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20200202的数学作业

20200202的数学作业

作者: the_Miracle | 来源:发表于2020-02-02 18:21 被阅读0次

典型例题

1.

解:

(1)

由题易知,P_3不在C\Leftrightarrow$$P_4不在C

此结论与已知条件矛盾,故P_3P_4都在C上,P_1不在C

则可得\frac{1}{a^2} + \frac{3}{4b^2} = 1

\begin{cases} \begin{aligned} & \frac{1}{a^2} + \frac{3}{4b^2} = 1 \\ & \frac{0}{a^2} + \frac{1}{b^2} = 1 \end{aligned} \end{cases}

解得
\begin{cases} \begin{aligned} & a^2 = 4 \\ & b^2 = 1 \end{aligned} \end{cases}
C: \frac{x^2}{4} + y^2 = 1

证明:

(2)

A(x_1, y_1)B(x_2, y_2)

  1. l的斜率存在

l: y = kx + b,代入C,整理得
(4k^2 + 1)x^2 + 8kbx + 4b^2 - 4 = 0
其中
\begin{aligned} \Delta & = (8kb)^2 - 4(4k^2 + 1)(4b^2 - 4) \\ & = 64k^2 - 16b^2 + 16 > 0 \Rightarrow 4k^2 + 1 > b^2 \end{aligned}
由题有
\begin{aligned} k_{P_2A} + k_{P_2B} & = -1 \\ \frac{y_1-1}{x_1} + \frac{y_2-1}{x_2} & = -1 \\ \frac{2kx_1x_2 + (b - 1)(x_1 + x_2)}{x_1x_2} & = -1 \end{aligned}
由韦达定理
\begin{cases} \begin{aligned} & x_1 + x_2 = -\frac{8kb}{4k^2 + 1} \\ & x_1x_2 = \frac{4b^2 - 4}{4k^2 + 1} \end{aligned} \end{cases}
代入整理得b = -2k - 1

l: y + 1 = k(x - 2)

即此时l过定点(2, -1)

  1. l的斜率不存在

l: x = t,此时x_1 = x_2 \neq 0y_1 = -y_2

则有k_{P_2A} + k_{P_2B} = \frac{y_1 - 1}{t} + \frac{-y_1 - 1}{t} = \frac{-2}{t} = -1

m = 2

\Rightarrow此时l过椭圆右顶点,不存在两个交点

\Rightarrow舍去


综上所述,直线l恒过定点(2, -1)<p align="right"> \blacksquare </p>


2.

证明:

P(x_0, y_0),则Q(-x_0, -y_0)

由题,有\frac{x^2_0}{4} + \frac{y^2_0}{2} = 1 \Leftrightarrow x^2_0 + 2y^2_0 = 4成立

A(-2, 0) \Rightarrow PA: y = \frac{y_0}{x_0 + 2} (x + 2)

M(0, \frac{2y_0}{x_0 + 2})PA: y = \frac{y_0}{x_0 - 2} (x + 2)

N(0, \frac{2y_0}{x_0 - 2})

则以MN为直径的圆为
x^2 + (y - \frac{2y_0}{x_0 + 2})(y - \frac{2y_0}{x_0 - 2}) = 0

代入x^2_0 + 2y^2_0 = 4并整理得
x^2 + y^2 + \frac{2x_0}{y_0} y - 2 = 0

y = 0,解得x = \pm \sqrt{2}

故以线段MN为直径的圆恒过顶点(\sqrt{2}, 0)(-\sqrt{2}, 0)<p align="right"> \blacksquare </p>

课堂练习

1.

解:

(2, 1)代入C,解得p = 2 \Rightarrow C: x^2 = 4y

A(x_1, y_1)B(x_2, y_2),则A'(-x_1, y_1)

由题,l的斜率一定存在

故可设l: y = kx - 1,代入C,整理得

x^2 -4kx +4 = 0

其中

\begin{aligned} \Delta & = (-4k)^2 - 4 \cdot 1 \cdot 4 \\ & = 16(k^2 - 1) > 0 \Rightarrow k^2 > 1 \end{aligned}

k_{A'B} = \frac{y_2 - y_1}{x_2 - (-x_1)} = \frac{\frac{x^2_2}{4} - \frac{x^2_1}{4}}{x_1 + x_2} = \frac{x_2 - x_1}{4}

A'B: y - \frac{x^2_2}{4} = \frac{x_2 - x_1}{4} (x - x_2)

A'B: y = \frac{x_2 - x_1}{4} x + 1

即直线A'B恒过定点(0, 1)<p align="right"> \blacksquare </p>


2.

解:

\vec{MS} = \vec{SN}\vec{PT} = \vec{TQ} \Rightarrow SMN中点,TPQ中点

P(x_1, y_1)Q(x_2, y_2)M(x_3, y_3)N(x_4, y_4)

  1. 若两直线斜率皆存在

l_1: y = k(x - 1),则l_2: y = -\frac{1}{k} (x - 1)

l_1代入C,整理得
(2k^2 + 1)x^2 - 4k^2x + 2k^2 - 4 = 0
其中
\begin{aligned} \Delta & = (-4k^2)^2 - 4(2k^2 + 1)(2k^2 - 4) \\ & = 8(3k^2 + 2) > 0 \end{aligned}

由韦达定理,
\begin{cases} \begin{aligned} & x_1 + x_2 = \frac{4k^2}{2k^2 + 1} \\ & x_1x_2 = \frac{2k^2 - 4}{2k^2 + 1} \end{aligned} \end{cases}

T(\frac{2k^2}{2k^2 + 1}, \frac{-k}{2k^2 + 1}),同理可得S(\frac{2}{k^2 + 2}, \frac{k}{k^2 + 2})

\Rightarrow$$k_{ST} = \frac{-3k}{2(k^2 - 1)}

\Rightarrow$$ST: y + \frac{k}{2k^2 + 1} = \frac{-3k}{2(k^2 - 1)} (x - \frac{2k^2}{2k^2 + 1})

ST: y = \frac{-3k}{2(k^2 - 1)} (x - \frac{2}{3})

即此时ST恒过定点(\frac{2}{3}, 0)

  1. 若两直线斜率分别为0和不存在

此时其中一条直线的方程为y = 0(\frac{2}{3}, 0)

综上所述,直线ST恒过定点(\frac{2}{3}, 0)

课后作业

1.

证明:

由题,l_1l_2的斜率一定存在且不为0

l_1: y = k(x - 12) + 8,则l_2: y = \frac{1}{k} (x - 12) + 8

l_1代入\Gamma,整理得

ky^2 - 4y - 48k + 32 = 0

其中
\begin{aligned} \Delta & = (-4)^2 - 4k(32 - 48k) \\ & = 16(2k - 1)(6k - 1) > 0 \Rightarrow k \in (-\infty, \frac{1}{6}) \cup (\frac{1}{2}, +\infty) \end{aligned}

同理有\frac{1}{k} \in (-\infty, \frac{1}{6}) \cup (\frac{1}{2}, +\infty)

综上有k \in (-\infty, 0) \cup (\frac{1}{2}, 2) \cup (6, +\infty)

由韦达定理,
\begin{cases} \begin{aligned} & y_C + y_D = \frac{4}{k} \\ & y_Cy_D = \frac{32 - 48k}{k} \end{aligned} \end{cases}
x_C + x_D = 24 + \frac{4}{k^2} - \frac{16}{k} \Rightarrow M(12 + \frac{2}{k^2} - \frac{8}{k}, \frac{2}{k})

同理可得N(12 + 2k^2 - 8k, 2k)

故可求得k_{MN} = \frac{1}{k + \frac{1}{k} - 4}

MN: y - 2k = \frac{1}{k + \frac{1}{k} - 4} [x - (2k^2 - 8k + 12)]

MN: (k + \frac{1}{k} - 4)y = x - 10

即直线MN恒过定点(10, 0)<p align="right"> \blacksquare </p>


2.

证明:

由题易知l的斜率一定存在

故可设l: y = kx + bM(x_1, y_1)N(x_2, y_2)

l代入椭圆方程,整理得

(2k^2 + 1)x ^ 2 + 4kbx + 2(b^2 - 1) = 0

其中
\begin{aligned} \Delta & = (4kb)^2 - 4(2k^2 + 1)(2b^2 - 2) \\ & = 8(2k^2 - b^2 + 1) > 0 \Rightarrow 2k^2 + 1 > b^2 \end{aligned}

由韦达定理,

\begin{cases} \begin{aligned} & x_1 + x_2 = -\frac{4kb}{2k^2 + 1} \\ & x_1x_2 = \frac{2(b^2 - 1)}{2k^2 + 1} \end{aligned} \end{cases}

\begin{cases} \begin{aligned} & y_1 + y_2 = \frac{2b}{2k^2 + 1} \\ & y_1y_2 = \frac{b^2 - 2k^2}{2k^2 + 1} \end{aligned} \end{cases}

由题可知\vec{AM} \cdot \vec{AN} = x_1x_2 + (y_1 - 1)(y_2 - 1) = 0 \Rightarrow (3b + 1)(b - 1) = 0

又直线不经过A$$\Rightarrow$$b = -\frac{1}{3}

故直线l恒过定点(0, -\frac{1}{3})<p align="right"> \blacksquare </p>


3.

解:

(1)

由题,F(\frac{p}{2}, 0),则AB: y = \sqrt{2} (x - \frac{p}{2})

AB代入C,整理得

x^2 - 2px + \frac{p^2}{4} = 0

其中\Delta = 4p^2 - p^2 = 3p^2 > 0

由韦达定理,x_1 + x_2 = 2px_1x_2 = \frac{p^2}{4}

\lvert AB \rvert = \sqrt{1 + 2} \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = 6 \Rightarrow p = 2

C: y^2 = 4x

(2)

D(x_1, y_1)E(x_2, y_2)

(1)M(4, 4),且DE斜率不为0

故可设DE: my = x + t,代入C,整理得

y^2 - 4my + 4t = 0

其中\Delta = 16(m^2 - t) > 0 \Rightarrow m^2 > t

由韦达定理,y_1 + y_2 = 4my_1y_2 = 4t

\vec{MD} \cdot \vec{ME} = x_1x_2 + (y_1 - 4)(y_2 - 4) = t^2 + 12t - 16m^2 - 16m + 32 = 0

(t + 6)^2 = 4(2m + 1)^2$$\Rightarrow$$t = -4m + 44m + 8

DE: m(y - 4) = x - 4(舍去)或DE: m(y + 4) = x - 8

即直线DE恒过定点(8, -4)

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