等效判别式

原理

$$\{\begin{array}{l}{\displaystyle \frac{x^2}{m}}+{\displaystyle \frac{y^2}{n}}=1\\ Ax+By+C=0\end{array}$$$${\mathrm{\Delta }}^{\prime }=(A^{2}m+B^{2}n-C^2)$$$$\mathrm{\Delta }=4mn{B}^2{\mathrm{\Delta }}^{\prime }$$

TIP

对于双曲线 ${\displaystyle \frac{x^2}{a^2}}-$${\displaystyle \frac{y^2}{b^2}}=$$1$，将 $n$ 视作 $-$$b^2$。

例

${\displaystyle \frac{x^2}{4}}+$${\displaystyle \frac{y^2}{3}}=$$1$ 和 $y=$$kx+$$2$ 有两个交点，求 $k$ 的范围。

解：

$$\{\begin{array}{l}{\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{y^2}{3}}=1\\ kx-y+2=0\end{array}$$$$\begin{array}{ccc}m=4& n=3& \\ A^2=k^2& B^2=1& C^2=4\end{array}$$$${\mathrm{\Delta }}^{\prime }=(4k^2+3-4)>0$$$$k^2>{\displaystyle \frac{1}{4}}$$$$k<-{\displaystyle \frac{1}{2}}\text{或}k>{\displaystyle \frac{1}{2}}$$

简化二次恒等式

原理

$$\{\begin{array}{l}{b}^2{x}^2+a^2{y}^2-a^2{b}^2=0\\ Ax+By+C=0\end{array}$$$$(a^2{A}^2+b^2{B}^2)x^2+(2a^{2}AC)x+(a^2{C}^2-a^2{b}^2{B}^2)=0$$

TIP

二倍爬山放中间，

直线平方交叉乘。

$$\left[\begin{array}{ccccc}{b}^2& & a^2& & -a^2{b}^2\\ & {\times }_1& & {\times }_2& \\ A& & B& & C\end{array}\right]$$$$({\times }_{1^2}^{+})x^2+(2╱╲)x+({\times }_{2^2}^{+})=0$$

例

$$\{\begin{array}{l}{x}^2+4y^2-4=0\\ kx-y+\sqrt{3}k=0\end{array}$$$$\left[\begin{array}{ccc}1& 4& -4\\ k& -1& \sqrt{3}k\end{array}\right]$$$$(4k^2+1)x^2+(8\sqrt{3}{k}^2)x+(12k^2-4)=0$$

简化韦达定理

原理

$$\{\begin{array}{l}\alpha x^2+\beta y^2=\gamma \\ Ax+By+C=0\end{array}$$$$ϵ=\alpha B^2+\beta A^2$$$$x_1+x_2={\displaystyle \frac{-2C\cdot \beta A}{ϵ}},y_1+y_2={\displaystyle \frac{-2C\cdot \alpha B}{ϵ}}$$$$x_1{x}_2={\displaystyle \frac{\beta C^2-\gamma B^2}{ϵ}},y_1{y}_2={\displaystyle \frac{\alpha C^2-\gamma A^2}{ϵ}}$$$$x_1{y}_2+x_2{y}_1={\displaystyle \frac{2\gamma AB}{ϵ}}$$

TIP

$$\left[\begin{array}{ccccc}\alpha & & \beta & & \gamma \\ & {\times }_1& & {\times }_2& \\ A& & B& & C\end{array}\right]$$$$x_1+x_2={\displaystyle \frac{-2C\cdot {╱}_1}{{\times }_{1^2}^{+}}},y_1+y_2={\displaystyle \frac{-2C\cdot {╲}_1}{{\times }_{1^2}^{+}}}$$$$x_1{x}_2={\displaystyle \frac{{\times }_{2^2}^{-}}{{\times }_{1^2}^{+}}},y_1{y}_2={\displaystyle \frac{{\times }_{{12}^2}^{-}}{{\times }_{1^2}^{+}}}$$$$x_1{y}_2+x_2{y}_1={\displaystyle \frac{2\cdot \underset{―}{66}╱}{{\times }_{1^2}^{+}}}$$

例

${\displaystyle \frac{x^2}{4}}+$${\displaystyle \frac{y^2}{3}}=$$1$ 和 $y=$$kx+$$2$ 有两个交点 $A(x_1,$$y_1),$$B(x_2,$$y_2)$。

$$\{\begin{array}{l}3x^2+4y^2=12\\ kx-y+2=0\end{array}$$$$\left[\begin{array}{ccc}3& 4& 12\\ k& -1& 2\end{array}\right]$$$$ϵ=3\times (-1{)}^2+4\times k^2=3+4k^2$$$$x_1+x_2={\displaystyle \frac{-2\times 2\times 4\times k}{3+4k^2}}={\displaystyle \frac{-16k}{3+4k^2}}$$$$x_1{x}_2={\displaystyle \frac{4\times 2^2-12\times (-1{)}^2}{3+4k^2}}={\displaystyle \frac{4}{3+4k^2}}$$$$x_1{y}_2+x_2{y}_1={\displaystyle \frac{2\times 12\times k\times (-1)}{3+4k^2}}={\displaystyle \frac{-24k}{3+4k^2}}$$

弦长公式

$$\{\begin{array}{l}{\displaystyle \frac{x^2}{m}}+{\displaystyle \frac{y^2}{n}}=1\\ Ax+By+C=0\end{array}$$

弦长长度为

$${\displaystyle \frac{2\sqrt{mn\cdot (A^2+B^2)\cdot (m{A}^2+n{B}^2-C^2)}}{m{A}^2+n{B}^2}}$$

TIP

马牛（$mn$）逼，装大方（$A^2+$$B^2$）

组团上街耍流氓（$m{A}^2+$$n{B}^2-$$C^2$）

露完上面露下面（$m{A}^2+$$n{B}^2$）

见人就递两根烟（$2\sqrt{\ast }$）