Course: 10.2 二次曲线极赤道

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum

二次曲线的极平方

allow you to extend your knowledge of conics in a new context. Calculators are an excellent tool for graphing polar conics. What settings do you need to know in order to properly use your calculator?
::允许您在新的上下文中扩展对二次曲线的知识。计算器是绘制极地二次曲线图表的极好工具。您需要知道哪些设置才能正确使用计算器?

Polar Equations of Conics
::二次曲线的极平方

Polar equations refer to the radius $\begin{aligned} r \end{aligned}$ as a function of the angle $\begin{aligned} θ \end{aligned}$ . There are a few typical polar equations you should be able to recognize and graph directly from their polar form .
::极方程式指半径 r 作为角度的函数。有一些典型的极方程式, 您应该能够直接从极形中识别和图形。

The following polar function is a circle of radius $\begin{aligned} \frac{a}{2} \end{aligned}$ passing through the origin with a center at angle $\begin{aligned} β \end{aligned}$ .
::下面的极函数是半径a2 的圆圈,它通过源,以角度β为中心。

$\begin{aligned} r = a \cdot \cos (θ - β) \end{aligned}$
:) ()

There are other ways of representing a circle like this using cofunction identities and .
::也有一些其他方法可以代表像这样的圆圈, 使用共同功能身份和。

, and have a common general polar equation. Just like with the circle, there are other ways of representing these relations using cofunction and coterminal angles; however, this general form is easiest to use because each parameter can be immediately interpreted in a graph. A parameter is a constant in a general equation that takes on a specific value in a specific equation.
::,并有一个共同的普通极方方程式。和圆一样,还有其他方法可以使用共函数和共界角度来表示这些关系;然而,这种一般形式最容易使用,因为每个参数都可以在图表中立即解释。一个参数是一个总方程式中的常数,在特定方程式中以特定值计算。

$\begin{aligned} r = \frac{k \cdot e}{1 - e \cdot \cos (θ - β)} \end{aligned}$
::r=ke1-ecos()

One of the focus points of a conic written in this way is always at the pole (the origin). The angle $\begin{aligned} β \end{aligned}$ indicates the angle towards the center if the conic is an ellipse, the opening direction if the conic is a parabola and the angle away from the center if the conic is a hyperbola. The eccentricity $\begin{aligned} e \end{aligned}$ should tell you what conic it is. The constant $\begin{aligned} k \end{aligned}$ is the distance from the focus at the pole to the nearest directrix. This directrix lies in the opposite direction indicated by $\begin{aligned} β \end{aligned}$ .
::以这种方式写成的二次曲线的一个焦点点总是在极( 源) 。角度 β 指向中心的角度, 如果二次曲线是椭圆, 则指向中心的方向; 如果二次曲线是抛光线, 则指向中心的方向; 如果二次曲线是超高波拉, 则指向中心。偏心度应该告诉你什么是二次曲线。常数 k 是极点与最近的直线之间的距离。这个直线指位于 β 表示的相反方向。

There are many opportunities for questions involving partial information with polar conics. A few relationships that are often useful for solving these questions are:
::有很多机会可以用极地二次曲线提供部分信息的问题。

$\begin{aligned} e = \frac{c}{a} = \frac{\bar{P F}}{\bar{P D}} \to \bar{P F} = e \cdot \bar{P D} \end{aligned}$
::\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

Ellipses: $\begin{aligned} k = \frac{a^{2}}{c} - c \end{aligned}$
::椭圆: k=a2c-c

Hyperbolas: $\begin{aligned} k = c - \frac{a^{2}}{c} \end{aligned}$
::超光谱: k=c-a2c

A great way to discover new types of graphs in polar coordinates is to experiment on your own with your calculator. Try to come up with equations and graphs that look similar to the following two polar functions.
::在极地坐标中发现新类型的图表的一个好方法就是用计算器进行自己的实验。尝试用类似于以下两个极函数的方程式和图形来做实验。

The circle in blue has a center at $\begin{aligned} 90^{\circ} \end{aligned}$ and has a diameter of 2. Its equation is $\begin{aligned} r = 2 \cos (θ - 90^{\circ}) . \end{aligned}$
::蓝色圆以90°C为中心,直径为2°C,方程式为 r=2cos(90°C)。

The red ellipse appears to have center at (2, 0) with $\begin{aligned} a = 4 \end{aligned}$ and $\begin{aligned} c = 2 \end{aligned}$ . This means the eccentricity is $\begin{aligned} e = \frac{1}{2} \end{aligned}$ . In order to write the equation in polar form you still need to find $\begin{aligned} k \end{aligned}$ .
::红色椭圆似乎在 2, 0) 和 a=4 和 c=2 的中心处, 这意味着偏心度是 e=12。要以极形写出方程式, 您仍然需要找到 k 。

$\begin{aligned} k = \frac{a^{2}}{c} - c = \frac{4^{2}}{2} - 2 = 8 - 2 = 6 \end{aligned}$
::k=a2c-c=422-2=8-2=6

Thus the equation for the ellipse is:
::因此,椭圆的方程是:

$\begin{aligned} r = \frac{6 \cdot \frac{1}{2}}{1 - \frac{1}{2} \cdot \cos (θ)} \end{aligned}$
::r=6121-12cos()

Examples
::实例

Example 1
::例1

Earlier, you were asked about how to use your calculator to . Most calculators have a polar coordinate mode. On the TI-84, the mode can be switched to polar in the mode menu. This changes the graphing features. You can choose to be in radians or degrees and graphs will look the same. When you graph a circle of the form $\begin{aligned} r = 8 \cdot \cos θ \end{aligned}$ . you should see the following on your calculator.
::早些时候,有人询问您如何使用计算器。大多数计算器都有极地坐标模式。在 TI-84 上,该模式可以在模式菜单中转换为极地。这可以改变图形特征。您可以选择以弧度或度表示,而图形将看起来相同。当您绘制表单 r=8cos\\\\ 的圆时,您应该在计算器上看到以下内容。

When you go to the window setting you should notice that in addition to $\begin{aligned} X_{m i n}, X_{m a x} \end{aligned}$ there are new settings called $\begin{aligned} θ_{m i n}, θ_{m a x} \end{aligned}$ and $\begin{aligned} θ_{s t e p} \end{aligned}$ .
::当您去窗口设置时, 您应该注意到, 除了 Xmin, Xmax 外, 还有新的设置, 叫做 min, max 和 step 。

If $\begin{aligned} θ_{m i n} \end{aligned}$ and $\begin{aligned} θ_{m a x} \end{aligned}$ do not span an entire period, you may end up missing part of your polar graph.
::如果 min 和 max 不跨越整个周期, 最终可能会丢失极图的一部分。

The $\begin{aligned} θ_{s t e p} \end{aligned}$ controls how accurate the graph should be. If you put $\begin{aligned} θ_{s t e p} \end{aligned}$ at a low number like 0.1 the graph will plot extremely slowly because the calculator is doing 3600 cosine calculations. On the other hand if $\begin{aligned} θ_{s t e p} = 30 \end{aligned}$ then the calculator will do fewer calculations producing a rough circle, but probably not accurate enough for your purposes.
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 2
::例2

Identify the center, foci, vertices and equations of the directrix lines for the following conic:
::确定下列二次曲线的直线的中心、角、顶点和方程式:

$\begin{aligned} r = \frac{20}{4 - 5 \cdot \cos (θ - \frac{3 π}{4})} \end{aligned}$
::r=204 - 5cos (34)

First the polar equation needs to be in graphing form. This means that the denominator needs to look like $\begin{aligned} 1 - e \cdot \cos (θ - β) . \end{aligned}$
::首先极方程式需要以图形形式显示。这意味着分母需要看起来像 1 - ecos 。

\begin{aligned} r = \frac{20}{4 - 5 \cdot \cos (θ - \frac{3 π}{4})} \cdot \frac{\frac{1}{4}}{\frac{1}{4}} = \frac{5}{1 - \frac{5}{4} \cdot \cos (θ - \frac{3 π}{4})} = \frac{4 \cdot \frac{5}{4}}{1 - \frac{5}{4} \cdot \cos (θ - \frac{3 π}{4})} \\ e = \frac{5}{4}, k = 4, β = \frac{3 π}{4} = 135^{\circ} \end{aligned}

34)1414=51-544)=4541 - 544)e=54,k=4,k=4,34=135

Using this information and the relationships you were reminded of in the guidance, you can set up a system and solve for $\begin{aligned} a \end{aligned}$ and $\begin{aligned} c \end{aligned}$ .
::使用此信息以及指南中提醒您的关系, 您可以设置一个系统, 并解决 a 和 c 。

\begin{aligned} 4 & = c - \frac{a^{2}}{c} \\ \frac{5}{4} & = \frac{c}{a} \to \frac{4}{5} = \frac{a}{c} \to \frac{4 c}{5} = a \\ 4 & = c - {(\frac{4 c}{5})}^{2} \cdot \frac{1}{c} \\ 4 & = c - \frac{16 c^{2}}{25 c} \\ 4 & = \frac{9 c}{25} \\ \frac{100}{9} & = c \\ \frac{80}{9} & = a \end{aligned}

::4=c-a2c54=ca45=ac4c5=a4=c-(4c5)21c4=c-16c225c4=9c251009=c809=a

The center is the point $\begin{aligned} (\frac{100}{9}, \frac{7 π}{4}) \end{aligned}$ which is much more convenient to write in polar coordinates. The closest directrix is the line $\begin{aligned} r = 4 \cdot \sec (θ - \frac{7 π}{4}) \end{aligned}$ . The other directrix is the line $\begin{aligned} r = (2 \cdot \frac{100}{9} - 4) \cdot \sec (θ - \frac{7 π}{4}) \end{aligned}$ . One focus is at the pole, the other focus is the point $\begin{aligned} (\frac{200}{9}, \frac{7 π}{4}) \end{aligned}$ . The vertices are at the center plus or minus $\begin{aligned} a \end{aligned}$ in the same angle: $\begin{aligned} (\frac{100}{9} \pm \frac{80}{9}, \frac{7 π}{4}) \end{aligned}$
::中心是极坐标( 109, 74) 比较方便以极坐标书写。最近的直线是 r= 4sec( 74) 。最近的直线是 r = ( 21009 - 4) sec( 74) 。一个焦点是极线, 另一个焦点是点(2009, 74)。顶部位于中心, 加上或减去一个相同角度: (1009+809, 74) 。

Putting all this information together, the graph of the conic is:
::将所有这些信息汇总在一起, 二次曲线的图示是:

Example 3
::例3

Convert the following conic from polar form to rectangular form .
::将以下的二次曲线从极形转换为矩形形。

$\begin{aligned} r = \frac{3}{2 - \cos θ} \end{aligned}$
::r=32-cos

There are many ways to convert from polar form to rectangular form. You should become comfortable with the algebra.
::从极形转换成矩形有许多方法。您应该对代数感到舒服。

\begin{aligned} r & = \frac{3}{2 - \cos θ} \\ r (2 - \cos θ) & = 3 \\ 2 r - r \cdot \cos θ & = 3 \\ 2 r & = 3 + r \cdot \cos θ = 3 + x \\ 4 r^{2} & = 9 + 6 x + x^{2} \\ 4 (x^{2} + y^{2}) & = 9 + 6 x + x^{2} \\ 4 x^{2} + 4 y^{2} & = 9 + 6 x + x^{2} \\ 3 x^{2} + 6 x + 4 y^{3} & = 9 \\ 3 (x^{2} + 2 x + 1) + 4 y^{2} & = 9 + 3 \\ 3 (x + 1)^{2} + 4 y^{2} & = 12 \\ \frac{(x + 1)^{2}}{4} + \frac{y^{2}}{3} & = 1 \end{aligned}

::r=32-cosr(2-cos)=32r-rcos3=3+rcos}3+x4r2=9+6xx+x24xxx24(x2+y2)=9+6x+x24x2+4y2=9+6x+x232+6x2+6x6x+4y3=93(x2+2x+4x+1)+4y2=9+33(x+1)2+4y2=12(x+124+y23=1)

Example 4
::例4

Graph the following conic.
::绘制以下二次曲线图。

$\begin{aligned} r = \frac{3}{2 - \cos (θ - 30^{\circ})} \end{aligned}$
::r=32-cos(30)

Convert to the standard conic form.
::转换为标准二次曲线窗体。

\begin{aligned} r = \frac{3}{2 - \cos (θ - 30^{\circ})} \\ r = \frac{3}{2 - \cos (θ - 30^{\circ})} \cdot \frac{\frac{1}{2}}{\frac{1}{2}} = \frac{\frac{3}{2}}{1 - \frac{1}{2} \cdot \cos (θ - 30^{\circ})} = \frac{3 \cdot \frac{1}{2}}{1 - \frac{1}{2} \cdot \cos (θ - 30^{\circ})} \\ k = 3, e = \frac{1}{2}, β = 30^{\circ} \end{aligned}

::r=32 -cos(30)r=32 -cos(30)r=32 -cos(30)r=121212=321 -12cos(30)=3121 -12cos(30)k=3,e=12,30

Example 5
::例5

Translate the following conic to polar form.
::将以下二次二次曲线转换为极形。

$\begin{aligned} (x - 3)^{2} + (y + 4)^{2} = 25 \end{aligned}$
:x-3)2+(y+4)2=25

Expand the original equation and then translate to polar coordinates:
::展开原始方程式, 然后翻译为极坐标 :

\begin{aligned} (x - 3)^{2} + (y + 4)^{2} = 25 \\ x^{2} - 6 x + 9 + y^{2} + 8 y + 16 = 25 \\ r^{2} - 6 x + 8 y = 0 \\ r^{2} - 6 r \cdot \cos θ + 8 r \cdot \sin θ = 0 \\ r - 6 \cos θ + 8 \sin θ = 0 \\ r = 6 \cos θ - 8 \sin θ \end{aligned}

x-3)2+2+(y+4)2=25x2-6x+9+y2+8y+16=25r2-6x+8y=0r2-6rcos=8r__sin0r-6cos=8sin0r=6cos_8sin=0r=6cos_8sin}

Summary

Polar equations refer to the radius $\begin{aligned} r \end{aligned}$ as a function of the angle $\begin{aligned} θ . \end{aligned}$
::极方程式指半径 r 作为角的函数。

The general polar equation for conics is given by: $\begin{aligned} r = \frac{k \cdot e}{1 - e \cdot \cos (θ - β)}, \end{aligned}$ where $\begin{aligned} k \end{aligned}$ is the distance from the focus to the nearest directrix, $\begin{aligned} e \end{aligned}$ is the eccentricity.
::二次曲线的一般极方程由 r=ke1-ecos() 给出, K 是焦点到最近的直线之间的距离, e 是偏心度。

Useful relationships for polar conics include:
- $\begin{aligned} e = \frac{c}{a} = \frac{\bar{P F}}{\bar{P D}} \to \bar{P F} = e \cdot \bar{P D} \end{aligned}$
  ::\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
- Ellipses: $\begin{aligned} k = \frac{a^{2}}{c} - c \end{aligned}$
  ::椭圆: k=a2c-c
- Hyperbolas: $\begin{aligned} k = c - \frac{a^{2}}{c} \end{aligned}$
  ::超光谱: k=c-a2c
::极地二次曲线的有用关系包括: e=ca=PFP =PP =PF =PF = @ e=PD = Ellipses: k=a2c-c Heperbolas: k=c-a2c

Review
::回顾

Convert the following conics from polar form to rectangular form. Then, identify the conic.
::将以下的二次曲线从极形转换为矩形。然后标明二次曲线。

1. $\begin{aligned} r = \frac{5}{3 - \cos θ} \end{aligned}$
::1. r=53-cos

2. $\begin{aligned} r = \frac{4}{2 - \cos θ} \end{aligned}$
::2. r=42-cos

3. $\begin{aligned} r = \frac{2}{2 - \cos θ} \end{aligned}$
::3. r=22-cos

4. $\begin{aligned} r = \frac{3}{2 - 4 \cos θ} \end{aligned}$
::4.r=32-4cos

5. $\begin{aligned} r = 5 \cos (θ) \end{aligned}$
::5. r=5cos___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Graph the following conics.
::绘制以下二次曲线图。

6. $\begin{aligned} r = \frac{5}{4 - 2 \cos (θ - 90^{\circ})} \end{aligned}$
::6. r=54-2cos(90)

7. $\begin{aligned} r = \frac{5}{3 - 7 \cos (θ - 60^{\circ})} \end{aligned}$
::7. r=53-7cos(60)

8. $\begin{aligned} r = \frac{3}{3 - 3 \cos (θ - 30^{\circ})} \end{aligned}$
::8 r= 33-3cos (30)

9. $\begin{aligned} r = \frac{1}{2 - \cos (θ - 60^{\circ})} \end{aligned}$
::9. r=12-cos(60)

10. $\begin{aligned} r = \frac{3}{6 - 3 \cos (θ - 45^{\circ})} \end{aligned}$
::10.r=36-33cos(45)

Translate the following conics to polar form.
::将以下二次二次曲线转换为极形。

11. $\begin{aligned} \frac{(x - 1)^{2}}{4} + \frac{y^{2}}{3} = 1 \end{aligned}$
::11. (x-1)24+y23=1

12. $\begin{aligned} (x - 5)^{2} + (y + 12)^{2} = 169 \end{aligned}$
::12. (x-5)2+(y+12)2=169

13. $\begin{aligned} x^{2} + (y + 1)^{2} = 1 \end{aligned}$
::13. x2+(y+1)2=1

14. $\begin{aligned} (x - 1)^{2} + y^{2} = 1 \end{aligned}$
::14. (x-1)2+y2=1

15. $\begin{aligned} - 3 x^{2} - 4 x + y^{2} - 1 = 0 \end{aligned}$
::15.-3x2-4x+y2-1=0

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

二次曲线的极平方

Polar Equations of Conics ::二次曲线的极平方

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)