$20200203$的数学作业

典型例题

1.

证明：

由题可知\(A(-2, 0)\)

设\(D(x_0, y_0)\ (x_0 \neq \pm 2)\)，则\(E(-x_0, -y_0)\)，且\(\frac{x_0^2}{4} + y_0^2 = 1\)

则由题有\(AD: y = \frac{y_0}{x_0 + 2} (x + 2) \Rightarrow M(0, \frac{2y_0}{x_0 + 2})\)

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

设以\(MN\)为直径的圆与\(x\)轴交于\(P(x', 0)\)和\(Q(-x', 0)\)，不妨令\(x' > 0\)

那么\(\vec{PM} = (-x', \frac{2y_0}{x_0 + 2})\)，\(\vec{PN} = (-x', \frac{2y_0}{x_0 - 2})\)

\(\vec{PM} \cdot \vec{PN} = x'^2 + \frac{4y_0^2}{x_0^2 - 4} = 0\)

代入\(\frac{x_0^2}{4} + y_0^2 = 1\)，整理可得\(x' = 1\) \(\Rightarrow\)\(\lvert PQ \rvert = 2\)

即以\(MN\)为直径的圆被\(x\)轴截的的弦长恒为定值\(2\)

\(\blacksquare\)

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2.

解：

设\(M(x_1, y_1)\)，\(N(x_2, y_2)\)

I若\(k = 0\)

由题易知\(m \in (-\sqrt{3}, \sqrt{3})\)

II若\(k \neq 0\) \(\Rightarrow l: y = -\frac{1}{k} x - \frac{1}{2}\)

将\(y = kx + m\)代入椭圆方程，整理得

\[(4k^2 + 3)x^2 + 8kmx + 4m^2 - 12 = 0 \]

其中

\[\begin{aligned} \Delta & = (8km)^2 - 4(4k^2 + 3)(4m^2 - 12) \\ & = 48(4k^2 - m^2 + 3) > 0 \Rightarrow 4k^2 + 3 > m^2 \end{aligned} \]

由韦达定理，

\[\begin{cases} \begin{aligned} & x_1 + x_2 = -\frac{8km}{4k^2 + 3} \\ & x_1x_2 = \frac{4m^2 - 12}{4k^2 + 3} \end{aligned} \end{cases} \]

设\(MN\)的中点为\(P\)，则\(P(-\frac{4km}{4k^2 + 3}, \frac{3m}{4k^2 + 3})\)，故有

\[\frac{3m}{4k^2 + 3} = \frac{4m}{4k^2 + 3} - \frac{1}{2} \]

整理得\(4k^2 + 3 = 2m\) \(\Rightarrow\) \(2m > m^2\) \(\Rightarrow\) \(0 < m < 2\)

上述方程又可化为\(2m - 3 = 4k^2 > 0\) \(\Rightarrow\) \(m > \frac{3}{2}\) \(\Rightarrow\) \(\frac{3}{2} < m < 2\)

此时，若\(y = kx + m\)过左顶点 \(\Rightarrow\) \(m = 2k\) \(\Rightarrow\) \((2k^2 - 1)^2 = -2\) \(\Rightarrow\) 无解

故\(y = kx + m\)不过左顶点，由对称性可得其不过右顶点

即此时\(m \in (\frac{3}{2}, 2)\)

综上所述，\(m \in (-\sqrt{3}, 2)\)

课堂练习

1.

解：

由题可知\(P(0, 2) \Rightarrow M(0, 1)\)

设\(A(x_1, y_1)\)，\(B(x_2, y_2)\)

I若\(l\)的斜率存在

设\(l: y = kx + 1\)，代入椭圆方程，整理得

\[(2k^2 + 1)x^2 + 4kx - 6 = 0 \]

其中

\[\begin{aligned} \Delta & = (4k)^2 - 4(2k^2 + 1) \cdot (-6) \\ & = 8(8k^2 + 3) > 0 \end{aligned} \]

由几何关系易知

\[\begin{aligned} k_{AC} + k_{BC} & = 0 \\ \frac{y_1}{x_1 + 3} + \frac{y_2}{x_2 + 3} & = 0 \\ 2kx_1x_2 + (3k + 1)(x_1 + x_2) + 6 = 0 \end{aligned} \]

由韦达定理，

\[\begin{cases} \begin{aligned} & x_1 + x_2 = -\frac{4k}{2k^2 + 1} \\ & x_1x_2 = -\frac{6}{2k^2 + 1} \end{aligned} \end{cases} \]

代入，整理得\(k = \frac{3}{8}\) \(\Rightarrow\) \(l: 3x - 8y + 8 = 0\)

II若\(l\)斜率不存在

此时显然满足\(\Rightarrow l: x = 0\)

综上所述，\(l: 3x - 8y + 8 = 0\)或\(l: x = 0\)

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2.

证明：

由几何关系易知\(PA \perp PB\)

I若\(PA \perp x\)轴

此时\(PA: x = 3\)或\(PA: x = -3\) \(\Rightarrow\) \(PB: y = 2\)或\(PB: y = -2\)

易知\(PB\)与\(C\)相切

II若\(PA \perp y\)轴

由I易知此时\(PB\)与\(C\)相切

III若\(PA\)与\(x\)轴和\(y\)均不垂直

设\(P(x_0, y_0)\)，\(PA: y - y_0 = k(x - x_0)\)，则\(PB: y - y_0 = -\frac{1}{k} (x - x_0)\)

将\(PA\)代入\(C\)，整理得

\[(9k^2 + 4)x^2 + 18k(y_0 - kx_0)x + 9(y_0 - kx_0)^2 - 36 = 0 \]

\(PA\)与\(C\)相切\(\Rightarrow\)\(\Delta_1 = [18k(y_0 - kx_0)]^2 - 4(9k^2 + 4)[9(y_0 - kx_0)^2 - 36] = 0\) \(\Rightarrow\) \((x_0^2 - 9)k^2 + 2x_0y_0k + y_0^2 - 4 = 0\)

将\(PB\)代入\(C\)，整理求得\(\Delta_2 = \frac{x_0^2 - 9}{k^2} + \frac{2x_0y_0}{k} + y_0^2 - 4\)

又\(x_0^2 + y_0^2 = 13\) \(\Rightarrow\) \(\Delta_2 = -\frac{144[(x_0^2 - 9)k^2 + 2x_0y_0k + y_0^2 - 4]}{k^2} = 0\)

故此时\(PB\)与\(C\)相切

综上所述，直线\(PB\)与椭圆\(C\)相切

\(\blacksquare\)

课后作业

1.

解：

若\(\Delta ABC\)为正三角形，则\(AB \perp OC\)且\(\sqrt{3} \lvert OA \rvert = \lvert OC \rvert\)

由几何关系易知\(AB\)的斜率存在且不为\(0\)

故可设\(AB: y = kx\)，则\(OC: y = -\frac{1}{k} x\)

将\(AB\)代入\(\Gamma\)，整理可得

\[\begin{cases} \begin{aligned} & x^2 = \frac{30}{5k^2 + 3} \\ & y^2 = \frac{30k^2}{5k^2 + 3} \end{aligned} \end{cases} \]

故\(\lvert OA \rvert = \sqrt{x^2 + y^2} = \sqrt{\frac{30(k^2 + 1)}{5k^2 + 3}}\)，同理\(\lvert OC \rvert = \sqrt{\frac{30(k^2 + 1)}{3k^2 + 5}}\)

又\(\sqrt{3} \lvert OA \rvert = \lvert OC \rvert\)

代入整理得\(k^2 = -3\) \(\Rightarrow\) 无实数解

故\(\Delta ABC\)不可能为正三角形

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2.

解：

由题，\(C(1, 0)\)，\(E(-m, n)\)

则\(CD: y = \frac{n}{m - 1} (x - 1)\) \(\Rightarrow\) \(M(0, \frac{n}{1 - m})\)；\(CE: y = -\frac{n}{m + 1} (x - 1)\) \(\Rightarrow\) \(N(0, \frac{n}{1 + m})\)

设\(Q(t, 0)\)，则\(\tan \angle OQM = \lvert \frac{n}{(m - 1)t} \rvert\)，\(\tan \angle ONQ = \lvert \frac{(m + 1)t}{n} \rvert\)

则由题有

\[\begin{aligned} \lvert \frac{n}{(1 - m)t} \rvert & = \lvert \frac{(1 + m)t}{n} \rvert \\ t^2 & = \frac{n ^ 2}{1 - m^2} = \frac{n^2}{\frac{n^2}{2}} = 2 \\ t & = \pm \sqrt{2} \end{aligned} \]

故存在\(Q\)，且\(Q(\sqrt{2}, 0)\)或\(Q(-\sqrt{2}, 0)\)

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3.

解：

由题可知，\(A(1, 0)\)，\(AP\)的斜率存在且不为\(0\)

故可设\(AP: my = x - 1\)，其中\(m \neq 0\)，代入\(l\)得\(P(-1, -\frac{2}{m})\)，\(Q(-1, \frac{2}{m})\)

将\(AP\)代入椭圆方程，解得\(y = 0\)或\(-\frac{6m}{3m^2 + 4}\)

\(B\)异于点\(A\)\(\Rightarrow\)\(B(\frac{4 - 3m^2}{3m^2 + 4}, -\frac{6m}{3m^2 + 4})\)

故\(BQ: \frac{8}{3m^2 + 4} (y - \frac{2}{m}) = -\frac{12m^2 + 8}{m(3m^2 + 4)} (x + 1)\) \(\Rightarrow\) \(D(\frac{2 - 3m^2}{3m^2 + 2}, 0)\)

\(\Rightarrow \lvert AD \rvert = \frac{6m^2}{3m^2 + 2}\)

故\(S_{\Delta APD} = \frac{6m^2}{3m^2 + 2} \cdot \frac{2}{\lvert m \rvert} \div 2 = \frac{\sqrt{6}}{2}\)

整理得\(3 \lvert m \rvert ^2 - 2 \sqrt{6} \lvert m \rvert + 2 = 0\) \(\Rightarrow\) \(\lvert m \rvert = \frac{\sqrt{6}}{3}\) \(\Rightarrow\) \(m = \pm \frac{\sqrt{6}}{3}\)

故\(AP: 3x + \sqrt{6} y - 3 = 0\)或\(3x - \sqrt{6} y - 3 = 0\)