解析几何
解析几何
直线
- 直线的倾斜角 $0\leqslant\theta<\pi$
- 斜率 $$k=tan\theta=\frac{y_2-y_1}{x_2-x_1}$$ (当$l$垂直于$x$轴时, $\theta=\frac{\pi}{2}$, $k$不存在)
直线方程表示
- 点斜式 $y-y_0=k(x-x_0)$
- 斜截式 $y=kx+b$
- 两点式 $$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$$
- 一般式 $ax+by+c=0$ (ab不同时为0) 其中一般式 $l$ 一个法向量为 $\vec{n}=(a,b)$
- 点法式 $a(x-x_0)+b(y-y_0)=0$
直线的位置关系
记两条直线:
$l_1: a_1x+b_1y+c_1z=0$ ($a_1$$b_1$不同时为0)
$l_2: a_2x+b_2y+c_2z=0$ ($a_2$$b_2$不同时为0)
- 重合: 存在 $\lambda$ 使得 $a_1=\lambda a_2$, $b_1=\lambda b_2$, $c_1=\lambda c_2$
- 平行: 存在 $\lambda$ 使得 $a_1=\lambda a_2$, $b_1=\lambda b_2$, $c_1\neq\lambda c_2$
- 相交: $a_1b_2\neq a_2b_1$
当 $a_2$ $b_2$ $c_2$ 不为 0 时
重合: $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$
平行: $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$$
相交: $$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$$
垂直: $a_1a_2+b_1b_2=0$, $k_1k_2=-1$
直线夹角
记$l_1$$l_2$夹角为$\theta$
$$cos\theta = \frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$
通过 $\vec{n_1}=(a_1,b_1), \vec{n_2}=(a_2,b_2)$ 的向量夹角得出直线夹角或其补角
$$tan\theta = |\frac{k_1-k_2}{1+k_1k_2}|$$
$$ tan\theta = tan(|\theta_1 - \theta_2|) = |\frac{tan\theta_1 - tan\theta_2}{1+tan\theta_1 tan\theta_2}| = |\frac{k_1-k_2}{1+k_1k_2}| $$
点到直线的距离
记 直线$l$: $ax+by+c=0$, 点$P(x_0,y_0)$
$$d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$
通过点到直线的垂足得出
当两直线平行时 $l_1: a_1x+b_1y+c_1z=0$ ($a_1$$b_1$不同时为0)
$l_2: a_2x+b_2y+c_2z=0$ ($a_2$$b_2$不同时为0)
两直线间距离
$d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$
直线解题技巧
当直线中含参数时, 直线有定点 例 含参数 $m$, $n$ 时 $l: mx+ny+m=0$ 将直线转化为 $m(x+1)+ny=0$
$$\begin{matrix} x+1=0 \y=0 \end{matrix}$$ 可得到直线过定点 $(-1,0)$
共点直线系方程: 若直线 $l_1: A_1 x+B_1 y+C_1=0$ 与直线 $l_2: A_2 x+B_2 y+$ $C_2=0$ 相交于点 $S$, 则过这两条直线交点的直线系的方程是 $m\left(A_1 x+B_1 y+C_1\right)+$ $n\left(A_2 x+B_2 y+C_2\right)=0$ (其中 $m, n$ 是不同时为零的实数), 该直线系表示了所有 过点 $S$ 的直线.
上述题可以用联立方程组求出交点解决,不必一定使用此方法
圆
圆的方程
- 标准方程 $(x-a)^2+(y-b)^2=r^2$ 其中圆心为 $(a,b)$, 半径为 $r$ 当圆与 $x$ 轴相切时, $|b|=r$ 当圆与 $y$ 轴相切时, $|a|=r$ 当圆与 $x$, $y$ 相切时, $|a|=|b|=r$
- 一般方程 $x^2+y^2+D x+E y+F=0$ 配方后得 $$\left(x+\frac{D}{2}\right)^2+\left(y+\frac{E}{2}\right)^2=\frac{D^2+E^2-4 F}{4}$$ 当 $D^2+E^2-4 F>0$ 时, 方程表示以 $\left(-\frac{D}{2},-\frac{E}{2}\right)$ 为圆心, $\frac{\sqrt{D^2+E^2-4 F}}{2}$ 为半径的圆 当 $D^2+E^2-4 F=0$ 时, 方程表示一个点 $\left(-\frac{D}{2},-\frac{E}{2}\right)$ 当 $D^2+E^2-4 F<0$ 时, 方程无图像
点与圆的位置关系
为点到圆心的位置关系与半径 $r$ 相比较 C $(a, b)$ 为圆心, $r$ 为半径, $P(x, y)$ 为点 $d=\sqrt{(x-a)^2+(y-b)^2}$ $d<r$ 在圆内 $d=r$ 在圆上 $d>r$ 在圆外
直线与圆的位置关系
| 位置关系 | 公共点情况 | 判別方法1 | 判別方法2 |
|---|---|---|---|
| 相离 | 没有公共点 | $d>r$ | $\Delta<0$ |
| 相切 | 唯一公法共 | $d=r$ | $\Delta=0$ |
| 相交 | 两个不同的公共点 | $d<r$ | $\Delta>0$ |
上表中, 直线 $l: A x+B y+C=0\left(A^2+B^2 \neq 0\right)$ 和圆 $C: (x-a)^2+$ $(y-b)^2=r^2$ $$d=\frac{|A a+B b+C|}{\sqrt{A^2+B^2}}$$ ( 圆心 $C$ 到直线 $l$ 的距离 )
当直线 $l$ 与坐标轴不垂直时, 将直线 $l$ 和圆 $C$ 的方程联立消去 $y$ 后得 到一个关于 $x$ 的一元二次方程, 其判别式记为 $\Delta$. 当直线与 $x$ 轴垂直时, 消去 $x$; 当直线与 $y$ 轴垂直时, 消去 $y$
圆与圆的位置关系
方法1
$\begin{aligned} & C_1:\left(x-a_1\right)^2+\left(y-b_1\right)^2=r_1^2 \ & C_2:\left(x-a_2\right)^2+\left(y-b_2\right)^2=r_2^2\end{aligned}$ 圆心距 $d=\sqrt{\left(a_1-a_2\right)^2+\left(b_1-b_2\right)^2}$
当 $d<\left|r_1-r_2\right|$ 时, 两圆内含; 当 $d=\left|r_1-r_2\right|$ 时，两圆内切； 当 $\left|r_1-r_2\right|<d<r_1+r_2$ 时, 两圆相交; 当 $d=r_1+r_2$ 时, 两圆外切; 当 $d>r_1+r_2$ 时, 两圆外离.
方法2
$\begin{aligned} & C_1: x^2+y^2+D_1 x+E_1 y+F_1=0 \ & C_2: x^2+y^2+D_2 x+E_2 y+F_2=0\end{aligned}$ 联立方程组 {$\begin{array}{l}x^2+y^2+D_1 x+E_1 y+F_1=0 \\ x^2+y^2+D_2 x+E_2 y+F_2=0\end{array}$ 得 $\left(D_1-D_2\right) x+\left(E_1-E_2\right) y+\left(F_1-F_2\right)=0$ 两组解, 相交 一组解, 内切或外切 若无解, 内含或外离
圆解题技巧
- 弦长公式 (适用于直线与任意曲线) 直线 $l$ 与 曲线 $c$ 交于 $A$ 和 $B$ {$\begin{matrix} l: y=k x + b \\ c: a x^2 + b y^2 + c x + d y + e = 0 \end{matrix}$ 带入消 $y$ 得 $ a x^2 + b x + c = 0$ 由一元二次方程计算得到 $\Delta$ 弦长 $$AB = \sqrt{1+k^2}|x_1-x_2| = \sqrt{1+k^2}\frac{\sqrt{\Delta}}{|a|}$$
椭圆 && 双曲线
| 名称 | 椭圆 | 双曲线 |
|---|---|---|
| 定义 | 到两定点 $F_1, F_2$ 的距离和为常数 $2a$ ( $2a> \lvert F_1F_2 \rvert $ ) | 到两定点 $F_1, F_2$ 的距离差的绝对值为常数 $2a$ ( $0<2a< \lvert F_1F_2 \rvert $ ) |
| $PF_1 + PF_2 = 2a$ | $\lvert PF_1 - PF_2 \rvert = 2a$ | |
| 当 $2 a>2 c$ 时, 轨迹是椭圆 | 当 $2 a<2 c$ 时, 轨迹是双曲线 | |
| 当 $2 a=2 c$ 时, 轨迹是线段$ \lvert F_1F_2 \rvert $ | 当 $2 a=2 c$ 时, 轨迹是两条射线 | |
| 当 $2 a<2 c$ 时, 轨迹不存在 | 当 $2 a>2 c$ 时, 轨迹不存在 | |
| 标准方程 | $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ( 焦点在 $x$ 轴上 ) | $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ( 焦点在 $x$ 轴上 ) |
| 标准方程 | $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ ( 焦点在 $y$ 轴上 ) | $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ ( 焦点在 $y$ 轴上 ) |
| 焦点在哪 | 分母的大小判断 | 项的正负判断 |
| a, b, c 关系 | $a^2=b^2 + c^2$ | $c^2=a^2+b^2$ |
| 长轴 2a 短轴 2b 焦距2c | 实轴 2a 虚轴 2b 焦距 2c | |
| 取值() | $-a \leqslant x \leqslant a $ ( 焦点x轴上 ) | $x \leqslant -a$ 或 $x\geqslant a $ ( 焦点x轴上 ) |
| 范围 | $-b \leqslant y \leqslant b $ | $y \in R$ |
| 对称性 | 关于 $x$, $y$ 轴, 原点对称 | 关于 $x$, $y$ 轴, 原点对称 |
| 顶点 | $(\pm a, 0)$ $(0, \pm b)$ ( 焦点x轴上 ) | $(\pm a, 0)$ ( 焦点x轴上 ) |
| 离心率 | $e=\frac{c}{a}\in(0,1)$ | $e=\frac{c}{a}=\sqrt{1+\frac{b^2}{a^2}} \in(1,+\infty)$ |
| $e\to 1$扁, $ e\to 0$ 圆 | $e\to 1$开口小(扁), $ e\to +\infty$ 开口大 | |
| 渐近线(双曲线) | 把 "1" 换为 "0" 即 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ (焦点x轴上) |
注意事项 a, b, c > 0
双曲线与直线位置关系
判断点在椭圆/双曲线内外
椭圆&&双曲线解题技巧
- 弦长公式 同圆
- 焦点三角形
| 椭圆 | 双曲线 |
|---|---|
| $PF_1 \cdot PF_2 = \frac{2b^2}{1+cos\theta}$ | $PF_1 \cdot PF_2 = \frac{2b^2}{1-cos\theta}$ |
| $cos\theta = \frac{2b^2}{PF_1 \cdot PF_2} - 1$ | $cos\theta = 1 - \frac{2b^2}{PF_1 \cdot PF_2}$ |
| $S_\triangle F_1PF_2 = b^2tan\frac{\theta}{2}$ | $S_\triangle F_1PF_2 = b^2cot\frac{\theta}{2}$ |
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