超双

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Hyperbolas can be oriented so that they open side to side or up and down.  One of the most common mistakes that you can make is to forget which way a given hyperbola should open.  What are some strategies to help?
::Hyperbolas可以调整方向,让它们从侧面向上或向下打开。最常见的错误之一是忘记某一双曲线应该以哪种方式打开。有什么策略可以帮助呢?

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Graphing Hyperbolas
::超分层图

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A hyperbola has two foci . For every point on the hyperbola, the difference of the distances to each foci is constant.  This is what defines a hyperbola.  The graphing form of a hyperbola that opens side to side is:
::双倍波拉有两个角度。 对于双倍波拉上的每一点, 与每个角度的距离差异是恒定的。 这就是双倍波拉的定义。 向侧打开的双倍波拉的图形形式是 :

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$\begin{align*}\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\end{align*}$
:x-h)2a2-(y-k)2b2=1

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A hyperbola that opens up and down is:
::一个向上和向下打开的双倍波拉是:

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$\begin{align*}\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\end{align*}$
:y-k)2a2-(x-h)2b2=1

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Notice that for hyperbolas,   $\begin{align*}a\end{align*}$ goes with the positive term and  $\begin{align*}b\end{align*}$ goes with the negative term.  It does not matter which constant is larger.
::请注意,对于超光子, 使用正值术语和 b 使用负值术语。 哪个恒定值更大并不重要 。

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When graphing, the constants  $\begin{align*}a\end{align*}$ and  $\begin{align*}b\end{align*}$ enable you to draw a rectangle around the center .  The transverse axis travels from vertex to vertex and has length $\begin{align*}2a\end{align*}$ .  The conjugate axis travels perpendicular to the transverse axis through the center and has length $\begin{align*}2b\end{align*}$ .  The foci lie beyond the vertices so the eccentricity, which is measured as $\begin{align*}e=\frac{c}{a}\end{align*}$ , is larger than 1 for all hyperbolas.  Hyperbolas also have two directrix lines that are  $\begin{align*}\frac{a^2}{c}\end{align*}$ away from the center (not shown on the image).
::当图形绘制时,常数 a 和 b 使您能够在中心周围绘制矩形。 横轴从顶部向顶部移动, 长度为 2a 。 共振轴通过中心垂直到横轴, 长度为 2b 。 顶部位于顶部之外, 因此以 e=ca 测量的偏心度大于 1 , 对所有双波las 来说, 超双波las 的偏心度大于 1 。 Hyperbolas 也有两条直径直线, 离中心有 A2c 距离( 未显示在图像上 ) 。

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The focal radius is $\begin{align*}a^2+b^2=c^2\end{align*}$ .
::焦半径为a2+b2=c2。

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked how to determine the direction that a hyperbola opens. The best strategy to remember which direction the hyperbola opens is often the simplest.  Consider the hyperbola $\begin{align*}x^2-y^2=1\end{align*}$ . This hyperbola opens side to side because  $\begin{align*}x\end{align*}$ can clearly never be equal to zero.  This is a basic case that shows that when the negative is with the  $\begin{align*}y\end{align*}$ value then the hyperbola opens up side to side.
::早些时候,有人问您如何确定超重波拉打开的方向。 记住超重波拉打开的方向的最好策略往往是最简单的。 想想超重波拉 x2- y2= 1 。 这个超重波拉打开侧侧面, 因为 x 显然永远不能等于零 。 这是一个基本的例子, 表明当负值与y值存在时, 超重波拉打开侧面。

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Example 2
::例2

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Put the following hyperbola into graphing form, list the components, and sketch it. 
::将以下的双倍波拉放入图形形式,列出部件并草图。

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$\begin{align*}9x^2-4y^2+36x-8y-4=0\end{align*}$
::9x2-4y2+36x-8y-4=0

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9 ( x 2 + 4 x ) − 4 ( y 2 + 2 y ) = 4 9 ( x 2 + 4 x + 4 ) − 4 ( y 2 + 2 y + 1 ) = 4 + 36 − 4 9 ( x + 2 ) 2 − 4 ( y + 1 ) 2 = 36 ( x + 2 ) 2 4 − ( y + 1 ) 2 9 = 1

$$\begin{align*}9(x^2+4x)-4(y^2+2y)&=4\\ 9(x^2+4x+4)-4(y^2+2y+1)&=4+36-4\\ 9(x+2)^2-4(y+1)^2&=36\\ \frac{(x+2)^2}{4}-\frac{(y+1)^2}{9}&=1\end{align*}$$
::9(x2+4x)-4(y2+2y)=49(x2+4x+4)-4(y2+2y+1)=4+36-49(x+2)-2-4(y+1)-2-4(y+1)=36(x+2)=36(x+2)24-(y+1)29=1

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Shape: Hyperbola that opens horizontally. 
::形状:水平开口的双波拉。

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Center: (-2, -1)
::中心-2,-1)

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$\begin{align*}a=2\end{align*}$
::a=2

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$\begin{align*}b=3\end{align*}$
::b=3 =

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$\begin{align*}c=\sqrt{13}\end{align*}$
::13

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$\begin{align*}e=\frac{c}{a}=\frac{\sqrt{13}}{2}\end{align*}$
::e=ca132

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$\begin{align*}d=\frac{a^2}{c}=\frac{4}{\sqrt{13}}\end{align*}$
::d=a2c=413

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Foci: $\begin{align*}(-2+\sqrt{13}, -1),~ ( -2-\sqrt{13} ,  -1)\end{align*}$
::成员: (-213,-1), (-213,-1)

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Vertices: (-4, -1), (0, -1)
::顶点4,4,-1),(0,-1)

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Equations of asymptotes: $\begin{align*}(y+1)=\pm \frac{3}{2}(x+2)\end{align*}$
::小数数的等同度y+1) 32(x+2)

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Note that it is easiest to write the equations of the asymptotes in point-slope form using the center and the slope.
::请注意,使用中间和斜坡以点窗体形式以点窗体写出小数方程是最容易的。

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Equations of directrices:  $\begin{align*}y=-2 \pm \frac{4}{\sqrt{13}}\end{align*}$
::电流平方:y2413

 

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Example 3
::例3

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Find the equation of the hyperbola with foci at (-3, 5) and (9, 5) and asymptotes with slopes of  $\begin{align*}\pm \frac{4}{3}\end{align*}$ . 
::查找超重波拉方位( 3, 5) 和( 9, 5) 的方位方位, 以及斜度为 +43 的单位方位方位 。

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The center is between the foci at (3, 5).  The focal radius is $\begin{align*}c=6\end{align*}$ .  The slope of the asymptotes is always the rise over run inside the box.  In this case since the hyperbola is horizontal and  $\begin{align*}a\end{align*}$ is in the  $\begin{align*}x\end{align*}$ direction the slope is $\begin{align*}\frac{b}{a}\end{align*}$ .  
::中心位于( 3, 5 ) 的角之间, 焦半径为 c=6 。 小微粒的斜坡总是在框内运行的上升。 在这种情况下, 双倍波拉是水平的, a 是向 x 方向的, 斜坡是 ba 。

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W e have the ratio $\begin{align*}\frac{b}{a}=\frac43.\end{align*}$   S o we'll say , $\begin{align*}b=b\end{align*}$  and $\begin{align*}a=\frac34b.\end{align*}$
::我们有Ba=43的比例,所以我们说,b=b和a=34b。

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We can substitute these into $\begin{align*}a^2+b^2=c^2\end{align*}$  and solve for $\begin{align*}b:\end{align*}$
::我们可以将这些替换为 a2+b2=c2, 并解决 b:

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$$\begin{align*}a^2+b^2&=c^2\\ \left(\frac34b\right)^2+(b)^2&=(6)^2\\ \frac{9}{16}b^2+b^2&=36\\ \frac{9}{16}b^2+\frac{16}{16}b^2&=36\\ \frac{25}{16}b^2&=36\\ b^2&=36\times\frac{16}{25}\\ b^2&=\frac{576}{25}\\ b&=\sqrt{\frac{576}{25}}\\ b&=\frac{24}{5}\\\end{align*}$$
::a2+b2=c2(34b)2+(b)2=(6)2916b2+b2=36916b2+1616b2=362516b2=36b2=36b2=361625b2=361625b2=57625b=57625b=57625b=57625b=245

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When you solve for $\begin{align*}a,\end{align*}$  you get:
::当你解决一个,你得到:

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$$\begin{align*}a^2+b^2&=c^2\\ a^2+\left(\frac{24}{5}\right)^2&=(6)^2\\  a^2+\frac{576}{25}&=36\\ a^2&=\frac{900}{25}-\frac{576}{25}\\ a&=\sqrt{\frac{324}{25}}\\ a&=\frac{18}{25}\end{align*}$$
::a2+b2=c2a2+(245)2=(6)2a2+57625=36a2=900025=57625a32425a=1825

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So,  $\begin{align*}a=\frac{18}{25}\end{align*}$  and  $\begin{align*}b=\frac{24}{5}.\end{align*}$  
::所以,a=1825和b=245。

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$$\begin{align*}\frac{(x-3)^2}{\left(\frac{18}{25}\right)^2}-\frac{(y-5)^2}{\left(\frac{24}{5}\right)^2}=1\\\end{align*}$$
:x-3)2(1825)2-(y-5)2(245)2=1

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Example 4
::例4

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Find the equation of the conic that has a focus point at (1, 2), a directrix at  $\begin{align*}x=5\end{align*}$ , and an eccentricity equal to $\begin{align*}\frac{3}{2}\end{align*}$ .  Use the property that the distance from a point on the hyperbola to the focus is equal to the eccentricity times the distance from that same point to the directrix:
::查找在(1, 2) 有焦点点的二次曲线的方程、 x=5 的直线和等于 32. 的偏心度。 使用以下属性,即从超博拉点到焦点点的距离等于从同一点到直线点的偏心度乘数:

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$\begin{align*}\overline{PF}=e \overline{PD}\end{align*}$
::你的父亲

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This relationship bridges the gap between which have eccentricity less than one and hyperbolas which have eccentricity greater than one.  When eccentricity is equal to one, the shape is a parabola .
::这种关系弥合了偏心性小于1和偏心性大于1的超高代谢之间的鸿沟。 当偏心性等于1时,形状就是一个抛物线。

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$\begin{align*}\sqrt{(x-1)^2+(y-2)^2}=\frac{3}{2}\sqrt{(x-5)^2}\end{align*}$
:x--1)2+(y-2)2=32(x-5)2

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Square both sides and rearrange terms so that it is becomes a hyperbola in graphing form. 
::方形两侧和重新排列的词义, 使其在图形形式中变成双波形 。

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$\begin{align*}x^2 -2x+1+(y-2)^2&=\frac94(x^2-10x+25)\\ x^2-2x+1+(y-2)^2&=\left(\frac94x^2-\frac{90}{4}x+\frac{255}{4}\right)\\ x^2-\frac94x^2-2x+\frac{90}{4}x+1-\frac{225}{4}+(y-2)^2&=0\\ -\frac54x^2+\frac{82}{4}x-\frac{221}{4}+(y-2)^2&=0\\ -5x^2+82x-221+4(y-2)^2&=0\\ -5x^2+82x+4(y-2)^2&=221\\ -5\left(x^2-\frac{82}{5}x\right)+4(y-2)^2&=221\\ \\ \\ -5\left(x^2-\frac{82}{5}x+\left(\frac{82}{5}\right)^2\right)+4(y-2)^2&=221+\left(-5\times\left(\frac{82^2}{5^2}\right)\right)\\ -5\left(x^2-\frac{82}{5}\right)^2+4(y-2)^2&=221-\frac{82^2}{5}\\ -5\left(x^2-\frac{82}{5}\right)^2+4(y-2)^2&=\frac{1105}{5}-\frac{6724}{5}\\ -5\left(x^2-\frac{82}{5}\right)^2+4(y-2)^2&=-\frac{5,619}{5}\\ \frac{-5\left(x^2-\frac{82}{5}\right)^2}{\left(-\frac{5,619}{5}\right)}+\frac{4(y-2)^2}{\left(-\frac{5,619}{5}\right)}&=1\\ \frac{\left(x^2-\frac{82}{5}\right)^2}{\frac{5,619}{25}}-\frac{(y-2)^2}{\frac{5,619}{20}}&=1\\\end{align*}$
::2-2-2-2-2-2x+1-2-2-2-2-2-2-2-2-2-2-2-2-2(94x2-904x+2554)x-2-2-2(94x2-904x+252-2)2=(94x2-94x2-904x+904x+1-2254xx2-942-292x-2-2x-2x-2904x+1-2254-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2(94x-2-904x-904x+1-2-2-2-2-2-2-2-2-2-2-25-2-202-2041414+(y-2-2)2-2-2-2-2-2-2-2-2,2-2-2-2-9-2,2-2-2-2-2-221-2-2-2-2-2-2-2-2-2-221(y-2-2-2)-221+5(-5(5(8(82,82,825252))-5(5,19(192,192,192,192,192,192,19,2,2,2,2,2,2,2,2,2,2,2,19,19,19,19,19,19,2,2,2,2,19,19,19,2,19,19,19,19,19,19,19,2,2,2,2,2,2,2,5,5,5,5,5,5,2,5,5,5,5,5,5,5,2,2,2,5,5,2,5,5,5,2,2,5,5,5,5,5,5,5,5,5,5,2,5,5,5,5,5,5,5,2,2,2,5,5,5,5,5,5,5,5,5,5,2,2,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,2,5,5,5,5,5,2

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Example 5
::例5

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Given the following graph, estimate the equation of the conic.
::根据下图,估计二次曲线的方程。

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Since exact points are not marked, you will need to estimate the slope of asymptotes to get an approximation for  $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ . The slope seems to be about $\begin{align*}\pm \frac{2}{3}\end{align*}$ .  The center seems to be at (-1, -2).  The transverse axis is 6 which means $\begin{align*}a=3\end{align*}$ .
::由于未标出精确点, 您需要估计小行星的斜度, 才能得出 a 和 b 的近似值。 斜度似乎是 + 23 。 中心似乎在 (-1 - 2) , 横轴是 6 , 这意味着 a= 3 。

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$\begin{align*}\frac{(x+1)^2}{9}-\frac{(y+2)^2}{4}=1\end{align*}$
:x+1)29-(y+2)24=1