课程： 10.11 二次曲线各区分类

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分类二次曲线区域

You and your friends are playing Name the Conic Section. Your friend pulls a card with the equation $\begin{aligned} x^{2} + 3 x y = - 5 y^{2} - 10 \end{aligned}$ written on it. What type of conic section is represented by the equation?
::你和你的朋友们正在玩名称二次曲线部分。您的朋友拉一张卡片, 上面写着方程式 x2+3xy5y2- 10 。方程式代表哪种二次曲线部分 ?

Classifying Conic Sections
::分类二次曲线区域

Another way to classify a conic section when it is in the general form is to use the discriminant, like from the Quadratic Formula. The discriminant is what is underneath the radical, $\begin{aligned} b^{2} - 4 a c \end{aligned}$ , and we can use this to determine if the conic is a parabola , circle , ellipse, or . If the general form of the equation is $\begin{aligned} A x^{2} + B x y + C y^{2} + D x + E y + F = 0 \end{aligned}$ , where $\begin{aligned} B = 0 \end{aligned}$ , then the discriminant will be $\begin{aligned} B^{2} - 4 A C \end{aligned}$ .
::当二次曲线部分处于一般形式时,另一种对二次曲线部分进行分类的方法是使用比方方方程式,如“二次曲线公式”中的比方方方程式。比方方方程式 b2 - 4ac 下方方方程式为“b2 - 4ac ” ,我们可以用此方法确定二次曲线是“parbola”、“圆”、“椭圆”还是“椭圆”。如果方程式的一般形式是Ax2+Bxy+Cy2+Dx+Ey+F=0,B=0,那么比方方方方程式将是“B2 - 4AC ” 。

Use the table below:
::使用下表:

$\begin{aligned} B^{2} - 4 A C = 0 \end{aligned}$ and $\begin{aligned} A = 0 \end{aligned}$ or $\begin{aligned} C = 0 \end{aligned}$	Parabola
$\begin{aligned} B^{2} - 4 A C < 0 \end{aligned}$ and $\begin{aligned} A = C \end{aligned}$	Circle
$\begin{aligned} B^{2} - 4 A C < 0 \end{aligned}$ and $\begin{aligned} A \neq C \end{aligned}$	Ellipse
$\begin{aligned} B^{2} - 4 A C > 0 \end{aligned}$	Hyperbola

Let's use the discriminant to determine the type of conic section for the following equations.
::让我们用对话者来决定下方方程式的二次曲线段类型。

$\begin{aligned} x^{2} - 4 y^{2} + 5 x - 8 y + 16 = 0 \end{aligned}$
::x2 - 4y2+5x-8y+16=0

$\begin{aligned} A = 1 \end{aligned}$ , $\begin{aligned} B = 0 \end{aligned}$ , and $\begin{aligned} C = - 4 \end{aligned}$
::A=1,B=0,C=4

$\begin{aligned} 0^{2} - 4 (1) (- 4) = 16 \end{aligned}$ This is a hyperbola.
::02-4(1)(-4)=16 这是超双曲线。

$\begin{aligned} 3 x^{2} + 3 y^{2} - 9 x - 12 y - 20 = 0 \end{aligned}$
::3x2+3y2-9x-12y-20=0

$\begin{aligned} A = 3 \end{aligned}$ , $\begin{aligned} B = 0 \end{aligned}$ , $\begin{aligned} C = 3 \end{aligned}$
::A=3,B=0,C=3 A=3,B=0,C=3

$\begin{aligned} 0^{2} - 4 (3) (3) = - 36 \end{aligned}$ Because $\begin{aligned} A = C \end{aligned}$ and the discriminant is less than zero, this is a circle.
::02-4(3)(3)36 因为A=C和共和党人小于零,这是一个圆。

Finally, let's use the discriminant to determine the type of conic. Then, we'll change the equation into standard form to verify our answer. We'll also find the center (or vertex, if it is a parabola). $\begin{aligned} x^{2} + y^{2} - 6 x + 14 y - 86 = 0 \end{aligned}$
::最后,让我们使用对话者来决定二次曲线的类型。然后, 我们将将方程式修改为标准格式来验证答案。我们还会找到中心( 或顶点, 如果是抛物线的话) x2+y2 - 6x+14y- 86=0

$\begin{aligned} A = 1, B = 0, C = 1 \end{aligned}$ This is a circle.
::A=1, B=0, C=1 这是圆形。

\begin{aligned} (x^{2} - 6 x + 9) + (y^{2} + 14 y + 49) & = 86 + 49 + 9 \\ (x - 3)^{2} + (y + 7)^{2} & = 144 \end{aligned}

x2-6x+9)+(y2+14y+49)=86+49+9(x-3)2+(y+7)2=144

The center is $\begin{aligned} (3, - 7) \end{aligned}$ .
::中心是 (3, - 7) 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine the type of conic section represented by the equation $\begin{aligned} x^{2} + 3 x y = - 5 y^{2} - 10 \end{aligned}$ .
::早些时候,您被要求确定方程式 x2+3xy5y2- 10 所代表的二次曲线段的类型。

First we need to rewrite the equation is standard form.
::首先我们需要重写方程式是标准格式

\begin{aligned} x^{2} + 3 x y = - 5 y^{2} - 10 x^{2} + 3 x y + 5 y^{2} + 10 = 0 \end{aligned}

::x2+3xy%5y2 - 10x2+3xy+5y2+10=0

Now we can use the discriminant to find the type of conic section represented by the equation.
::现在我们可以使用对立方程式找到方程式所代表的二次曲线段的类型。

$\begin{aligned} A = 1, B = 3, C = 5 \end{aligned}$
::A=1,B=3,C=5 A=1,B=3,C=5

$\begin{aligned} 3^{2} - 4 (1) (5) = - 11 \end{aligned}$ Because $\begin{aligned} A \neq C \end{aligned}$ and the discriminant is less than zero, this equation represents an ellipse.
::32-4(1)(5)11 因为AC和对立方体小于零,这个方程式代表椭圆。

For Examples 2 & 3, use the discriminant to determine the type of conic.
::对于例2和例3,使用歧见来决定二次曲线的类型。

Example 2
::例2

$\begin{aligned} 2 x^{2} + 5 y^{2} - 8 x + 25 y + 115 = 0 \end{aligned}$
::2x2+5y2-8x+25y+115=0

$\begin{aligned} 0^{2} - 4 (2) (5) = - 40 \end{aligned}$ , this is an ellipse.
::02 -4(2)(5)40,这是一个椭圆。

Example 3
::例3

$\begin{aligned} 5 y^{2} - 9 x - 10 y - 14 = 0 \end{aligned}$
::5y2-9x-10y-14=0

$\begin{aligned} 0^{2} - 4 (0) (5) = 0 \end{aligned}$ , this is a parabola.
::02-4(0)(5)=0,这是一个抛物线。

Example 4
::例4

Use the discriminant to determine the type of conic. Then, change the equation into standard form to verify your answer. Find the center or vertex, if it is a parabola.
::使用 dispriminant 来确定二次曲线的类型。然后, 将方程式更改为标准格式以验证您的答复。如果它是抛物线, 请找到中心或顶点。

$\begin{aligned} - 4 x^{2} + 3 y^{2} - 8 x + 24 y + 32 = 0 \end{aligned}$
::− 4x2+3y2-8x+24y+32=0

$\begin{aligned} 0^{2} - 4 (- 4) (3) = 48 \end{aligned}$ , this is a hyperbola. Changing it to standard form, we have:
::02-4(-4)(3)=48,这是一个双倍波拉。将其改为标准形式,我们有:

\begin{aligned} (- 4 x^{2} - 8 x) + (3 y^{2} + 24 y) & = - 32 \\ - 4 (x^{2} + 2 x + 1) + 3 (y^{2} + 8 y + 16) & = - 32 + 48 - 4 \\ - 4 (x + 1)^{2} + 3 (y + 4)^{2} & = 12 \\ \frac{- (x + 1)^{2}}{3} + \frac{(y + 4)^{2}}{4} & = 1 \end{aligned}

-4x2-8x)+(3y2+24y)+(3y2+24y)\32-4(x2+2x+1)+3(y2+8y+16)+3(y2+8y+16)+32+48-4-4(x+1)+3+3(y+4)2+3(y+4)2=12-(x+1)23+(y+4)24=1

Usually, we write the negative term second, so the equation is $\begin{aligned} \frac{(y + 4)^{2}}{4} - \frac{(x + 1)^{2}}{3} = 1 \end{aligned}$ . The center is $\begin{aligned} (- 1, - 4) \end{aligned}$ .
::通常,我们写负数第二行,所以方程式是(y+4)24-(x+1)23=1.中心是(y+1)-4。

Review
::回顾

Use the discriminant to determine the type of conic each equation represents.
::使用对比度来确定每个方程代表的二次曲线类型。

$\begin{aligned} 2 x^{2} + 2 y^{2} + 16 x - 8 y + 25 = 0 \end{aligned}$
::2x2+2y2+16x-8y+25=0

$\begin{aligned} x^{2} - y^{2} - 2 x + 5 y - 12 = 0 \end{aligned}$
::x2 - y2 - 2x+5y - 12=0

$\begin{aligned} 6 x^{2} + y^{2} - 12 x + 7 y + 35 = 0 \end{aligned}$
::6x2+y2 - 12x+7y+35=0

$\begin{aligned} 3 x^{2} - 15 x + 9 y - 18 = 0 \end{aligned}$
::3x2 - 15x+9y - 18=0

$\begin{aligned} 10 y^{2} + 6 x - 40 y + 253 = 0 \end{aligned}$
::10y2+6x-40y+253=0

$\begin{aligned} 4 x^{2} + 4 y^{2} + 32 x + 48 y + 465 = 0 \end{aligned}$
::4x2+4y2+32x+48y+465=0

Match the equation with the correct graph.
::方程式与正确的图形匹配。

$\begin{aligned} x^{2} + 10 x + 4 y + 41 = 0 \end{aligned}$
::x2+10x+4y+41=0

$\begin{aligned} 4 y^{2} + x + 56 y + 188 = 0 \end{aligned}$
::4y2+x+56y+188=0

$\begin{aligned} x^{2} + y^{2} + 10 x - 14 y + 65 = 0 \end{aligned}$
::x2+y2+10x-14y+65=0

$\begin{aligned} 25 x^{2} + y^{2} - 200 x - 10 y + 400 = 0 \end{aligned}$
::25x2+y2-200x-10y+400=0

$\begin{aligned} x^{2} - 12 x + 6 y + 66 = 0 \end{aligned}$
::x2 - 12x+6y+66=0

$\begin{aligned} x^{2} + y^{2} + 2 x + 2 y - 2 = 0 \end{aligned}$
::x2+y2+2x+2y-2y=0

$\begin{aligned} x^{2} - y^{2} - 10 x - 10 y - 10 = 0 \end{aligned}$
::x2 - y2 - 10x - 10y - 10=0

$\begin{aligned} y^{2} - 10 x + 8 y + 46 = 0 \end{aligned}$
::y2 - 10x+8y+46=0

Find the Area of an Ellipse Graph $\begin{aligned} x^{2} + y^{2} = 36 \end{aligned}$ and find its area.
1. Then, graph $\begin{aligned} \frac{x^{2}}{36} + \frac{y^{2}}{25} = 1 \end{aligned}$ and $\begin{aligned} \frac{x^{2}}{25} + \frac{y^{2}}{36} = 1 \end{aligned}$ on the same axes.
  ::然后,在同一轴上的图x236+y225=1和x225+y236=1。
2. Do these ellipses have the same area? Why or why not?
  ::这些椭圆有相同的区域吗?
3. If the equation of the area of a circle is $\begin{aligned} A = π r^{2} \end{aligned}$ , what do you think the area of an ellipse is? Use $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ as in the standard form, $\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ .
  ::如果圆区域的方程式是 Ar2, 您认为椭圆区域是什么? 使用标准窗体中的 a 和 b , x2a2+y2b2=1 。
4. Find the areas of the ellipses from part a. Are the areas more or less than the area of the circle? Why or why not?
  ::查找 a 部分的省略号区域。区域是否或多或少于圆区域? 为什么或为何不?
::查找 Ellipse Graph x2+y2=36 的区域, 并查找其区域。然后, 在同一轴上查找图 x236+y225=1 和 x225+y236=1 。这些椭圆是否具有相同的区域? 为什么? 如果圆区域方程式是 Ar2, 您认为椭圆区域是什么? 使用标准格式的 a 和 b , x2a2+y2b2=1 。查找部分a 的椭圆区域。区域是否大于或小于圆区域? 为什么或不?

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。

章节大纲

常规

分类二次曲线区域

Classifying Conic Sections ::分类二次曲线区域

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)