{b^2},\]
其上有一点 `P(x_0,y_0,z_0)`，令 `x=x_0+lt`, `y=y_0+mt`, `z=z_0+nt`，代入方程中化简得
\[\left(\frac{l^2}{a^2}-\frac{m^2}{b^2}\right)t^2+\left(\frac{2lx_0}{a^2}-\frac{2my_0}{b^2}-n\right)t=0,\]
令
\[\frac{l^2}{a^2}-\frac{m^2}{b^2}=\frac{2lx_0}{a^2}-\frac{2my_0}{b^2}-n=0,\]
得到
\[l:m:n=a:b:2\left(\frac{x_0}a-\frac{y_0}b\right)~\text{或}~a:-b:2\left(\frac{x_0}a+\frac{y_0}b\right),\]
由此可见，以下两条直线
\[
\led
x&=x_0+at,\\
y&=y_0+bt,\\
z&=z_0+2\left(\frac{x_0}a-\frac{y_0}b\right)t,
\endled\quad\led
x&=x_0+at,\\
y&=y_0-bt,\\
z&=z_0+2\left(\frac{x_0}a+\frac{y_0}b\right)t,
\endled
\quad\text{（$t$ 为参数）}\]
都在双曲抛物面上。

kuing

看了一下书上证明单叶双曲面为直纹面的方法，这里应该也可以这样做。

将双曲抛物面写为
\[\left( \frac xa-\frac yb \right)\left( \frac xa+\frac yb \right)=z,\]
那么，对于任意实数 `\lambda`, `\mu`，以下两条直线
\[\led
\frac xa-\frac yb&=\lambda,\\
\lambda\left(\frac xa+\frac yb\right)&=z,
\endled\quad\led
\mu\left(\frac xa-\frac yb\right)&=z,\\
\frac xa+\frac yb&=\mu,
\endled\]
均在双曲抛物面上，于是，若 `P(x_0,y_0,z_0)` 在双曲抛物面上，则过 `P` 的两条直母线就是
\[\led
\frac xa-\frac yb&=\frac{x_0}a-\frac{y_0}b,\\
\left(\frac{x_0}a-\frac{y_0}b\right)\left(\frac xa+\frac yb\right)&=z,
\endled\quad\led
\left(\frac{x_0}a+\frac{y_0}b\right)\left(\frac xa-\frac yb\right)&=z,\\
\frac xa+\frac yb&=\frac{x_0}a+\frac{y_0}b,
\endled\]
利用 `z_0=(x_0/a-y_0/b)(x_0/a+y_0/b)` 还可将上式写成
\[\led
\frac xa-\frac yb&=\frac{x_0}a-\frac{y_0}b,\\
\left( \frac xa+\frac yb \right)z_0&=\left( \frac{x_0}a+\frac{y_0}b \right)z,
\endled\quad\led
\left( \frac xa-\frac yb \right)z_0&=\left( \frac{x_0}a-\frac{y_0}b \right)z,\\
\frac xa+\frac yb&=\frac{x_0}a+\frac{y_0}b.
\endled\]
多漂亮的结果。

青青子衿

设双曲抛物面 \[z=\frac{x^2}{a^2}-\frac{y^2}{b^2},\]
...
kuing 发表于 2018-10-27 21:42

回复 2# kuing
谢谢kk，但是好像大部分介绍“双曲抛物面”的书中都是将它的标准方程记作“\(2z=\frac{x^2}{a^2}-\frac{y^2}{b^2}\)”，可能这样规定有同样的效果吧，如同规定平面直角坐标系下抛物线标准方程为“\(y^2=2px\)”

\(\overline{\hspace{15cm}}\)
\[\frac{l^2}{a^2}-\frac{m^2}{b^2}=\frac{2lx_0}{a^2}-\frac{2my_0}{b^2}-n=0,\]
\[
\left\{\begin{array}{r}
\begin{split}
\displaystyle\frac{l}{a}&=\frac{m}{b}\\
\displaystyle\frac{2x_0}{a}\frac{l}{a}-\frac{2y_0}{b}\frac{m}{b}&=n\\
\end{split}
\end{array}\right.
\quad \text{or}\quad
\left\{\begin{array}{r}
\begin{split}
\frac{l}{a}&=-\frac{m}{b}\\
\frac{2x_0}{a}\frac{l}{a}-\frac{2y_0}{b}\frac{m}{b}&=n\\
\end{split}
\end{array}\right.\\
\Rightarrow
\left\{\begin{array}{r}
\begin{split}
\frac{l}{a}&=\frac{m}{b}\\
\left(\frac{2x_0}{a}-\frac{2y_0}{b}\right)\frac{l}{a}&=n\\
\end{split}
\end{array}\right.
\quad \text{or}\quad
\left\{\begin{array}{r}
\begin{split}
\frac{l}{a}&=-\frac{m}{b}\\
\left(\frac{2x_0}{a}+\frac{2y_0}{b}\right)\frac{l}{a}&=n\\
\end{split}
\end{array}\right.\\
\Rightarrow
\left\{\begin{array}{r}
\begin{split}
\frac{l}{a}&=\frac{m}{b}\\
\frac{l}{a}&=\dfrac{n}{\left(\frac{2x_0}{a}-\frac{2y_0}{b}\right)}\\
\end{split}
\end{array}\right.
\quad \text{or}\quad
\left\{\begin{array}{r}
\begin{split}
\frac{l}{a}&=-\frac{m}{b}\\
\frac{l}{a}&=\dfrac{n}{\left(\frac{2x_0}{a}+\frac{2y_0}{b}\right)}\\
\end{split}
\end{array}\right.
\]

kuing

回复 4# 青青子衿

嗯，的确，如果原方程有个2，那么2#后面的表达式少个2，确实会好看一些。
PS、我修改了一下2#，不要分母，原点也适用了。

Czhang271828

根据您的代码, 结果都 almost there 了...

在点 $(3,1,4)$ 局部, 曲面 $S$ 的 (局部) 参数化表示可以为 $S:r(u,v)=(u+v,u-v,2uv)$, 记 $p:=r(2,1)=(3,1,4)$.

考虑从 $r(2,1)$ 出发的正则曲线 $\gamma$, 即满足 $\gamma:[0,L]\to S$, $\gamma(0)=p$, $|\gamma'|\equiv 1$. 从而
$$
\gamma(s)=(u(s)+v(s),u(s)-v(s),2u(s)v(s)).
$$
随后考察 $\gamma''(s)=0$ 之情形, 得 $u''+v''=u''-v''=(uv)''=0$. 解得 $u=a_1s+b_1$, $v=a_2s+b_2$, $a_1a_2=0$. 因此 $u$ 或 $v$ 为常数.

直母线随便写写就行: $u$ 为常数时选 $(3+s,1-s,4+4s)$; $v$ 为常数时选 $(3+s,1+s,4+2s)$.

注: 一般双曲抛物面上的推论也是简单的, 就是写成类似 $(u+v,u-v,kuv)$ 的参数形式然后 $u$ 或 $v$ 等于常数就行了.

单叶双曲面 (如 $x^2+y^2=1+z^2$) 也有类似结论.

hbghlyj

相关帖子 动直线产生的曲面方程
The locus of lines meeting three given skew lines is called a regulus. Gallucci's theorem shows that the lines meeting the generators of the regulus (including the original three lines) form another "associated" regulus, such that every generator of either regulus meets every generator of the other. The two reguli are the two systems of generators of a ruled quadric.

...

Robert J. T. Bell described how the regulus is generated by a moving straight line. First, the hyperboloid $\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}-{\frac {z^{2}}{c^{2}}}=1$ is factored as\[\left({\frac {x}{a}}+{\frac {z}{c}}\right)\left({\frac {x}{a}}-{\frac {z}{c}}\right)\ =\ \left(1+{\frac {y}{b}}\right)\left(1-{\frac {y}{b}}\right).\]
Then two systems of lines, parametrized by $λ$ and $μ$ satisfy this equation:
\[{\frac {x}{a}}+{\frac {z}{c}}\ =\ \lambda \left(1+{\frac {y}{b}}\right),\quad {\frac {x}{a}}-{\frac {z}{c}}\ =\ {\frac {1}{\lambda }}\left(1-{\frac {y}{b}}\right)\] and
\[{\frac {x}{a}}-{\frac {z}{c}}\ =\ \mu \left(1+{\frac {y}{b}}\right),\quad {\frac {x}{a}}+{\frac {z}{c}}\ =\ {\frac {1}{\mu }}\left(1-{\frac {y}{b}}\right).\]
No member of the first set of lines is a member of the second. As $λ$ or $μ$ varies, the hyperboloid is generated. The two sets represent a regulus and its opposite. Using analytic geometry, Bell proves that no two generators in a set intersect, and that any two generators in opposite reguli do intersect and form the plane tangent to the hyperboloid at that point. (page 155).[5]

hbghlyj

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