中期测试（教师版）

$3x - 4y + 5 = 0 \implies y = \dfrac{3}{4}x + \dfrac{5}{4}$，斜率 $k = \dfrac{3}{4}$

配方：$(x-1)^2 + (y+2)^2 = 4 + 4 + 4 = 12$？不对，重新算：$(x-1)^2 - 1 + (y+2)^2 - 4 - 4 = 0 \implies (x-1)^2 + (y+2)^2 = 9$

圆心 $(1, -2)$，半径 $r = 3$

$a^2 = 16$，$b^2 = 9$，$c^2 = 16 - 9 = 7$，$c = \sqrt{7}$

$e = \dfrac{c}{a} = \dfrac{\sqrt{7}}{4}$

短轴长 $= 2b$，焦距 $= 2c$，所以 $b = c$

$a^2 = b^2 + c^2 = 2c^2$，$a = \sqrt{2}c$

$e = \dfrac{c}{a} = \dfrac{c}{\sqrt{2}c} = \dfrac{\sqrt{2}}{2}$

$a^2 = 4$，$b^2 = 1$，$k = 1$，$m = 1$

$a^2k^2 + b^2 - m^2 = 4 + 1 - 1 = 4 > 0$，所以 $\Delta > 0$，相交

$a = 5$，$|PF_1| + |PF_2| = 2a = 10$

$|PF_2| = 10 - 6 = 4$

焦点在 $y$ 轴上，需要 $4 > m > 0$，即 $0 < m < 4$

$b^2 = 4$，面积公式 $S = b^2 \tan\dfrac{\theta}{2} = 4 \tan 30° = 4 \times \dfrac{\sqrt{3}}{3} = \dfrac{4\sqrt{3}}{3}$

$k_1 = 2$，$k_2 = -\dfrac{1}{2}$，$k_1 \cdot k_2 = -1$，两直线垂直

化为标准形式：$\dfrac{x^2}{4} + \dfrac{y^2}{16} = 1$

$a^2 = 16$，$b^2 = 4$，$a = 4$，$b = 2$

长轴长 $= 2a = 8$，短轴长 $= 2b = 4$？不对，焦点在 $y$ 轴上，长轴在 $y$ 方向，长轴长 $= 2a = 8$，短轴长 $= 2b = 4$

$e = \dfrac{c}{a} = \dfrac{\sqrt{3}}{2}$，$c = \dfrac{\sqrt{3}}{2}a$

$b^2 = a^2 - c^2 = a^2 - \dfrac{3}{4}a^2 = \dfrac{1}{4}a^2$

$\dfrac{b}{a} = \dfrac{1}{2}$

联立：$(1 + 5k^2)x^2 + 10kx = 0$

$x_1 + x_2 = -\dfrac{10k}{1 + 5k^2}$，$x_1 x_2 = 0$

$|AB|^2 = (1+k^2) \cdot \dfrac{100k^2}{(1+5k^2)^2} = \dfrac{50}{9}$

解得 $k^2 = 1$，$k = \pm 1$

（1）$2a = 10$，$a = 5$

$e = \dfrac{c}{a} = \dfrac{3}{5}$，$c = 3$

$b^2 = a^2 - c^2 = 25 - 9 = 16$

题目未指明焦点位置，分两种情况：

焦点在 $x$ 轴：$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

焦点在 $y$ 轴：$\dfrac{y^2}{25} + \dfrac{x^2}{16} = 1$

（注：若只写一种情况给5分，两种都写给6分）

（2）以焦点在 $x$ 轴为例：

焦点：$F_1(-3, 0)$，$F_2(3, 0)$

顶点：$A_1(-5, 0)$，$A_2(5, 0)$，$B_1(0, -4)$，$B_2(0, 4)$

（4分）

（1）圆心在 $y = x$ 上，设圆心为 $(a, a)$

与 $x$ 轴相切于 $(2, 0)$，所以圆心的 $x$ 坐标为 2，即 $a = 2$

圆心 $(2, 2)$，半径 $r = |2 - 0| = 2$

圆的标准方程：$(x - 2)^2 + (y - 2)^2 = 4$

（5分）

（2）直线 $kx - y + 1 = 0$ 与圆相切，圆心到直线距离等于半径：

$\dfrac{|2k - 2 + 1|}{\sqrt{k^2 + 1}} = 2$

$\dfrac{|2k - 1|}{\sqrt{k^2 + 1}} = 2$

$(2k - 1)^2 = 4(k^2 + 1)$

$4k^2 - 4k + 1 = 4k^2 + 4$

$-4k = 3$，$k = -\dfrac{3}{4}$

（5分）

（1）$a = 4$，$|PF_1| + |PF_2| = 2a = 8$

$|PF_2| = 8 - 6 = 2$

（4分）

（2）$b^2 = 4$，$c^2 = 16 - 4 = 12$，$c = 2\sqrt{3}$，$|F_1F_2| = 4\sqrt{3}$

在 $\triangle PF_1F_2$ 中，三边为 $|PF_1| = 6$，$|PF_2| = 2$，$|F_1F_2| = 4\sqrt{3}$

用余弦定理求 $\angle F_1PF_2$：

$\cos\angle F_1PF_2 = \dfrac{36 + 4 - 48}{2 \times 6 \times 2} = \dfrac{-8}{24} = -\dfrac{1}{3}$

$\sin\angle F_1PF_2 = \sqrt{1 - \dfrac{1}{9}} = \dfrac{2\sqrt{2}}{3}$

$S = \dfrac{1}{2} \times 6 \times 2 \times \dfrac{2\sqrt{2}}{3} = 4\sqrt{2}$

（6分）

（1）联立：$\dfrac{x^2}{9} + \dfrac{(x+m)^2}{4} = 1$

$4x^2 + 9(x+m)^2 = 36$

$13x^2 + 18mx + 9m^2 - 36 = 0$

$\Delta = 324m^2 - 4 \times 13 \times (9m^2 - 36) = 324m^2 - 468m^2 + 1872 = -144m^2 + 1872$

$x_1 + x_2 = -\dfrac{18m}{13}$，$x_1 x_2 = \dfrac{9m^2 - 36}{13}$

$|AB| = \sqrt{1+1} \cdot \dfrac{\sqrt{\Delta}}{13} = \sqrt{2} \cdot \dfrac{\sqrt{1872 - 144m^2}}{13}$

$= \sqrt{2} \cdot \dfrac{12\sqrt{13 - m^2}}{13} = \dfrac{12\sqrt{2}\sqrt{13-m^2}}{13}$

令 $\dfrac{12\sqrt{2}\sqrt{13-m^2}}{13} = \dfrac{12\sqrt{2}}{5}$

$\sqrt{13 - m^2} = \dfrac{13}{5}$

$13 - m^2 = \dfrac{169}{25}$，$m^2 = 13 - \dfrac{169}{25} = \dfrac{325 - 169}{25} = \dfrac{156}{25}$

$m = \pm\dfrac{2\sqrt{39}}{5}$

（5分）

（2）设中点 $M(x_0, y_0)$

$x_0 = \dfrac{x_1 + x_2}{2} = -\dfrac{9m}{13}$

$y_0 = x_0 + m = -\dfrac{9m}{13} + m = \dfrac{4m}{13}$

消去 $m$：$m = -\dfrac{13x_0}{9}$，代入 $y_0 = \dfrac{4m}{13}$：

$y_0 = \dfrac{4}{13} \times \left(-\dfrac{13x_0}{9}\right) = -\dfrac{4x_0}{9}$

轨迹方程：$y = -\dfrac{4}{9}x$

还需满足 $\Delta > 0$，即 $m^2 < 13$，$x_0^2 < \dfrac{9 \times 13}{13} = 9$，$-3 < x_0 < 3$

轨迹方程：$y = -\dfrac{4}{9}x$（$-3 < x < 3$）

（5分）