20200202的数学作业

典型例题

1.

解：

((1))

由题易知，(P_3)不在(C)上(Leftrightarrow)(P_4)不在(C)上

此结论与已知条件矛盾，故(P_3)，(P_4)都在(C)上，(P_1)不在(C)上

则可得(frac{1}{a^2} + frac{3}{4b^2} = 1)

[egin{cases} egin{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

[egin{cases} egin{aligned} & a^2 = 4 \ & b^2 = 1 end{aligned} end{cases} ]

故(C: frac{x^2}{4} + y^2 = 1)

证明：

((2))

设(A(x_1, y_1))，(B(x_2, y_2))

i. 若(l)的斜率存在

设(l: y = kx + b)，代入(C)，整理得

[(4k^2 + 1)x^2 + 8kbx + 4b^2 - 4 = 0 ]

其中

[egin{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

[egin{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

[egin{cases} egin{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)

则(l: y + 1 = k(x - 2))

即此时(l)过定点((2, -1))

ii. 若(l)的斜率不存在

设(l: x = t)，此时(x_1 = x_2 eq 0)，(y_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)

即(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))

(lacksquare)

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2.

证明：

设(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

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

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

故(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 ]

令(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))和((-sqrt{2}, 0))

(lacksquare)

课堂练习

1.

解：

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

设(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

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

则

[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} ]

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

即(A'B: y = frac{x_2 - x_1}{4} x + 1)

即直线(A'B)恒过定点((0, 1))

(lacksquare)

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2.

解：

(vec{MS} = vec{SN})，(vec{PT} = vec{TQ}) (Rightarrow) (S)为(MN)中点，(T)为(PQ)中点

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

i. 若两直线斜率皆存在

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

将(l_1)代入(C)，整理得

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

其中

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

由韦达定理，

[egin{cases} egin{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} ]

故(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}))

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

即此时(ST)恒过定点((frac{2}{3}, 0))

ii. 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)过((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_1)与(l_2)的斜率一定存在且不为(0)

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

将(l_1)代入(Gamma)，整理得

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

其中

[egin{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，

[egin{cases} egin{aligned} & y_C + y_D = frac{4}{k} \ & y_Cy_D = frac{32 - 48k}{k} end{aligned} end{cases} ]

则(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})

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

即(MN: (k + frac{1}{k} - 4)y = x - 10)

即直线(MN)恒过定点((10, 0))

(lacksquare)

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2.

证明：

由题易知(l)的斜率一定存在

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

将(l)代入椭圆方程，整理得

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

其中

[egin{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，

[egin{cases} egin{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} ]

则

[egin{cases} egin{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}))

(lacksquare)

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3.

解：

((1))

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

将(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 = 2p)，(x_1x_2 = frac{p^2}{4})

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

故(C: y^2 = 4x)

((2))

设(D(x_1, y_1))，(E(x_2, y_2))

由((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 = 4m)，(y_1y_2 = 4t)

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

即((t + 6)^2 = 4(2m + 1)^2)(Rightarrow)(t = -4m + 4)或(4m + 8)

故(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))