Course: 12.6 椭圆 | The Way To Learn

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椭圆

Occasionally, stones in the gallbladder (gallstones) and stones in the kidney (kidney stones) are treated with a procedure called lithotripsy, by which a shock wave is applied not at the site of the stone, but at another related point. These points are determined by an ellipse ¹ . I n this section, w e determine equations of ellipses and show how to graph them.
::有时,胆囊中的石块(石块)和肾中的石块(孩子的石块)被用一种叫做闪石状的程序处理,根据这种程序,休克波不是在石块的所在地,而是在另一个相关点。这些点由椭圆确定1。在本节中,我们确定椭圆的方程式,并展示如何用图表绘制它们。

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Geometry of Ellipses
::椭圆的几何

In the previous section, we discussed circles. A circle is created by the intersection of a plane parallel to the base of the cone with the surface, or nappe, of a cone. An ellipse is formed when that plane is not parallel to the base, and is at an angle greater than the vertex angle. If we generalize and say that an ellipse is the shape created when the plane is at an angle greater than the vertex angle, then a circle is a type of ellipse.
::在前一节中,我们讨论了圆圈。一个圆圈是由与锥体底部平行的平面交叉点与锥体表面或环形相平行的平面所创建的。当平面与底部不平行时,就形成椭圆,其角度大于顶端角。如果我们概括并说,椭圆是当平面的角大于顶角时所创造的形状,那么,一个圆就是椭圆的一种类型。

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Like the previous we've discussed, an ellipse can be defined in terms of a geometric property.
::和以前我们讨论过的一样椭圆可以用几何属性来定义

Geometry of an Ellipse
::椭圆的几何

An ellipse is the set of all points such that the sum of the distances from two fixed points, called foci (the plural of focus ), is constant.

Let's consider how this works with an investigation.
::让我们来考虑一下这和调查有什么关系

Investigation: Drawing an Ellipse
::调查:绘制椭圆

1. On a piece of graph paper, draw a set of axes and plot $\begin{aligned} (- 2, 0) \end{aligned}$ and $\begin{aligned} (2, 0) \end{aligned}$ . These will be the foci.
::1. 在一张图纸上,绘制一组轴和绘图(2.0)和(2.0)。

2. From the definition, we can conclude a point $\begin{aligned} (x, y) \end{aligned}$ is on an ellipse if the sum of the distances is always constant. In the picture, $\begin{aligned} d_{1} + d_{2} = r \end{aligned}$ and $\begin{aligned} g_{1} + g_{2} = r \end{aligned}$ .
::2. 从定义中,我们可以得出一个点(x,y)在椭圆上,如果距离总和总是不变的话。在图片中,是 d1+d2=r和 g1+g2=r。

3. Determine how far apart the foci are. Then find $\begin{aligned} d_{1} \end{aligned}$ and $\begin{aligned} d_{2} \end{aligned}$ .
::3. 确定介质之间的距离,然后找到 d1 和 d2 。

4. Determine if the point $\begin{aligned} (- 2, 3) \end{aligned}$ is on the ellipse.
::4. 确定点(-2-3)是否在椭圆上。

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There are always two foci, and they are on the major axis . The major axis is the longer of the two axes that pass through the center of an ellipse. Also on the major axis are the vertices , which are its endpoints and are the furthest two points away from each other on an ellipse. The shorter axis that passes through the center is called the minor axis , with endpoints called co-vertices . The midpoint of both axes is the center .
::始终有两个角, 它们位于主轴上。主要轴是穿过椭圆中心的两个轴的长度。主要轴上还有顶端, 它们是它的终点, 也是在椭圆上距离对方最远的两个点。通过中间的较短轴被称为小轴, 端点称为共垂直点。两个轴的中点是中点。

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Equations of Ellipses
::椭圆平

Like , the equations of ellipses have different standard forms, based on whether the major axis is horizontal or vertical.
::例如,省略号等式的标准形式不同,以主轴是水平轴还是垂直轴为基础。

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

Note that $\begin{aligned} a \end{aligned}$ is ALWAYS greater than $\begin{aligned} b \end{aligned}$ . If they are equal, we have a circle.
::请注意, a 的 Al-WAYS 大于 b. 如果等值,我们有一个圆。

For , we have the following:
::如下:

Equation of an Ellipse, Centered at the Origin
$\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{aligned}$	HORIZONTAL ::洪都拉斯 Major axis is the $\begin{aligned} x \end{aligned}$ -axis with length $\begin{aligned} 2 a . \end{aligned}$ ::主轴是长度为2a的x轴轴。 Minor axis is the $\begin{aligned} y \end{aligned}$ -axis with length $\begin{aligned} 2 b . \end{aligned}$ ::小轴为Y轴,长度为2b。
$\begin{aligned} \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \end{aligned}$	VERTICAL ::核查、核查、核查、核查、 Major axis is the $\begin{aligned} y \end{aligned}$ -axis with length $\begin{aligned} 2 a . \end{aligned}$ ::主轴是长度为 2a 的 Y 轴。 Minor axis is the $\begin{aligned} x \end{aligned}$ -axis with length $\begin{aligned} 2 b . \end{aligned}$ ::小轴是长度为 2b 的 X 轴。

Example 1
::例1

Write an equation for the ellipse with the given characteristics below and centered at the origin.
::为椭圆写一个方程, 以下面给定的特性为中心, 以源代码为中心。

a. vertex: $\begin{aligned} (- 6, 0) \end{aligned}$ ; co-vertex: $\begin{aligned} (0, 4) \end{aligned}$
::a. 顶点-6,0);共同顶点0,4)

b. vertex: $\begin{aligned} (0, 9) \end{aligned}$ ; focus: $\begin{aligned} (0, - 5) \end{aligned}$
::b. 顶部: (0,9);焦点: (0,5)

Solution: For part a, we can conclude that $\begin{aligned} a = 6 \end{aligned}$ and $\begin{aligned} b = 4 \end{aligned}$ . The ellipse is horizontal because the larger value, $\begin{aligned} a \end{aligned}$ , is part of the $\begin{aligned} x \end{aligned}$ -value of the vertex. The equation is $\begin{aligned} \frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 \end{aligned}$ .
::解答: 对于 a 部分, 我们可以得出 a= 6 和 b= 4 的结论。椭圆是水平的, 因为较大的值是顶点 x 值的一部分。方程式是 x236+y216=1 。

For part b, we know that $\begin{aligned} a = 9 \end{aligned}$ and $\begin{aligned} c = 5 \end{aligned}$ and the ellipse is vertical. Solve for $\begin{aligned} b \end{aligned}$ using $\begin{aligned} c^{2} = a^{2} - b^{2} \end{aligned}$ .
::对于 b 部分, 我们知道 a= 9 和 c= 5 以及椭圆是垂直的。使用 c2= a2 - b2 解决 b 。

\begin{aligned} 5^{2} & = 9^{2} - b^{2} \\ 25 & = 81 - b^{2} \\ b^{2} & = 56 \to b = 2 \sqrt{14} \end{aligned}

::52=92-b225=81-b2b2=56b=214

The equation is $\begin{aligned} \frac{x^{2}}{56} + \frac{y^{2}}{81} = 1 \end{aligned}$
::方程式是 x256+y281=1

Example 2
::例2

Find the equation of the ellipse with co-vertex $\begin{aligned} (0, - 7) \end{aligned}$ , focus $\begin{aligned} (15, 0) \end{aligned}$ , and centered at the origin.
::查找椭圆的方程,以共脊椎(0,-7)为焦点(15,0),以原点为中心。

Solution: Because the co-vertex is $\begin{aligned} (0, - 7), b = 7 \end{aligned}$ and the ellipse is horizontal. From the foci, we know that $\begin{aligned} c = 15 \end{aligned}$ . Find $\begin{aligned} a \end{aligned}$ .
::解答: 因为共同顶点是 (0, -7, b=7) , 椭圆是水平的。从顶部, 我们知道 c=15, 找到一个。

\begin{aligned} 15^{2} & = a^{2} - 7^{2} \\ a^{2} & = 225 + 49 = 274 & The equation is \frac{x^{2}}{274} + \frac{y^{2}}{49} = 1. \\ a & = \sqrt{274} \end{aligned}

::152=a2-72a2=225+49=274 方程式为 x2274+y249=1.a=274

by CK-12 demonstrates how to find the equation of an ellipse centered at the origin.
::由 CK-12 演示如何找到以原产物为主的椭圆方程。

Example 3
::例3

For a lithotripsy, a signal is sent from one focus and reflects around an elliptical shape to break up the stone at the other focus ¹ . (For more information, see the feature below.) If the equation that models a particular lithotripsy is $\begin{aligned} 16 x^{2} + 9 y^{2} = 144, \end{aligned}$ and the stone is located above the focus where the signal is sent from, where is the stone?
::对于闪石突变,信号从一个焦点发出,并反射到一个椭圆形状,以拆分另一个焦点的石头1。 (更多信息,请见下文特征。 )如果模型特定的闪石突变的方程式是 16x2+9y2=144, 并且石头位于发出信号的焦点之上, 石头在哪里?

Solution: This equation is not in standard form. For the equation to be in standard form, the right side of the equation must be 1. Divide everything by 144.
::解答: 此方程式不是标准形式。方程式要以标准形式出现, 方程式的右侧必须是 1. 1. 将一切除以 144 。

\begin{aligned} \frac{16 x^{2}}{144} + \frac{4 y^{2}}{144} & = \frac{144}{144} \\ \frac{x^{2}}{9} + \frac{y^{2}}{36} & = 1 \end{aligned}

::16x2144+4y2144=144144x29+y236=1

Now we can see that this is a vertical ellipse, where $\begin{aligned} b = 3 \end{aligned}$ and $\begin{aligned} a = 6 \end{aligned}$ .
::现在我们可以看到这是一个垂直的椭圆, b=3和a=6。

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

\begin{aligned} c^{2} & = 36 - 9 = 27 \\ c & = \sqrt{27} = 3 \sqrt{3} \end{aligned}

::c2=36-9=27c=27=33

The foci are therefore $\begin{aligned} (0, 3 \sqrt{3}), \end{aligned}$ where the stone is located, and $\begin{aligned} (0, - 3 \sqrt{3}), \end{aligned}$ where the signal is located.
::因此,方位是(0,33),石头所在,(0,33),信号所在。

Example 4
::例4

Find the center, vertices, co-vertices, and foci of $\begin{aligned} \frac{{(x + 4)}^{2}}{81} + \frac{{(y - 7)}^{2}}{16} = 1 \end{aligned}$ .
::查找 (x+4) 281+(y-7) 216=1 的中心、顶部、共同顶部和角。

Solution: The center is $\begin{aligned} (- 4, 7), a = \sqrt{81} = 9, \end{aligned}$ and $\begin{aligned} b = \sqrt{16} = 4 \end{aligned}$ , making the ellipse horizontal. The vertices are $\begin{aligned} (- 4 \pm 9, 7) \end{aligned}$ or $\begin{aligned} (- 13, 7) \end{aligned}$ and $\begin{aligned} (5, 7) \end{aligned}$ . The co-vertices are $\begin{aligned} (- 4, 7 \pm 4) \end{aligned}$ or $\begin{aligned} (- 4, 3) \end{aligned}$ and $\begin{aligned} (- 4, 11) \end{aligned}$ . Use $\begin{aligned} c^{2} = a^{2} - b^{2} \end{aligned}$ to find $\begin{aligned} c \end{aligned}$ .
::解答: 中心是 (4, 7, a=81=9, 和 b=16=4) , 使椭圆水平水平。脊椎是 (4, 9, 9, 7) 或 (13, 7) 和 (5, 7) 。共垂直是 (4, 7, 4, 4, 3) 或 (4, 4, 3) 和 (4, 11) , 使用 c2 =a2 - b2 查找 c 。

\begin{aligned} c^{2} & = 81 - 16 = 65 \\ c & = \sqrt{65} \end{aligned}

::c2=81-16=65c=65

The foci are $\begin{aligned} (- 4 - \sqrt{65}, 7) \end{aligned}$ and $\begin{aligned} (- 4 + \sqrt{65}, 7) . \end{aligned}$
::方向是(4-65,7)和(4+65,7)。

Example 5
::例5

Find the equation of the ellipse with co-vertices $\begin{aligned} (- 3, - 6) \end{aligned}$ and $\begin{aligned} (5, - 6) \end{aligned}$ and focus $\begin{aligned} (1, - 2) \end{aligned}$ .
::查找椭圆的等离子体方程式,加上共圆(3,6)和(5,6)以及焦点(1,2)。

Solution: The co-vertices $\begin{aligned} (- 3, - 6) \end{aligned}$ and $\begin{aligned} (5, - 6) \end{aligned}$ are the endpoints of the minor axis. It is $\begin{aligned} | - 3 - 5 | = 8 \end{aligned}$ units long, making $\begin{aligned} b = 4 \end{aligned}$ . The midpoint between the co-vertices is the center.
::解决方案 : 共白( 3, 6) 和 (5, 6) 是小轴的终点。共白( 3, 6) 是 & 3 - 5 - 5 - 5+8 单位的长度, 使 b= 4 。共白之间的中点是中点。

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

The focus is $\begin{aligned} (1, - 1) \end{aligned}$ and the distance between it and the center is 4 units, or $\begin{aligned} c \end{aligned}$ . Find $\begin{aligned} a \end{aligned}$ .
::焦点是(1,-1),它与中心之间的距离是4个单位,或 c. 查找一个。

\begin{aligned} 16 & = a^{2} - 16 \\ 32 & = a^{2} \\ a & = \sqrt{32} = 4 \sqrt{2} \end{aligned}

::16=a2 - 1632=a2a=32=42

The equation of the ellipse is $\begin{aligned} \frac{{(x - 1)}^{2}}{16} + \frac{{(y + 6)}^{2}}{32} = 1 \end{aligned}$ .
::椭圆的方程式是 (x-1) 216+(y+6)232=1。

by CK-12 demonstrates how to find the equation of an ellipse when you know the foci and the length of the major axis.
::由 CK-12 显示当您知道主轴的角和长度时如何找到椭圆的方程。

Graphing Ellipses
::绘制椭圆图

The process for graphing an ellipse is similar to the process for graphing a circle.
::图形化椭圆的过程与图形化圆的过程相似。

How to Graph an Ellipse
::如何绘制椭圆

1. Find the center and note whether the ellipse is oriented horizontally or vertically.
::1. 寻找中心并注意椭圆是横向方向还是纵向方向。

2. Using the center, determine the vertices and the co-vertices.
::2. 利用中心,确定脊椎和共白。

3. Draw an ellipse through the vertices and co-vertices.
::3. 在脊椎和共同脊椎中绘制椭圆。

Example 6
::例6

Find the vertices, co-vertices, and foci of $\begin{aligned} \frac{x^{2}}{64} + \frac{y^{2}}{25} = 1 \end{aligned}$ . Then graph the ellipse.
::查找 x264+y225=1 的顶部、共垂直和角。然后绘制椭圆图。

Solution: The center of this ellipse is at the origin. Next, b ecause $\begin{aligned} 64 > 25 \end{aligned}$ , we know that the ellipse will be horizontal.
::解决方案 : 此椭圆的中心在源头。接下来, 因为 64> 25, 我们知道椭圆是水平的。

$\begin{aligned} a^{2} = 64, \end{aligned}$ making $\begin{aligned} a = \sqrt{64} = 8, \end{aligned}$ and $\begin{aligned} b^{2} = 25 \end{aligned}$ , making $\begin{aligned} b = \sqrt{25} = 5 \end{aligned}$ . The vertices will be $\begin{aligned} (8, 0) \end{aligned}$ and $\begin{aligned} (- 8, 0), \end{aligned}$ and the co-vertices will be $\begin{aligned} (0, 5) \end{aligned}$ and $\begin{aligned} (0, - 5) \end{aligned}$ .
::a2=64, 制成 a=64=8, 和 b2=25, b=25=5, 使 b=25=5. 顶部为 (8,0) 和 (8,0) , 双面为 (0, 5) 和 (0, 5) 。

To graph the ellipse, plot the vertices and co-vertices and connect the four points to make the closed curve.
::绘制椭圆图,绘制脊椎和共同垂直图,并连接四点,以形成封闭曲线。

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To find the foci, we need to use the equation $\begin{aligned} c^{2} = a^{2} - b^{2} \end{aligned}$ and solve for $\begin{aligned} c \end{aligned}$ .
::要找到方块,我们需要使用方程式 c2=a2 -b2 并解决 c。

\begin{aligned} c^{2} & = 64 - 25 = 39 \\ c & = \sqrt{39} \end{aligned}

::c2=64-25=39c=39

The foci are $\begin{aligned} (\sqrt{39}, 0) \end{aligned}$ and $\begin{aligned} (- \sqrt{39}, 0) \end{aligned}$ .
::底部是(39,0)和(-39,0)。

Example 7
::例7

Graph $\begin{aligned} 49 x^{2} + 9 y^{2} = 441 \end{aligned}$ . Identify the foci.
::图49x2+9y2=441. 识别角。

Solution: This equation is not in standard form. To be in standard form, the right side of the equation must be 1. Divide everything by 441.
::解答: 此方程式不是标准形式。要以标准形式出现, 方程式的右侧必须是 1. 将每方除以441 。

\begin{aligned} \frac{49 x^{2}}{441} + \frac{9 y^{2}}{441} & = \frac{441}{441} \\ \frac{x^{2}}{9} + \frac{y^{2}}{49} & = 1 \end{aligned}

::49x2441+9y2441=441441x29+y249=1

Now we can see that this is a vertical ellipse, where $\begin{aligned} b = 3 \end{aligned}$ and $\begin{aligned} a = 7 \end{aligned}$ . The vertices are (0,7), (0,-7), (3,0), and (-3,0).
::现在我们可以看到,这是一个垂直的椭圆, b=3和a=7。脊椎是(0,7, 0, -7, (3,0) 和 (3,0) 。

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

\begin{aligned} c^{2} & = 49 - 9 = 40 \\ c & = \sqrt{40} = 2 \sqrt{10} \end{aligned}

::c2=49-9=40c=40=210

The foci are $\begin{aligned} (0, 2 \sqrt{10}) \end{aligned}$ and $\begin{aligned} (0, - 2 \sqrt{10}) . \end{aligned}$
::角为(0,210)和(0,210)。

Example 8
::例8

Graph $\begin{aligned} \frac{{(x - 3)}^{2}}{16} + \frac{{(y + 1)}^{2}}{4} = 1 \end{aligned}$ . Then find the vertices, co-vertices, and foci.
::图(x-3) 216+(y+1)24=1. 然后找到顶部、共同顶部和顶部。

Solution: First, we know this is a horizontal ellipse because $\begin{aligned} 16 > 4 \end{aligned}$ . The center is $\begin{aligned} (3, - 1) \end{aligned}$ and $\begin{aligned} a = 4 \end{aligned}$ and $\begin{aligned} b = 2 \end{aligned}$ .
::解决方案: 首先, 我们知道这是一个水平椭圆, 因为 16> 4。中心是 (3, 1) 和 a= 4 和 b= 2 。

To graph, plot the center and then go 4 units to the right and left and then up and down 2 units. This is also how you can find the vertices and co-vertices. The vertices are $\begin{aligned} (3 \pm 4, - 1) \end{aligned}$ or $\begin{aligned} (7, - 1) \end{aligned}$ and $\begin{aligned} (- 1, - 1) \end{aligned}$ . The co-vertices are $\begin{aligned} (3, - 1 \pm 2) \end{aligned}$ or $\begin{aligned} (3, 1) \end{aligned}$ and $\begin{aligned} (3, - 3) \end{aligned}$ .
::图形、绘制中心, 然后向右、左、上和向下移动 4 个单位。这也是如何找到顶点和共同的顶点。顶点是 (3+4、 1- 1 或 (7、 1 和 1- 1 ) 。双点是 (3, 1+ 2) 或 (3, 1) 和 (3, 3) 。

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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foci, 我们需要用 c2=a2 -b2 找到 c。

\begin{aligned} c^{2} & = 16 - 4 = 12 \\ c & = 2 \sqrt{3} \end{aligned}

::c2=16-4=12c=23

Therefore, the foci are $\begin{aligned} (3 \pm 2 \sqrt{3}, - 1) . \end{aligned}$
::因此,焦点是(323)-1。

by TucsonMathDoc shows how to graph an ellipse.
::TucsonMathDoc展示了如何绘制椭圆图。

Example 9
::例9

Graph $\begin{aligned} 49 (x - 5)^{2} + 25 (y + 2)^{2} = 1, 225 \end{aligned}$ and find the foci.
::图49(x-5)2+25(y+2)2=1,225,并找到角。

Solution: First we have to get this into standard form, like the equations above. To make the right side 1, we need to divide everything by 1,225.
::解决方案: 首先我们必须把它变成标准的形式, 像上面的方程式一样。为了右侧1, 我们需要将所有东西除以 1 225 。

\begin{aligned} \frac{49 {(x - 5)}^{2}}{1, 225} + \frac{25 {(y + 2)}^{2}}{1, 225} & = \frac{1, 225}{1, 225} \\ \frac{{(x - 5)}^{2}}{25} + \frac{{(y + 2)}^{2}}{49} & = 1 \end{aligned}

::49(x-5)21 225+25(y+2)21 225=1,2251,225(x-55)225+(y+2)249=1

Now we know the ellipse will be vertical because $\begin{aligned} 25 < 49 \end{aligned}$ . $\begin{aligned} a = 7, b = 5, \end{aligned}$ and the center is $\begin{aligned} (5, - 2) \end{aligned}$ .
::现在我们知道椭圆是垂直的因为25<49. a=7,b=5,中心是5,2。

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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 = 24 \\ c & = \sqrt{24} = 2 \sqrt{6} \end{aligned}

::c2=49-25=24c=24=26

The foci are $\begin{aligned} (5, - 2 \pm 2 \sqrt{6}) \end{aligned}$ or $\begin{aligned} (5, - 6.9) \end{aligned}$ and $\begin{aligned} (5, 2.9) \end{aligned}$ .
::方块是(5,-226)或(5,-6.9)和(5,2.9)。

How to Graph an Ellipse Using Desmos
::如何使用 Desmos 绘制椭圆

Enter the equation as it is. To get an exponent, use the caret key ^ (Shift 6), and to divide, use the slash key /.
::输入正方程。要获得一个引号, 请使用 {( Shift 6) } 和分割, 请使用 scrap 键 / 。

How to Graph an Ellipse Using a TI-83/84
::如何使用TI-83/84来绘制椭圆

1. Press APPS and choose Conics . (Conics is an application that does not come with all TI calculators. You may need to load it from the TI website.)
::1. 按 APPS 键并选择二次曲线。 (二次曲线是一个不是所有TI计算器一起产生的应用程序,可能需要从TI网站上加载。 )

2. Choose Ellipse.
::2. 选择椭圆。

3. Choose the form that matches your equation—that is, horizontal major axis or vertical major axis.
::3. 选择与您的方程(即水平主轴或垂直主轴)相匹配的窗体。

4. Input your values for $\begin{aligned} a, b, h, \end{aligned}$ and $\begin{aligned} k \end{aligned}$ .
::4. 输入a、b、h和k的数值。

5. Press GRAPH .
::5. 出版《GRAPH》。

Feature: Ouch! Fix It With an Ellipse
::特写:哎哟!用椭圆来修补它

by Meredith Beaton
::梅雷迪思·贝顿(签名)

Gallstones, or stones in the common bile duct usually originating in the gallbladder, have been described as one of the most painful acute illnesses—even worse than childbirth! Gallstones form because the bile (filled with salt crystals) sits in the gallbladder for too long. The crystals in the bile have the time to come together and form larger crystals that then try to leave the gallbladder through the cystic duct and enter the common bile duct.
::石块,或通常产自胆囊的常见卵石管中的石块,被描述为最痛苦的急性疾病之一——甚至比分娩更糟糕!石块的形式是因为(装满盐晶体的)卵石在胆囊中坐太久了。胆囊中的晶体有时间聚集在一起,形成更大的晶体,然后试图通过胆囊管离开胆囊,进入共同的胆囊管。

News You Can Use
::您可以使用新闻新闻

The pain of gallstones can be excruciating, and the backup of bile and other digestive juices when the duct is blocked can be painful and also make patients nauseous. Unfortunately, gallstones can affect anyone. But medical science has many effective methods for treating gallstones, including lithotripsy, which uses special properties of ellipses to destroy the stones!
::石碑的痛苦可能令人痛苦,而管道被堵时的肉汁和其他消化液的备份可能令人痛苦,也使病人恶心。不幸的是,石碑可以影响任何人。但医学有多种有效的方法处理石碑,包括使用石浆特殊特性摧毁石块的石碑。

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One important property of the ellipse is its reflective property. If an energy wave is emitted from one focus of an ellipse, it will pass through the 2nd focus of an ellipse. Doctors use this property to send shock waves to a gallstone to break it up without having to perform surgery. If the shock-wave generator is placed at one focus, and the gallstone is at the other focus (its position can be found on an ultrasound), the lithotripter (shaped as a half-ellipse) can generate a shock wave that will destroy just the stone without disrupting other tissues of the body!
::椭圆的一个重要属性是它的反射特性。如果能量波从椭圆的一个焦点排放出来, 它会通过椭圆的第二个焦点。医生们使用这种特性将冲击波送入一个石碑, 以将其碎裂而不必进行手术。如果冲击波发电机放在一个焦点, 而石块则在另一个焦点( 其位置可以在超声波上找到) , 光电波( 半椭圆形) 可以产生冲击波, 震波将摧毁石头, 而不会干扰身体的其他组织 !

by Howard County General Hospital describes the gallbladder and stones.
::霍华德县总医院描述了胆囊和石头。

Summary
::摘要

An ellipse is the set of all points such that the sum of the distances from two fixed points, called foci (the plural of focus), is constant.
::椭圆是所有点的组合,使两个固定点(称为焦点复数)的距离总和保持不变。

The standard forms of an equation of an ellipse are $\begin{aligned} \frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1 \end{aligned}$ (horizontal major axis) and $\begin{aligned} \frac{(x - h)^{2}}{b^{2}} + \frac{(y - k)^{2}}{a^{2}} = 1 \end{aligned}$ (vertical major axis). We assume $\begin{aligned} a > b . \end{aligned}$
::椭圆方程式的标准形式是 (x-h) 2a2+(y-k) 2b2=1 (横向主要轴) 和 (x-h) 2b2+(y-k) 2a2=1 (垂直主轴)。

To graph an ellipse, we find the center, the vertices, and the co-vertices. The ellipse goes through the vertices and co-vertices.
::要绘制椭圆图,我们找到中心, 脊椎, 和共圆。椭圆穿过顶和共圆。

Review
::回顾

Find the equation of the ellipse with the given information.
::用给定的信息查找椭圆的方程。

1. vertex: $\begin{aligned} (- 3, 0) \end{aligned}$ ; co-vertex: $\begin{aligned} (0, 1) \end{aligned}$
::1. 顶点3,30);共同顶点0,1)

2. co-vertex: $\begin{aligned} (7, 0) \end{aligned}$ ; centered at the origin; major axis: 18 units
::2. 共同顶点: (7,0),以原点为中心;主轴: 18个单位

3. vertex: $\begin{aligned} (0, 5) \end{aligned}$ ; centered at the origin; minor axis: 4 units
::3. 顶点: (0,5); 以原点为中心;小轴: 4个单位

4. co-vertex: $\begin{aligned} (17, 0) \end{aligned}$ ; focus: $\begin{aligned} (0, 17) \end{aligned}$
::4. 共同顶点: (17,0); 重点: (0,17)

5. vertex: $\begin{aligned} (4, 0) \end{aligned}$ ; focus: $\begin{aligned} (- 3, 0) \end{aligned}$
::5. 顶点: (4,0); 重点3,0)

6. focus: $\begin{aligned} (0, - 9) \end{aligned}$ ; minor axis: 16 units
::6. 重点0,-9);小轴:16个单位

7. vertices: $\begin{aligned} (- 2, - 3) \end{aligned}$ and $\begin{aligned} (8, - 3) \end{aligned}$ ; co-vertex: $\begin{aligned} (3, - 5) \end{aligned}$
::7. 顶点2-3)和(8.3);共同顶点3-5)

8. vertices: $\begin{aligned} (5, 6) \end{aligned}$ and $\begin{aligned} (5, - 12) \end{aligned}$ ; focus: $\begin{aligned} (5, - 7) \end{aligned}$
::8. 顶点5,6)和(5,12);重点5,7)

9. co-vertices: $\begin{aligned} (0, 4) \end{aligned}$ and $\begin{aligned} (14, 4) \end{aligned}$ ; focus: $\begin{aligned} (7, 1) \end{aligned}$
::9. 共同垂直数0,4)和(14,4);重点7,1)

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

11.

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12.

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13.

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Find the vertices, co-vertices, and foci of each ellipse below. Then graph.
::查找下方每个椭圆的顶部、共同的顶部和角。然后图形。

14. $\begin{aligned} \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \end{aligned}$
::14. X29+y216=1

15. $\begin{aligned} 4 x^{2} + 25 y^{2} = 100 \end{aligned}$
::15. 4x2+25y2=100

16. $\begin{aligned} \frac{x^{2}}{64} + y^{2} = 1 \end{aligned}$
::16. x264+y2=1

17. $\begin{aligned} 81 x^{2} + 100 y^{2} = 8, 100 \end{aligned}$
::17. 81x2+100y2=8 100

15. $\begin{aligned} \frac{x^{2}}{49} + \frac{y^{2}}{16} = 1 \end{aligned}$
::15. x249+y216=1

16. $\begin{aligned} 121 x^{2} + 9 y^{2} = 1, 089 \end{aligned}$
::16. 121x2+9y2=1 089

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

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

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

20. $\begin{aligned} 25 x^{2} + 64 (y - 6)^{2} = 1, 600 \end{aligned}$
::20. 25x2+64(y-6)2=1 600

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

22. $\begin{aligned} 81 (x + 4)^{2} + 4 (y + 5)^{2} = 324 \end{aligned}$
::22. 81(x+4)2+4(y+5)2=324

Explore More
::探索更多

1. A portion of the backyard of the White House is called The Ellipse. The major axis is 1,058 feet and the minor axis is 903 feet ² . Find the equation of the horizontal ellipse, assuming it is centered at the origin.
::1. 白宫后院的一部分称为 " 椭圆 " ,主轴为1 058英尺,小轴为903英尺。

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2. a. Find the equations of an ellipse and a parabola that are confocal. b. Find the equations of two ellipses that are confocal.
::2. a. 查找椭圆和抛物线的等式,它们是锥形。 b. 查找两个椭圆的等式,它们是锥形。

3. If the equation for an ellipse is $\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ , then find the equation of a circle contained in the ellipse that intersects the ellipse at two points.
::3. 如果椭圆的方程式是 x2a2+y2b2=1,那么在两个点上找到椭圆内圆的方程式,该方程式将椭圆相交。

4. Assume a racetrack is elliptical in shape. If the major axis is $\begin{aligned} \frac{9}{10} \end{aligned}$ of a mile, and the minor axis is $\begin{aligned} \frac{3}{10} \end{aligned}$ of mile, find an equation to describe the racetrack if we assume the center is the origin.
::4. 假设赛道是椭圆形的,如果主轴为每英里910,小轴为每英里310,如果我们假设中枢为原轴,请找到一个方程式来描述赛道。

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 interactives to explore the concepts in this section:
::尝试以下互动来探索本节中的概念 :

References
::参考参考资料

1. "Gallstones Non-Surgical Treatment: Lithotripsy, Contact Dissolution Therapy,"
::1. “石块非治疗性治疗:立石裂,接触性溶解疗法”,

2. "The Ellipse," last edited February 15, 2017,
::2017年2月15日经编辑

Section outline

General

椭圆

Geometry of Ellipses ::椭圆的几何

Geometry of an Ellipse ::椭圆的几何

Investigation: Drawing an Ellipse ::调查:绘制椭圆

Equations of Ellipses ::椭圆平

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Graphing Ellipses ::绘制椭圆图

How to Graph an Ellipse ::如何绘制椭圆

Example 6 ::例6

Example 7 ::例7

Example 8 ::例8

Example 9 ::例9

How to Graph an Ellipse Using Desmos ::如何使用 Desmos 绘制椭圆

How to Graph an Ellipse Using a TI-83/84 ::如何使用TI-83/84来绘制椭圆

Feature: Ouch! Fix It With an Ellipse ::特写:哎哟!用椭圆来修补它

News You Can Use ::您可以使用新闻新闻

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

References ::参考参考资料

Geometry of Ellipses
::椭圆的几何

Geometry of an Ellipse
::椭圆的几何

Investigation: Drawing an Ellipse
::调查:绘制椭圆

Equations of Ellipses
::椭圆平

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Graphing Ellipses
::绘制椭圆图

How to Graph an Ellipse
::如何绘制椭圆

Example 6
::例6

Example 7
::例7

Example 8
::例8

Example 9
::例9

How to Graph an Ellipse Using Desmos
::如何使用 Desmos 绘制椭圆

How to Graph an Ellipse Using a TI-83/84
::如何使用TI-83/84来绘制椭圆

Feature: Ouch! Fix It With an Ellipse
::特写:哎哟!用椭圆来修补它

News You Can Use
::您可以使用新闻新闻

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

References
::参考参考资料