节: 超双 | 12.7 超双螺旋体

章节大纲

Two friends who live one mile apart see a lightning strike. After a brief period, one friend hears the thunder, and three seconds later, the other friend hears the thunder. If the lightning struck directly south of the friend who heard the thunder 1st, where did the lightning strike?
::相隔一英里的两位朋友看到闪电袭击。一段短暂的时间后,一位朋友听到雷声,三秒钟后,另一位朋友听到雷声。如果闪电直接击中听到1号雷声的朋友的南面,闪电在哪里击中?

The relationship between the two friends' views of the thunder can be modeled with a hyperbola. We discuss hyperbolas in this section.
::两位朋友对雷声的看法之间的关系可以用双倍波拉模拟。我们在本节讨论双倍波拉。

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Geometry of Hyperbolas
::超高功率的几何

We have considered hyperbolas before in the chapter on rational functions. In this section, we will consider hyperbolas that are not necessarily functions. Like the conic section, a hyperbola can be thought of in terms of a geometric relationship.
::我们以前在关于理性功能的一章中曾考虑过超光波;在本节中,我们将考虑不一定具有功能的超光波。与二次曲线部分一样,超光波可被视为几何关系。

Geometry of a Hyperbola
::超波拉的几何

A hyperbola is the set of all points such that the differences of the distances from the foci are constant.

In the picture below, any point, $\begin{aligned} (x, y), \end{aligned}$ on a hyperbola has the property $\begin{aligned} d_{1} - d_{2} = P \end{aligned}$ , where $\begin{aligned} P \end{aligned}$ is a constant.
::如下图所示,(x,y)任何点,在超高波拉上,其属性为d1-d2=P,其中P为常数。

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Just as with an ellipse, there are two vertices and two foci on the hyperbola. Here they are the two points that are closest to each other on the graph. The line through the vertices and foci is called the transverse axis . Its midpoint is the center of the hyperbola. In this concept, the center will be the origin. There will always be two branches and two asymptotes for any hyperbola .
::和椭圆一样, 双曲线上有两个顶点和两个顶点。这是图形上两个最接近的点。通过顶点和顶点的线被称为横轴。中间点是双曲线的中心。在这个概念中, 中心将是源。任何双曲线总是有两个分支和两个微粒。

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Equations of Hyperbolas
::超重波拉斯平方

The equations for a hyperbola look similar to the equations for an ellipse, but be careful about the operation in between the two terms.
::双倍波拉的方程与椭圆的方程相似,但要注意两个词之间的操作。

Standard Form of an Equation of a Hyperbola
Orientation	Equation	Vertices	Foci	Asymptotes
Horizontal	$\begin{aligned} \frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1 \end{aligned}$	$\begin{aligned} (h \pm a, k) \end{aligned}$	$\begin{aligned} (h \pm c, k) \end{aligned}$	$\begin{aligned} y - k = \pm \frac{b}{a} (x - h) \end{aligned}$
Vertical	$\begin{aligned} \frac{(y - k)^{2}}{a^{2}} - \frac{(x - h)^{2}}{b^{2}} = 1 \end{aligned}$	$\begin{aligned} (h, k \pm a) \end{aligned}$	$\begin{aligned} (h, k \pm c) \end{aligned}$	$\begin{aligned} y - k = \pm \frac{a}{b} (x - h) \end{aligned}$

For a hyperbola centered at the origin, then, the equation will be $\begin{aligned} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ or $\begin{aligned} \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \end{aligned}$ . Notice in the vertical orientation of a hyperbola, the $\begin{aligned} y^{2} \end{aligned}$ term is 1st.
::对于以原产物为主的双倍波拉,则方程式为 x2a2-y2b2=1或 y2a2-x2b2=1. 双倍波拉垂直方向的通知, y2 术语为 1 。

Example 1
::例1

The equation of a hyperbola is $\begin{aligned} 9 y^{2} - 4 x^{2} = 36 \end{aligned}$ . What are the asymptotes and foci of your graph?
::双倍波拉的方程式是 9y2 - 4x2=36。您的图表中的小数和角是什么 ?

Solution: First we need to get the equation in the form $\begin{aligned} \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \end{aligned}$ , so divide by 36.
::解答:首先我们需要以 y2a2-x2b2=1 的形式获得方程式, 所以除以 36 。

\begin{aligned} 9 y^{2} - 4 x^{2} = 36 \\ \frac{9 y^{2}}{36} - \frac{4 x^{2}}{36} = \frac{36}{36} \\ \frac{y^{2}}{4} - \frac{x^{2}}{9} = 1 \end{aligned}

.
::9y2-4x2=369y236-4x236=3636y24-x29=1。

Now we can see that $\begin{aligned} a^{2} = 4 \end{aligned}$ and $\begin{aligned} b^{2} = 9 \end{aligned}$ , so $\begin{aligned} a = 2 \end{aligned}$ and $\begin{aligned} b = 3 \end{aligned}$ . Also, because the y -term comes 1st, the hyperbola is vertically oriented. Therefore, the asymptotes are $\begin{aligned} y = - \frac{a}{b} x \end{aligned}$ and $\begin{aligned} y = \frac{a}{b} x \end{aligned}$ .
::现在我们可以看到 a2=4 和 b2=9, 所以 a=2 和 b=3。另外, 因为y- term 即将到来, 双波拉是垂直方向的。因此, y=- abx 和 y=abx 的数位数是 y=- abx 和 y=abx 。

Substituting for a and b , we get $\begin{aligned} y = - \frac{2}{3} x \end{aligned}$ and $\begin{aligned} y = \frac{2}{3} x \end{aligned}$ .
::替换a和b, 我们得到y=23x和y=23x。

Finally, to find the foci, use $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ .
::最后,要找到角,请使用 c2=a2+b2。

\begin{aligned} c^{2} & = 4 + 9 = 13 \\ c & = \sqrt{13} \end{aligned}

::c2=4+9=13c=13

The foci are $\begin{aligned} (0, \sqrt{13}) \end{aligned}$ and $\begin{aligned} (0, - \sqrt{13}) \end{aligned}$ .
::角度是(0,13)和(0,13)。

Example 2
::例2

Find the vertices, foci, and asymptotes of $\begin{aligned} y^{2} - \frac{x^{2}}{25} = 1 \end{aligned}$ .
::查找 y2 - x225=1 的顶部、福西和零位数。

Solution: First, let's rewrite the equation like this: $\begin{aligned} \frac{y^{2}}{1} - \frac{x^{2}}{25} = 1 \end{aligned}$ . We know that the transverse axis is vertical because the $\begin{aligned} y \end{aligned}$ -term is 1st, making $\begin{aligned} a = 1 \end{aligned}$ and $\begin{aligned} b = 5 \end{aligned}$ . Therefore, the vertices are $\begin{aligned} (0, - 1) \end{aligned}$ and $\begin{aligned} (0, 1) \end{aligned}$ . The asymptotes are $\begin{aligned} y = \frac{1}{5} x \end{aligned}$ and $\begin{aligned} y = - \frac{1}{5} x \end{aligned}$ . Lastly, let's find the foci using $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ .
::解析度: 首先, 让我们重写这样的方程 : y21- x225=1. 我们知道横轴是垂直的, 因为 Y- term 是 1st, 使 a=1 和 b= 5。因此, 顶部是 (0, 1) 和 (0, 1) 。等式是 y= 15x 和 y= 15x 。最后, 我们使用 c2= a2+b2 来查找 foci 。

\begin{aligned} c^{2} & = 1 + 25 = 26 \\ c & = \sqrt{26} \end{aligned}

::c2=1+25=26c=26

The foci are $\begin{aligned} (0, - \sqrt{26}) \end{aligned}$ and $\begin{aligned} (0, \sqrt{26}) \end{aligned}$ .
::角为(0,26)和(0,26)。

by CK-12 shows how to identify components of a hyperbola from the equation.
::CK-12 显示如何从方程式中识别双波拉的元件。

Example 3
::例3

Find the equation of the hyperbola, centered at the origin, with a vertex of $\begin{aligned} (- 4, 0) \end{aligned}$ and a focus of $\begin{aligned} (- 6, 0) \end{aligned}$ .
::查找双波拉的方程式, 以原点为中心, 顶点为( 4, 0) , 焦点为( 6, 0) 。

Solution: Because the vertex and focus are on the $\begin{aligned} x \end{aligned}$ -axis, we know that the transverse axis is horizontal. Therefore, the equation will be $\begin{aligned} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ . From the vertex, we know that $\begin{aligned} a = 4 \end{aligned}$ and $\begin{aligned} c = 6 \end{aligned}$ . Let's solve for $\begin{aligned} b^{2} \end{aligned}$ using $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ .
::解析度: 因为顶点和焦点在 x 轴上, 我们知道横轴是水平的。因此, 方程将是 x2a2- y2b2=1. 从顶点上, 我们知道 a=4 和 c=6. 使用 c2=a2+b2 来解析 b2 。

\begin{aligned} 6^{2} & = 4^{2} + b^{2} \\ 36 & = 16 + b^{2} \to b^{2} = 20 \end{aligned}

::62=42+b236=16+b2=20

The equation of the hyperbola is $\begin{aligned} \frac{x^{2}}{16} - \frac{y^{2}}{20} = 1 \end{aligned}$ .
::双倍波拉的方程式是x216-y220=1。

Example 4
::例4

Find the equation of the hyperbola, centered at the origin, with an asymptote of $\begin{aligned} y = \frac{2}{3} x \end{aligned}$ and vertex of $\begin{aligned} (0, 12) \end{aligned}$ .
::查找双波拉的方程式,以原点为中心,以y=23x和顶点(0,12)为起点,以y=23x和顶点(0,12)。

Solution: We know that $\begin{aligned} a = 12 \end{aligned}$ , making the transverse axis vertical and the general equation of the asymptote $\begin{aligned} y = \frac{a}{b} x \end{aligned}$ . Therefore, $\begin{aligned} \frac{2}{3} = \frac{12}{b} \end{aligned}$ , making $\begin{aligned} b = 18 \end{aligned}$ . Therefore, the equation of the hyperbola is $\begin{aligned} \frac{y^{2}}{144} - \frac{x^{2}}{324} = 1 \end{aligned}$ .
::解答: 我们知道 a=12, 使横轴垂直垂直, 以及无ymptote y=abx的一般方程式。因此, 23=12b, making b=18。因此, 双倍方程式的方程式是 y2144 - x2324=1 。

In this example, we showed that the slope of the asymptote can be reduced to something that is not always $\begin{aligned} \frac{a}{b} \end{aligned}$ , but $\begin{aligned} c (\frac{m}{n}) = \frac{a}{b} \end{aligned}$ , where $\begin{aligned} c \end{aligned}$ is some constant by which we can reduce the fraction.
::在这个例子中,我们显示,小行星的斜坡可以缩小为不总是 ab,而是 c(mn)=ab, c 是某种常数, 我们可以通过这些常数来减少小行星的分数。

Example 5
::例5

Find the equations of two hyperbolas with an asymptote of $\begin{aligned} y = - \frac{5}{9} x \end{aligned}$ .
::查找两个超光子的方程, 以y=- 59x 的零星状态。

Solution: This asymptote can be for either a vertical or horizontal hyperbola. $\begin{aligned} - \frac{5}{9} \end{aligned}$ can also be a reduced fraction of $\begin{aligned} \frac{a}{b} \end{aligned}$ , as in the previous example. For instance, the asymptote $\begin{aligned} y = - \frac{10}{18} x \end{aligned}$ reduces to $\begin{aligned} y = - \frac{5}{9} x \end{aligned}$ .
::解答 : 此无症状可以是垂直或水平双曲线的。 -59 也可以是 ab 的减缩分数, 如前一例。例如, y=- 1018x 的衰减分数会降低到 y=- 59x 。

If the hyperbola is horizontal, then the equation of the asymptote is $\begin{aligned} y = - \frac{b}{a} x . \end{aligned}$ That would make $\begin{aligned} a = 9 \end{aligned}$ and $\begin{aligned} b = 5, \end{aligned}$ and the equation would be $\begin{aligned} \frac{x^{2}}{81} - \frac{y^{2}}{25} = 1 \end{aligned}$ . If the equation is vertical, then the asymptote is $\begin{aligned} y = - \frac{a}{b} x \end{aligned}$ and $\begin{aligned} a = 5 \end{aligned}$ and $\begin{aligned} b = 9 \end{aligned}$ . The equation would be $\begin{aligned} \frac{y^{2}}{25} - \frac{x^{2}}{81} = 1 \end{aligned}$ . If the slope is reduced from a larger fraction, we could also have $\begin{aligned} \frac{x^{2}}{324} - \frac{y^{2}}{100} = 1 \end{aligned}$ or $\begin{aligned} \frac{y^{2}}{100} - \frac{x^{2}}{324} = 1 \end{aligned}$ as a possible answer.
::如果双波拉是水平的,那么小数方程式的方程式是y=bax。这样可以使 a=9 和 b=5, 方程式是 x281 -y225=1. 如果方程式是垂直的, 那么, 小数方程式是 y=-abx 和 a=5 和 b=9. 。方程式是 y225 - x281=1. 如果斜度从较大部分缩小, 我们也可以使用 x2324 -y2100=1 或 y2100 - x2324=1 作为可能的答案。

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There are infinitely many hyperbolic equations with this asymptote.
::无限多的双曲方程式与这个无症状的方程式。

by CK-12 shows how to find the equation of a hyperbola centered at the origin.
::的 CK-12 显示如何找到以原位为主的双波拉的方程式。

Example 6
::例6

Find the center, vertices, foci, and asymptotes of $\begin{aligned} \frac{(y - 1)^{2}}{81} - \frac{(x + 5)^{2}}{16} = 1 \end{aligned}$ .
::查找(y- 1) 281 - (x+5) 216=1 的中心、顶部、福西和小数点的(y- 1) 281 - (x+5) 216=1 。

Solution: The center is $\begin{aligned} (- 5, 1) \end{aligned}$ , $\begin{aligned} a = \sqrt{81} = 9 \end{aligned}$ and $\begin{aligned} b = \sqrt{16} = 4 \end{aligned}$ , and the hyperbola is horizontal because the $\begin{aligned} y \end{aligned}$ -term is 1st. The vertices are $\begin{aligned} (- 5, 1 \pm 9) \end{aligned}$ or $\begin{aligned} (- 5, 10) \end{aligned}$ and $\begin{aligned} (- 5, - 8) \end{aligned}$ . Use $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ to find $\begin{aligned} c \end{aligned}$ .
::解答: 中心是 (5, 1, a= 81= 9 和 b= 16= 4) , 双曲线是水平的, 因为 y- term 是 1 。顶端是 ( 5, 1+9) 或 ( 5, 10) 和 ( 5, 8) 。使用 c2= a2+b2 来查找 c 。

\begin{aligned} c^{2} & = 81 + 16 = 97 \\ c & = \sqrt{97} \end{aligned}

::c2=81+16=97c=97

The foci are $\begin{aligned} (- 5, 1 + \sqrt{97}) \end{aligned}$ and $\begin{aligned} (- 5, 1 - \sqrt{97}) . \end{aligned}$
::重点领域是(5,1+97)和(5,1+97)。

The asymptotes are $\begin{aligned} y - 1 = \pm \frac{9}{4} (x + 5) \end{aligned}$ or $\begin{aligned} y = \frac{9}{4} x + 12 \frac{1}{4} \end{aligned}$ and $\begin{aligned} y = - \frac{9}{4} x - 10 \frac{1}{4} \end{aligned}$ .
::符号是 y - 1\\\ 94( x+5) 或 y= 94x+1214 和 y\ 94x- 1014 。

Example 7
::例7

Find the equation of the hyperbola with vertices $\begin{aligned} (- 6, - 3) \end{aligned}$ and $\begin{aligned} (- 6, 5) \end{aligned}$ and focus $\begin{aligned} (- 6, 7) \end{aligned}$ .
::查找双波拉的方程式,以顶部(-6,3)和(-6,5)和焦点(-6,7)为单位(-6,7)。

Solution: The vertices are $\begin{aligned} (- 6, - 3) \end{aligned}$ and $\begin{aligned} (- 6, 5) \end{aligned}$ and the focus is $\begin{aligned} (- 6, 7) \end{aligned}$ . The transverse axis is going to be vertical because the $\begin{aligned} x \end{aligned}$ -value does not change between these three points. The distance between the vertices is $\begin{aligned} | - 3 - 5 | = 8 \end{aligned}$ units, making $\begin{aligned} a = 4 \end{aligned}$ . The midpoint between the vertices is the center.
::解答 : 顶点是 (-6, 3) 和 (-6, 5) , 焦点是 (-6, 7) 。反向轴将是垂直的, 因为 X 值在上述三点之间没有变化。顶点之间的距离是 - 3 - 58 单位, 得出 a= 4。顶点之间的中点是中点。

\begin{aligned} (- 6, \frac{- 3 + 5}{2}) = (- 6, \frac{2}{2}) = (- 6, 1) \end{aligned}

The focus is $\begin{aligned} (- 6, 7) \end{aligned}$ and the distance between it and the center is 6 units, or $\begin{aligned} c \end{aligned}$ . Find $\begin{aligned} b \end{aligned}$ .
::焦点是 (-6,7 ) , 中心与中心之间的距离是 6 个单位, 或 c. 查找 b 。

\begin{aligned} 36 & = b^{2} + 16 \\ 20 & = b^{2} \\ b & = \sqrt{20} = 2 \sqrt{5} \end{aligned}

::36=b2+1620=b2b=20=25

The equation of the hyperbola is $\begin{aligned} \frac{(y - 1)^{2}}{16} - \frac{(x + 6)^{2}}{20} = 1 \end{aligned}$ .
::双倍波拉的方程式为(y-1)216-(x+6)220=1。

by CK-12 shows how to find the equations of the asymptotes when the center is not the origin.
::的 CK-12 显示当中心不是源代码时如何查找小数方程式的方程式。

Example 8
::例8

Find the equation of the hyperbola with vertices $\begin{aligned} (- 3, 2) \end{aligned}$ and $\begin{aligned} (7, 2) \end{aligned}$ and focus $\begin{aligned} (- 5, 2) \end{aligned}$ .
::查找双波拉的方程式,其中含有脊椎(3,2)和(7,2)以及焦点(5,2)和焦点(5,2)。

Solution: These two vertices create a horizontal transverse axis, making the hyperbola horizontal. If you are unsure, plot the given information on a set of axes. To find the center, use the midpoint formula with the vertices.
::解决方案 : 这两个顶端创建了横向横轴, 使双波拉水平。如果您不确定, 请在一组轴上绘制给定信息。要找到中心, 请使用中点公式和顶端。

\begin{aligned} (\frac{- 3 + 7}{2}, \frac{2 + 2}{2}) = (\frac{4}{2}, \frac{4}{2}) = (2, 2) \end{aligned}

The distance from one of the vertices to the center is $\begin{aligned} a \end{aligned}$ , $\begin{aligned} | 7 - 2 | = 5 \end{aligned}$ . The distance from the center to the given focus is $\begin{aligned} c \end{aligned}$ , $\begin{aligned} | - 5 - 2 | = 7 \end{aligned}$ . Use $\begin{aligned} a \end{aligned}$ and $\begin{aligned} c \end{aligned}$ to solve for $\begin{aligned} b \end{aligned}$ .
::从一个顶端到中心之间的距离是 7 -2 5 。从中心到给定焦点的距离是 c, 5 -2 。使用 a 和 c 解答 b 。

\begin{aligned} 7^{2} & = 5^{2} + b^{2} \\ b^{2} & = 24 \to b = 2 \sqrt{6} \end{aligned}

::72=52+b2b2=24b=26

Therefore, the equation is $\begin{aligned} \frac{(x - 2)^{2}}{25} - \frac{(y - 2)^{2}}{24} = 1 \end{aligned}$ .
::因此,等式是(x-2)225-(y-2)224=1。

Graphing Hyperbolas
::超分层图

How to Graph Hyperbolas
::如何绘制超分形图

1. Identify the center and the transverse axis.
::1. 确定中轴和横轴。

2. Count $\begin{aligned} a \end{aligned}$ units from the center along the transverse axis. Count $\begin{aligned} b \end{aligned}$ units from the center perpendicular to the center.
::2. 沿着横轴从中枢点数一个单位,从中枢直向中心点数B单位。

3. Draw a dashed rectangle that contains these four points, so that one pair of sides is parallel to the transverse axis.
::3. 绘制一个包含这四个点的破折矩形,使一对面与横轴平行。

4. The asymptotes are lines that include the diagonals of this rectangle.
::4. 微粒是包括该矩形对角线的线条。

5. Plot the vertices.
::5. 绘制顶点。

6. Draw a branch through the vertex and approaching the asymptotes.
::6. 通过顶部绘制分支并接近小粒子。

Important Note: The asymptotes and square are not a part of the function. They are included in graphing a hyperbola because it makes it easier to do so.
::重要注意: 微粒和正方形不是函数的一部分。它们被包含在超大波拉的图形中, 因为这样更容易做到。

Also, when graphing hyperbolas, we are sketching each branch. We did not make a table of values to find certain points and then connect.
::此外,在绘制超光谱图时,我们正在绘制每个分支的草图。我们没有绘制一个数值表来查找某些点,然后连接。

Example 9
::例9

Graph $\begin{aligned} \frac{x^{2}}{64} - \frac{y^{2}}{25} = 1 \end{aligned}$ .
::图x264-y225=1。

Solution: First, this hyperbola has a horizontal transverse axis because the $\begin{aligned} x^{2} \end{aligned}$ term is 1st. Also, with hyperbolas, the $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ terms stay in place, but the $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ terms switch. $\begin{aligned} a \end{aligned}$ is not always greater than $\begin{aligned} b \end{aligned}$ .
::解析度: 首先, 此双曲线有一个横向反向轴, 因为 x2 术语为 1 。另外, 在双曲线下, a 和 b 术语保持不变, 但 x 和 y 术语开关不总是大于 b 。 a 不一定要大于 b 。

Therefore, $\begin{aligned} a = \sqrt{64} = 8 \end{aligned}$ and $\begin{aligned} b = \sqrt{25} = 5 \end{aligned}$ . To graph this hyperbola, go out 8 units to the left and right of the center, and 5 units up and down to make a rectangle. The diagonals of this rectangle are the asymptotes.
::因此, a= 64=8 和 b= 25= 5 。要绘制此双曲线图, 请在中间的左边和右边取出 8 个单位, 向上和向下取出 5 个单位, 以形成矩形。此矩形的对角是 asymptotes 。

Draw the hyperbola branches with the vertices on the transverse axis and the rectangle. Sketch the branches to get close to the asymptotes, but not touch them.
::绘制横轴和矩形上的双波形树枝。将树枝拉伸以接近小行星, 但不触动它们。

The vertices are $\begin{aligned} (\pm 8, 0) \end{aligned}$ and the asymptotes are $\begin{aligned} y = \pm \frac{5}{8} x \end{aligned}$ .
::脊椎为(8,0),小数为y58x。

Example 10
::例10

Graph $\begin{aligned} 36 y^{2} - 9 x^{2} = 324 \end{aligned}$ .
::图36y2-9x2=324。

Solution: This equation is not in standard form. For the equation to be in standard form, the right side must be 1. Divide everything by 324.
::解答: 此方程式不是标准形式。方程式要达到标准形式, 右方必须是 1. 将所有方程式除以 324 。

\begin{aligned} \frac{36 y^{2}}{324} - \frac{9 x^{2}}{324} & = \frac{324}{324} \\ \frac{y^{2}}{9} - \frac{x^{2}}{36} & = 1 \end{aligned}

::36y2324-9x2324=324324y29-x236=1

Now we can see this is a vertical hyperbola, where $\begin{aligned} a = 3 \end{aligned}$ and $\begin{aligned} b = 6 \end{aligned}$ . Draw the rectangle and asymptotes, and plot the vertices on the $\begin{aligned} y \end{aligned}$ -axis.
::现在我们可以看到这是一个垂直的双曲线, a=3和b=6。绘制矩形和小状图, 在 Y 轴上绘制顶点。

Example 11
::例例11

Graph $\begin{aligned} \frac{x^{2}}{4} - \frac{y^{2}}{4} = 1 \end{aligned}$ . Identify the asymptotes.
::图 x24-y24=1. 识别小行星。

Solution: This will be a horizontal hyperbola, because the $\begin{aligned} x \end{aligned}$ -term is first. $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ will both be 2, because $\begin{aligned} \sqrt{4} = 2 \end{aligned}$ . Draw the square and diagonals to form the asymptotes.
::解析度 : 这将是一个水平双曲线, 因为 x 期为先。 a 和 b 将同时是 2 , 因为 4 = 2 。绘制正方形和对角, 以形成 asymptotes 。

The asymptotes are $\begin{aligned} y = \pm \frac{2}{2} x \end{aligned}$ or $\begin{aligned} y = x \end{aligned}$ and $\begin{aligned} y = - x \end{aligned}$ .
::22x 或 y=x 和 y=-x 等值为 y22x 或 y=x 和 y=-x 。

by Mathispower4u shows how to graph a hyperbola when the center is not at the origin.
::Mathispower4u 展示了当中心不是原点时如何绘制双曲线图。

Example 12
::例例12

Graph $\begin{aligned} \frac{(x - 2)^{2}}{16} - \frac{(y + 1)^{2}}{9} = 1 \end{aligned}$ . Then find the vertices, foci, and asymptotes.
::图(x-2) 216-(y+1)29=1. 然后找到脊椎、福西和微粒。

Solution: First, we know this is a horizontal hyperbola because the x- term is 1st. Therefore, the center is $\begin{aligned} (2, - 1) \end{aligned}$ and $\begin{aligned} a = 4 \end{aligned}$ and $\begin{aligned} b = 3 \end{aligned}$ . Use this information to graph the hyperbola.
::解决方案: 首先, 我们知道这是一个水平双曲线, 因为 x- 期为 1 。因此, 中心是 (2, 1) 和 a= 4 和 b= 3 。使用此信息来绘制双曲线。

To graph, plot the center, go 4 units to the right and left, and then up and down 3 units. Draw the box and asymptotes.
::绘制图表,绘制中心图,向右和向左走4个单位,然后向上和向下移动3个单位。绘制盒子和微粒。

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This is also how you can find the vertices. The vertices are $\begin{aligned} (2 \pm 4, - 1) \end{aligned}$ or $\begin{aligned} (6, - 1) \end{aligned}$ and $\begin{aligned} (- 2, - 1) \end{aligned}$ .
::您也可以通过这个方式找到顶点。顶点是 (2+4, 1) 或 (6, 1) 和 (2, 2) -1 。

To find the foci, we need to find $\begin{aligned} c, \end{aligned}$ using $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ .
::要找到角,我们需要找到 c,使用 c2=a2+b2。

\begin{aligned} c^{2} & = 16 + 9 = 25 \\ c & = 5 \end{aligned}

::c2=16+9=25c=5

Therefore, the foci are $\begin{aligned} (2 \pm 5, - 1) \end{aligned}$ or $\begin{aligned} (7, - 1) \end{aligned}$ and $\begin{aligned} (- 3, - 1) \end{aligned}$ .
::因此,介面是(2+5,-1)或(7,-1)和(3,-1)。

To find the asymptotes, we have to do a little work to find the $\begin{aligned} y \end{aligned}$ -intercepts. We know that the slope is $\begin{aligned} \pm \frac{b}{a} \end{aligned}$ or $\begin{aligned} \pm \frac{3}{4} \end{aligned}$ and they pass through the center. Let's write each asymptote in point-slope form using the center and each slope.
::要找到微粒, 我们需要做一点工作才能找到 Y 界面。我们知道斜坡是 ba 或 34 , 它们会穿过中心。我们用中间和每个斜坡来写每个微粒。

$\begin{aligned} y - 1 = \frac{3}{4} (x + 2) \end{aligned}$ and $\begin{aligned} y - 1 = - \frac{3}{4} (x + 2) \end{aligned}$
::y- 1=34(x+2) y- 1=-34(x+2) y- 1=-34(x+2)

Simplifying each equation, the asymptotes are
::简化每个方程式, 空数是

$\begin{aligned} y = \frac{3}{4} x - \frac{5}{2} \end{aligned}$ and $\begin{aligned} y = - \frac{3}{4} x + \frac{1}{2} \end{aligned}$ .
::y=34x-52和y=-34x+12。

Example 13
::例13

Graph $\begin{aligned} 49 (y - 3)^{2} - 25 (x + 4)^{2} = 1, 225 \end{aligned}$ and find the foci.
::图49(y-3)2 - 25(x+4)2=1,225,并找到角。

Solution: First we have to get the equation into standard form, like the equations above. To make the right side 1, we need to divide everything by 1,225.
::解答: 首先我们必须把方程式变成标准的形式, 如上面的方程式。要右侧一, 我们需要将所有方程式除以 1 225 。

\begin{aligned} \frac{49 (y - 3)^{2}}{1, 225} - \frac{25 (x + 4)^{2}}{1, 225} & = \frac{1, 225}{1, 225} \\ \frac{(y - 3)^{2}}{25} - \frac{(x + 4)^{2}}{49} & = 1 \end{aligned}

::49(y-33)21 225-225-225(x+4)21 225=1,2251,225(y-33)225-(x+4)249=1

Now we know the hyperbola will be vertical because the $\begin{aligned} y \end{aligned}$ -term is 1st. $\begin{aligned} a = 5 \end{aligned}$ , $\begin{aligned} b = 7, \end{aligned}$ and the center is $\begin{aligned} (- 4, 3) \end{aligned}$ .
::现在我们知道超博拉是垂直的因为y-term是 1st. a=5, b=7, 中心是 (4, 3) 。

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To find the foci, we 1st need to find $\begin{aligned} c \end{aligned}$ by using $\begin{aligned} c^{2} = a^{2} + b^{2} \end{aligned}$ .
::要找到foci, 我们首先需要通过使用 c2=a2+b2 来找到 c。

\begin{aligned} c^{2} & = 49 + 25 = 74 \\ c & = \sqrt{74} \end{aligned}

::c2=49+25=74c=74

The foci are $\begin{aligned} (- 4, 3 \pm \sqrt{74}) \end{aligned}$ or $\begin{aligned} (- 4, 11.6) \end{aligned}$ and $\begin{aligned} (- 4, - 5.6) \end{aligned}$ .
::反面为(4,374)或(4,11.6)和(4,5.6)。

How to Graph Hyperbolas Using Desmos
::如何使用脱模图

Enter the equation as it is. Remember to use the caret key ^ (Shift 6) for exponents, and the slash key / for division.
::输入正方程。记住要使用 {( Shift 6)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

How to Graph Hyperbolas Using a TI-83/84
::如何使用TI-83/84来绘制超高波拉斯图

1. Press APPS and choose Conics.
::1. 按APPS键并选择二次曲线。

2. Choose Hyperbola.
::2. 选择超重波拉。

3. Choose the form of an equation that matches the equation you are trying to graph.
::3. 选择与您试图绘制的方程式相匹配的方程式形式。

4. Enter a,b, h, and k.
::4. 输入 a、b、h和k。

5. Press GRAPH .
::5. 出版《GRAPH》。

Feature: Land Ho!
::何国!

by Meredith Beaton
::梅雷迪思·贝顿(签名)

Navigating the oceans used to be treacherous, and having a skilled navigator on board was literally the difference between life and death for a ship's crew. Early navigators often stayed close to shore and used the stars (which they had for only half the day) to determine their location.
::海洋的航行过去是险恶的,船上有一位熟练的航海家,这实际上是船员生死之间的差别。早期航海家往往离岸很近,利用恒星(他们只有半天的时间)决定他们的位置。

Why It Matters
::为何重要

In the 1940s a long-range navigation system called LORAN was developed to determine a ship's position. The LORAN system was developed at the Massachusetts Institute of Technology during World War II for military ships. The LORAN system works by using properties of hyperbolas to determine a ship's location.
::1940年代,开发了一个称为LORAN的远程导航系统,以确定船舶的位置。LORAN系统是马萨诸塞州理工学院在二战期间为军用船舶开发的。LORAN系统通过使用超光层特性来确定船舶的位置。

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LORAN relies on a transmitting station on land to determine positioning. The land-based transmitting stations send out signals traveling at the speed of light. They then use information about how long it takes those signals to reach the LORAN receiver on the ship to determine the distance of the ship from the stations. The distance the LORAN is from each station is subtracted. If we think of each transmitting station as a focus of a hyperbola, then the ship lies somewhere along the hyperbola. To determine exactly where the ship is along the hyperbola, a 3rd transmitting station is needed to create another hyperbola with one of the original stations. The intersection of the two hyperbolas indicates the exact location of the ship!
::LORAN 依靠陆地上的一个传输站来确定定位。陆基传输站以光速发送信号, 然后它们使用这些信号到达船上的LORAN接收器需要多长时间的信息来确定船舶与站点之间的距离。 LORAN 与每个站点的距离被减去。如果我们把每个传输站作为超重波的焦点, 那么该船就停在超重波拉沿线的某处。为了确定船舶沿超重波拉的确切位置, 需要一个第三传送站来创造另一个与原站站点的超重波拉。两个超重波拉的交叉点显示该船的确切位置 !

by the Smithsonian National Air and Space Museum demonstrates how LORAN works in a 1947 training video.
::Smithsonian National Air and Space博物馆展示了LORAN如何在1947年培训录像中工作。

Summary
::摘要

A hyperbola is the set of all points such that the differences of the distances from the foci are constant.
::双倍波拉是所有点的组合, 以便与方位距离的距离差异是恒定的。

::双倍波拉是所有点的组合, 以便与方位距离的距离差异是恒定的。

The standard forms of an equation of a hyperbola are $\begin{aligned} \frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1 \end{aligned}$ (horizontal orientation) and $\begin{aligned} \frac{(y - k)^{2}}{a^{2}} - \frac{(x - h)^{2}}{b^{2}} = 1 \end{aligned}$ (vertical orientation).
::超重波拉方程式的标准方程式是 (x-h)2a2-(y-k)2b2=1(横向方向)和 (y-k)2a2-(x-h)2b2=1(纵向方向)。

::超重波拉方程式的标准方程式是 (x-h)2a2-(y-k)2b2=1(横向方向)和 (y-k)2a2-(x-h)2b2=1(纵向方向)。

To graph a hyperbola, draw a box using the center, $\begin{aligned} a \end{aligned}$ , and $\begin{aligned} b \end{aligned}$ . Then draw the asymptotes and the vertices. Last, graph the branches of the hyperbola.
::要绘制双曲线图,请用中心( a) 和( b) 绘制一个框。然后绘制小行星和脊椎。最后,请绘制双曲线的分支。

Review
::回顾

Using the information below, find the equation of each hyperbola.
::使用以下信息,找到每个超重波的方程式。

1. vertex: $\begin{aligned} (- 2, 0) \end{aligned}$ ; focus: $\begin{aligned} (- 5, 0) \end{aligned}$ ; center: $\begin{aligned} (0, 0) \end{aligned}$
::1. 顶点2,0);重点5,5,0);中心0,0)

2. $\begin{aligned} b = 8 \end{aligned}$ ; focus: $\begin{aligned} (- 15, 0) \end{aligned}$ ; center: $\begin{aligned} (0, 0) \end{aligned}$
::2. b=8; 焦点15,50); 中心: (0,0)

3. vertex: $\begin{aligned} (- 6, 0) \end{aligned}$ ; asymptote: $\begin{aligned} y = \frac{4}{3} x \end{aligned}$
::3. 顶点-6,0); 停止: y=43x

4. $\begin{aligned} b = 6 \end{aligned}$ ; focus: $\begin{aligned} (0, 11) \end{aligned}$ ; center: $\begin{aligned} (0, 0) \end{aligned}$
::4. b=6; 焦点: (0,11); 中心: (0,0)

5. vertex: $\begin{aligned} (0, 5) \end{aligned}$ ; asymptote: $\begin{aligned} y = x \end{aligned}$
::5. 顶点: (0,5); 暂时: y=x

6. asymptote: $\begin{aligned} y = - \frac{1}{2} x \end{aligned}$ ; vertex: $\begin{aligned} (6, 0) \end{aligned}$
::6. 空点 : Y= 12x; 顶部 : (6,0)

7. asymptote: $\begin{aligned} y = 3 x \end{aligned}$ ; $\begin{aligned} b = 9 \end{aligned}$ ; vertical transverse axis
::7. 空点: y=3x; b=9; 垂直横轴

8. vertex: $\begin{aligned} (0, 8) \end{aligned}$ ; focus: $\begin{aligned} (0, 6 \sqrt{2}) \end{aligned}$ ; center: $\begin{aligned} (0, 0) \end{aligned}$
::8. 顶点: (0,8); 焦点: (0,62); 中心: (0,0)

9. vertices: $\begin{aligned} (- 2, - 3) \end{aligned}$ and $\begin{aligned} (8, - 3) \end{aligned}$ ; $\begin{aligned} b = 7 \end{aligned}$
::9. 顶点2-3)和(8.3);b=7

10. vertices: $\begin{aligned} (5, 6) \end{aligned}$ and $\begin{aligned} (5, - 12) \end{aligned}$ ; focus: $\begin{aligned} (5, - 15) \end{aligned}$
::10. 顶点5,6)和(5,12);重点5,15)

11. asymptote: $\begin{aligned} y + 3 = \frac{4}{9} (x + 1) \end{aligned}$ ; horizontal transverse axis
::11. 空点 : y+3=49( x+1); 水平反向轴

12. foci: $\begin{aligned} (- 11, - 4) \end{aligned}$ and $\begin{aligned} (1, - 4) \end{aligned}$ ; vertex: $\begin{aligned} (- 8, - 4) \end{aligned}$
::12.foci11,4)和(1,4);顶部8,4)

13.

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14.

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15.

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16.

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17.

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18.

lesson content

Find the center, vertices, foci, and asymptotes of each hyperbola below. Graph the hyperbolas.
::查找下方每双倍波拉的中心、脊椎、福西和微粒。绘制双倍波拉。

19. $\begin{aligned} \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \end{aligned}$
::19. x29-y216=1

20. $\begin{aligned} 4 y^{2} - 25 x^{2} = 100 \end{aligned}$
::20. 4y2-25x2=100

21. $\begin{aligned} \frac{x^{2}}{81} - \frac{y^{2}}{64} = 1 \end{aligned}$
::21. x281-y264=1

22. $\begin{aligned} x^{2} - y^{2} = 16 \end{aligned}$
::22. x2-y2=16

23. $\begin{aligned} 121 y^{2} - 9 x^{2} = 1, 089 \end{aligned}$
::23. 121y2-9x2=1 089

24. $\begin{aligned} y^{2} - x^{2} = 1 \end{aligned}$
::24. y2-x2=1

25. $\begin{aligned} \frac{y^{2}}{4} - \frac{x^{2}}{64} = 1 \end{aligned}$
::25. y24-x264=1

26. $\begin{aligned} \frac{(x + 5)^{2}}{25} - \frac{(y + 1)^{2}}{36} = 1 \end{aligned}$
::26. (x+5)225-(y+1)236=1

27. $\begin{aligned} (y + 2)^{2} - 16 (x - 6)^{2} = 16 \end{aligned}$
::27. (y+2)2-16(x-6)2=16

28. $\begin{aligned} \frac{(y - 2)^{2}}{9} - \frac{(x - 3)^{2}}{49} = 1 \end{aligned}$
::28. (y-2)29-(x-33)249=1

29. $\begin{aligned} 25 x^{2} - 64 (y - 6)^{2} = 1, 600 \end{aligned}$
::29. 25x2-64(y-6)2=1 600

30. $\begin{aligned} (x - 8)^{2} - \frac{(y - 4)^{2}}{9} = 1 \end{aligned}$
::30. (x-8)2-(y-4)29=1

31. $\begin{aligned} 81 (y + 4)^{2} - 4 (x + 5)^{2} = 324 \end{aligned}$
::31. 81(y+4)2-4(x+5)2=324

Explore More
::探索更多

a. Find the equation of two hyperbolas that have the same $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ values, have an asymptote with the equation $\begin{aligned} y = \frac{4}{5} x \end{aligned}$ , and are centered at the origin.
::a. 查找两个高角方程式的方程式,它们具有相同的 a 和 b 值,与 y= 45x 等式同,且以原值为中心。

b. Find the equation of two hyperbolas that have different $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ values, are both horizontal, have an asymptote with the equation $\begin{aligned} y = - \frac{2}{3} x \end{aligned}$ , and are centered at the origin.
::b. 查找两个具有不同a值和b值、均为水平值、与 y23x 等式有零点且以原值为核心的双极黄体的方程。

c. Find the equation of two hyperbolas that have different $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ values, are both vertical, have an asymptote with the equation $\begin{aligned} y = 6 x \end{aligned}$ , and are centered at the origin.
::c. 查找两个具有不同 a 和 b 值、均为垂直、与 y = 6x 等式同音、以源居于原位的双光体的方程。

d. Find the equation of two hyperbolas that have the same $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ values, have an asymptote with the equation $\begin{aligned} y = - \frac{10}{7} x \end{aligned}$ , and are centered at the origin.
::d. 找到两个高角体的方程式,它们具有相同的 a和 b 值,与 y 107x 等式同,以原值为中心。

a. Find the equations of a parabola and a hyperbola that are confocal.
::a. 找出抛物线和超重波的组合方程。

b. Find the equations of an ellipse and a hyperbola that are confocal.
::b. 查找椭圆形和双曲线方程式的方程。

c. Find the equations of two hyperbolas that are confocal.
::c. 找到两个组合的双螺旋体的方程。

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

PLIX
::PLIX

Try the following interactive to explore the concepts in this section.
::尝试以下互动来探索本节的概念。

章节大纲

Geometry of Hyperbolas ::超高功率的几何

Geometry of a Hyperbola ::超波拉的几何

Equations of Hyperbolas ::超重波拉斯平方

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 8 ::例8

Graphing Hyperbolas ::超分层图

How to Graph Hyperbolas ::如何绘制超分形图

Example 9 ::例9

Example 10 ::例10

Example 11 ::例例11

Example 12 ::例例12

Example 13 ::例13

How to Graph Hyperbolas Using Desmos ::如何使用脱模图

How to Graph Hyperbolas Using a TI-83/84 ::如何使用TI-83/84来绘制超高波拉斯图

Feature: Land Ho! ::何国!

Why It Matters ::为何重要

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

PLIX ::PLIX

Geometry of Hyperbolas
::超高功率的几何

Geometry of a Hyperbola
::超波拉的几何

Equations of Hyperbolas
::超重波拉斯平方

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 8
::例8

Graphing Hyperbolas
::超分层图

How to Graph Hyperbolas
::如何绘制超分形图

Example 9
::例9

Example 10
::例10

Example 11
::例例11

Example 12
::例例12

Example 13
::例13

How to Graph Hyperbolas Using Desmos
::如何使用脱模图

How to Graph Hyperbolas Using a TI-83/84
::如何使用TI-83/84来绘制超高波拉斯图

Feature: Land Ho!
::何国!

Why It Matters
::为何重要

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

PLIX
::PLIX