Course: 16.11 答复----第11章:二次曲线

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答案 - Ch 11:二次曲线
Section 11.2: General Form of a Conic
::第11.2节:二次曲线的一般形式

Ellipse
::椭圆

Circle
::圆圆

Ellipse
::椭圆

Sideways parabola
::侧道抛物线

Hyperbola
::超重波

Parabola
::帕拉波拉

$\begin{align*}(x+2)^2-4\end{align*}$
:x+2)2-4

$\begin{align*}(y-4)^2-16\end{align*}$
:y-4)2--16

$\begin{align*}3(x+1)^2 + 1\end{align*}$
::3(x+1)2+1

$\begin{align*}3(y+\tfrac{3}{2})^2+\tfrac{33}{4}\end{align*}$
::3(y+322)+334

$\begin{align*}2(x-3)^2 - 17\end{align*}$
::2(x-3-3)2-17

$\begin{align*}4(x-2)^2+(y+1)^2=16\end{align*}$
::4 (x- 2)2+(y+1)2=16

$\begin{align*}9(x-3)^2+(y-1)^2=1\end{align*}$
::9(x-3)2+(y-1)2=1

$\begin{align*}3(x-1)^2+4y^2=12\end{align*}$
::3(x-1)2+4y2=12

$\begin{align*}y=(x+2)^2-3\end{align*}$
::y= (x+2) 2-2- 3

Section 11.3: Parabolas
::第11.3节:准军事部队

$\begin{align*}(x-1)^2=4\cdot3(y-1)\end{align*}$
:x-1)2=43(y-1)

$\begin{align*}(y+1)^2=\text{-}4\cdot3(x-1)\end{align*}$
:y+1)2=-43(x-1)

$\begin{align*}(y-4)^2=4\cdot3(x-5)\end{align*}$
:y-4)2=43(x-5)

$\begin{align*}(x-1)^2 = \text{-}4\cdot4(y-8)\end{align*}$
:x-1)2=-44(y-8)

$\begin{align*}(y-4)^2=\text{-}4\cdot3(x-1)\end{align*}$
:y-4)2=-43(x-1)

$\begin{align*}(x-6)^2=\text{-}4\cdot5(y-4)\end{align*}$
:x-6)2=-45(y-4)

$\begin{align*}(x-1)^2=4(y-11)\end{align*}$
:x-1)2=4(y-11)

$\begin{align*}(y+1)^2=4\cdot2(x-4)\end{align*}$

:y+1)2=4(8)2(x-4)

$\begin{align*}(x-4)^2=\text{-}4\cdot5(y+1)\end{align*}$

:x-4)2=-45(y+1)

$\begin{align*}(x+3)^2=4 \cdot 3(y+2)\end{align*}$

:x+3)2=43(y+2)

$\begin{align*}(y-6)^2=\text{-}4\cdot2(x-4)\end{align*}$

:y-6)2=-4(8)2(x-4)

$\begin{align*}(x+3)^2=4(y-3)\end{align*}$

:x+3)2=4(y-3)

$\begin{align*}(y+7)^2=4\cdot \tfrac{1}{2}(x-5)\end{align*}$

:y+7)2=412(x-5)

$\begin{align*}(x+6)^2=4\cdot \tfrac{2}{3}(y+2)\end{align*}$

:x+6)2=4.23(y+2)

"> $\begin{align*}(x-1)^2=4 \cdot \tfrac{1}{5}(y)\end{align*}$

:x-1)2=415

Section 11.4: Circles
::第11.4节:圆圈

$\begin{align*}(x-2)^2+(y+5)^2=11\end{align*}$
:x-2)2+(y+5)2=11

$\begin{align*}(x+1)^2+(y-4)^2=16\end{align*}$
:x+1)2+(y-4)2=16

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

$\begin{align*}(x+1)^2+(y+7)^2=25\end{align*}$
:x+1)2+(y+7)2=25

$\begin{align*}(x-1)^2+(y-1)^2=2\end{align*}$
:x-1)2+(y-1)2=2

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

$\begin{align*}(x+7)^2+(y-1)^2=16\end{align*}$
:x+7)2+(y-1)2=16

$\begin{align*}x^2+(y+5)^2=25\end{align*}$
::x2+(y+5)2=25

$\begin{align*}x^2 + y^2 =4\end{align*}$
::x2+y2=4 x2+y2=4

$\begin{align*}(x+2)^2+y^2=1\end{align*}$
:x+2)2+y2=1

Section 11.5: Ellipses
::第11.5节:椭圆

Vertices: (1, -1) and (1, -9)
Foci: (1, $\begin{align*}\text{-}5 + \sqrt{12}\end{align*}$ ), (1, $\begin{align*}\text{-}5-\sqrt{12}\end{align*}$ )
Eccentricity: $\begin{align*}\tfrac{\sqrt{12}}{4}\end{align*}$
::顶点1,-1)和(1,-9)

Vertices: (2, -2) and (-4, 2)
Foci: ( $\begin{align*}\text{-}1+\sqrt{5}\end{align*}$ , -2), ( $\begin{align*}\text{-}1-\sqrt{5}\end{align*}$ , -2)
Eccentricity: $\begin{align*}\tfrac{\sqrt{5}}{3}\end{align*}$
::顶点2, 2) 和 (4, 2) 准点1+5, 2) , (1+5, 2) , (1-5, 5) , (2) 偏心度: 53

Vertices: (2, 3) and (2, -1)
Foci: (2, $\begin{align*}1 + \sqrt{3}\end{align*}$ ), (2, $\begin{align*}1- \sqrt{3}\end{align*}$ )
Eccentricity: $\begin{align*}\tfrac{\sqrt{3}}{2}\end{align*}$
::顶点2,3)和(2,-1) 准点2,1+3),(2,1+3),(2,1-3) 偏心度:32

$\begin{align*}\frac{(x+1)^2}{16}+\frac{(y+7)^2}{4}=1\end{align*}$
:x+1)216+(y+7)24=1

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

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

$\begin{align*}\frac{(x-4)^2}{9}+\frac{(y-3)^2}{25}=1\end{align*}$
:x-4)29+(y-3)225=1

$\begin{align*}\frac{(x-4)^2}{36}+\frac{(y-1)^2}{4}=1\end{align*}$
:x-4)236+(y-1)24=1

$\begin{align*}\frac{(x+2)^2}{25}+\frac{(y-4)^2}{9}=1\end{align*}$
:x+2)225+(y-4)29=1

$\begin{align*}\frac{(x-6)^2}{25}+\frac{(y+2)^2}{169}=1\end{align*}$
:x-6)225+(y+2)2169=1

Possible answer: $\begin{align*}\tfrac{x^2}{50^2} +\tfrac{y^2}{25^2}=1\end{align*}$
::可答: x2502+y2252=1

17.86 feet
::17.86英尺

Section 11.6: Hyperbolas
::第11.6节:超黄体

$\begin{align*}\frac{{(x+1)}^{2}}{16}-\frac{(y+3)^2}{4}=1\end{align*}$ . This is a hyperbola because the coefficients of $\begin{align*}{x}^{2}\end{align*}$ and $\begin{align*}{y}^{2}\end{align*}$ have opposite signs.
:x+1)216-(y+3)24=1.这是一个双倍波拉,因为 x2 和 y2 的系数有相反的符号。

This hyperbola opens side to side.
::这双重波拉打开侧面的侧面。

The vertices are at (-5, -3) and (-3, -3).
::顶点为(5,5,3)和(3,3)。

The asymptotes are at $\begin{align*}y= \pm \tfrac{1}{2}(x+1)-3.\end{align*}$
::asymptotes 位于 y12( x+1) - 3 。

$\begin{align*}\text{-}\frac{(x+2)^2}{16}+\frac{(y-1)^2}{9}=1\end{align*}$ . This is a hyperbola because the coefficients of $\begin{align*}{x}^{2}\end{align*}$ and $\begin{align*}{y}^{2}\end{align*}$ have opposite signs.
::- (x+2)216+(y-1)29=1. 这是一个双倍波拉,因为 x2 和 y2 的系数有相反的符号。

Opens up and down.
::向上和向下开放。

(-2, 4) and (-2, -2)
:-2,4个)和(-2,2个)

$\begin{align*}y=\pm \tfrac{3}{4}(x+2)+1\end{align*}$
::y34(x+2)+1

$\begin{align*}\frac{(x-3)^2}{9}-\frac{(y+3)^2}{1}=1\end{align*}$ . This is a hyperbola because the coefficients of $\begin{align*}{x}^{2}\end{align*}$ and $\begin{align*}{y}^{2}\end{align*}$ have opposite signs.
:x-3)29-(y+3)21=1. 这是一个双倍波拉,因为 x2 和 y2 的系数有相反的符号。

Opens side to side
::将侧对侧打开

Vertices are at (6, -3) and (0, -3).
::顶点在(6,3)和(0,3)之间。

$\begin{align*}y = \pm \tfrac{1}{3}(x-3)-3\end{align*}$
::y13(x-3)-3-3

Section 11.7: Degenerate Conics
::第11.7节:离子二次曲线

The degenerate form of a hyperbola is two intersecting lines, of a circle is a point, and of a parabola is a line or two parallel lines.
::双倍波拉的退化形式是两条交叉线,圆是一个点,抛物线是一条线或两条平行线。

$\begin{align*}\frac{(x-3)^2}{9}-\frac{(y+3)^2}{1}=0;\end{align*}$ degenerate hyperbola
:x-3)29-(y+3)21=0;退化双曲线

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

$\begin{align*}\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=0;\end{align*}$ point
:x+2)24+(y-3)29=0;点

$\begin{align*}\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=1;\end{align*}$ ellipse
:x+2)24+(y-3)29=1;椭圆

$\begin{align*}\tfrac{5}{2}x+\tfrac{1}{2}=y\end{align*}$ ; line
::52x+12=y; 行

$\begin{align*}y = (x+2)^2+4;\end{align*}$ parabola
::y= (x+2) 2+4; 抛物线

$\begin{align*}(x-1)^2+(y-3)^2= 4;\end{align*}$ circle
:x-1)2+(y-3);2=4;圆

$\begin{align*}\frac{(x-1)^2}{4}-(y-3)^2=0;\end{align*}$ degenerate hyperbola
:x-1)24-(y-3)2=0;衰变双曲线

$\begin{align*}\frac{(x-1)^2}{4}+(y-3)^2=1;\end{align*}$ ellipse
:x-1)24+(y-3);2=1;椭圆

Section outline

General

答案 - Ch 11:二次曲线

Section 11.2: General Form of a Conic ::第11.2节:二次曲线的一般形式

Section 11.3: Parabolas ::第11.3节:准军事部队

Section 11.4: Circles ::第11.4节:圆圈

Section 11.5: Ellipses ::第11.5节:椭圆

Section 11.6: Hyperbolas ::第11.6节:超黄体

Section 11.7: Degenerate Conics ::第11.7节:离子二次曲线

Section 11.2: General Form of a Conic
::第11.2节:二次曲线的一般形式

Section 11.3: Parabolas
::第11.3节:准军事部队

Section 11.4: Circles
::第11.4节:圆圈

Section 11.5: Ellipses
::第11.5节:椭圆

Section 11.6: Hyperbolas
::第11.6节:超黄体

Section 11.7: Degenerate Conics
::第11.7节:离子二次曲线