$20200203$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 eq 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 vert = 2)

即以(MN)为直径的圆被(x)轴截的的弦长恒为定值(2)

(lacksquare)

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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 eq 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

[egin{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，

[egin{cases} egin{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

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

由几何关系易知

[egin{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，

[egin{cases} egin{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

(lacksquare)

课后作业

1.

解：

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

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

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

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

又(sqrt{3} lvert OA vert = lvert OC vert)

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( an angle OQM = lvert frac{n}{(m - 1)t} vert)，( an angle ONQ = lvert frac{(m + 1)t}{n} vert)

则由题有

[egin{aligned} lvert frac{n}{(1 - m)t} vert & = lvert frac{(1 + m)t}{n} vert \ 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 eq 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 vert = frac{6m^2}{3m^2 + 2})

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

整理得(3 lvert m vert ^2 - 2 sqrt{6} lvert m vert + 2 = 0) (Rightarrow) (lvert m vert = 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)