求解圆锥曲线证明

nttz

求解，用平面去截取圆锥，然后截面按轴线与平面的夹角分类讨论，证明它们分别是什么曲线，能否从立体几何的角度说明满足它们各自的定义

kuing

自行百度“Dandelin双球”

hbghlyj

For the analytic treatment, see M4Lectures21.pdf.zip (1020.02 KB, Downloads: 57)

2022-6-15 06:07 Upload

Click on the file to download the attachment

Chapter 4 Conics: start from page 23

Intersection of the cone with the plane: start from bottom of page 27

Example 52 Let $0<\theta, \alpha<\pi / 2$. Show that the intersection of the cone $x^{2}+y^{2}=z^{2} \cot ^{2} \alpha$ with the plane $z=\tan \theta(x-1)$ is an ellipse, parabola or hyperbola. Determine which type of conic arises in terms of $\theta$.

Solution We will denote as $C$ the intersection of the cone and plane. In order to properly describe $C$ we need to set up coordinates in the plane $z=\tan \theta(x-1)$. Note that
$$
\mathbf{e}_{1}=(\cos \theta, 0, \sin \theta), \quad \mathbf{e}_{2}=(0,1,0), \quad \mathbf{e}_{3}=(\sin \theta, 0,-\cos \theta),
$$
are mutually perpendicular unit vectors in $\mathbb{R}^{3}$ with $\mathbf{e}_{1}, \mathbf{e}_{2}$ being parallel to the plane and $\mathbf{e}_{3}$ being perpendicular to it. Any point $(x, y, z)$ in the plane can then be written uniquely as
$$
(x, y, z)=(1,0,0)+X \mathbf{e}_{1}+Y \mathbf{e}_{2}=(1+X \cos \theta, Y, X \sin \theta)
$$
for some $X, Y$, so that $X$ and $Y$ then act as the desired coordinates in the plane. Substituting the above expression for $(x, y, z)$ into the cone's equation $x^{2}+y^{2}=z^{2} \cot ^{2} \alpha$ gives
$$
(1+X \cos \theta)^{2}+Y^{2}=(X \sin \theta)^{2} \cot ^{2} \alpha
$$
This rearranges to
$$
\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right) X^{2}+2 X \cos \theta+Y^{2}=-1
$$
If $\theta \neq \alpha$ then we can complete the square to arrive at
$$
\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)^{2}}{\sin ^{2} \theta \cot ^{2} \alpha}\left(X+\frac{\cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha}\right)^{2}+\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)}{\sin ^{2} \theta \cot ^{2} \alpha} Y^{2}=1
$$
If $\theta<\alpha$ then the coefficients of the squares are positive and we have an ellipse while if $\theta>\alpha$ we have a hyperbola as the second coefficient is negative. If $\theta=\alpha$ then our original equation has become
$$
2 X \cos \theta+Y^{2}=-1
$$
which is a parabola. Further calculation shows that the eccentricity of the conic is $\sin \theta / \sin \alpha$.$■$

hbghlyj

The size limit for attachment (1MB) is a bit annoying. The file is 1061 KB, which means I have to zip it before uploading :(

hbghlyj

Further calculation shows that the eccentricity of the conic is $\sin\theta\over\sin\alpha$

My calculation:$\require{physics}$$${b^2\over a^2}=\flatfrac{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)^{2}}{\sin ^{2} \theta \cot ^{2} \alpha}}{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)}{\sin ^{2} \theta \cot ^{2} \alpha}}=\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha$$$$e=\sqrt{1-{b^2\over a^2}}=\sqrt{\sin^{2} \theta+\sin ^{2} \theta \cot ^{2} \alpha}=\sin\theta\sqrt{1+\cot^2α}=\frac{\sinθ}{\sin\alpha}$$
The result of the macro $\verb|\flatfrac|$ provided by MathJax extension $\verb|physics|$ looks nice.

nttz

hbghlyj 发表于 2022-6-15 06:02
For the analytic treatment, see

Chapter 4 Conics: start from page 23

全部英文啊，，有中文版么

nttz

kuing 发表于 2022-6-14 22:29
自行百度“Dandelin双球”

看了很多Dandelin双球 的解释，这些都是基于它存在的情况下，如何知道任意截面下与圆锥之间一定存在同时相切的球体呢

nttz

hbghlyj 发表于 2022-6-15 06:28
My calculation:$\require{physics}$$${b^2\over a^2}=\flatfrac{\frac{\left(\cos ^{2} \theta-\sin ^{2} ...

能不能用中文啊