直线与曲线联立速算
定理推算
椭圆/双曲线
已知椭圆$C:\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1(a > b > 0)$与直线$l:y = kx + m$交于$E$、$F$
联立:
消去$x$得$(a^2k^2 + b^2)x^2 + 2a^2kmx + a^2(m^2 - b^2) = 0$
消去$y$得$(a^2k^2 + b^2)y^2 + 2b^2my + b^2(m^2 - a^2k^2) = 0$
$
\Rightarrow
\Delta_x = 4a^2b^2(a^2k^2 + b^2 - m^2) \\
\Delta_y = 4a^2b^2k^2(a^2k^2 + b^2 - m^2)
$
韦达定理得:
$
x_1 + x_2 = -\frac{2a^2km}{a^2k^2 + b^2} \\
x_1x_2 = \frac{a^2m^2 - a^2b^2}{a^2k^2 + b^2}
$
$
y_1 + y_2 = \frac{2b^2m}{a^2k^2 + b^2} \\
y_1y_2 = \frac{b^2m^2 - a^2b^2k^2}{a^2k^2 + b^2}$
(用$x$的公式推$y$的公式的方法:$a^2$、$b^2$互换,$k$、$1$互换,$m$不变)
$
|EF| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\
= \sqrt{(1 + k^2)[(x_2 + x_2)^2 - 4x_1x_2]}\\
= \sqrt{k^2 + 1}\frac{\sqrt{\Delta_x}}{|a^2k^2 + b^2|} \\
= \sqrt{k^2 + 1}\frac{\sqrt{4a^2b^2(a^2k^2 + b^2 - m^2)}}{|a^2k^2 + b^2|}
$
| 曲线方程 | 方程变形 | a变形 | b变形 |
|---|---|---|---|
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ | $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ | $a^2$ | $b^2$ |
| $\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1$ | $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ | $b^2$ | $a^2$ |
| $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ | $\frac{x^2}{a^2} + \frac{y^2}{-b^2} = 1$ | $a^2$ | $-b^2$ |
| $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ | $\frac{x^2}{-b^2} + \frac{y^2}{a^2} = 1$ | $-b^2$ | $a^2$ |
| $x^2 + y^2 = r^2$ | $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$ | $r^2$ | $r^2$ |
抛物线
已知抛物线$C:y^2 = nx$和直线$l:y = kx + m$交于$E$、$F$
联立:
消去$y$得$k^2x^2 + (2km - n)x + m^2 = 0$
消去$x$得$y^2 - \frac{n}{k}y+\frac{nm}{k} = 0$
$
\Rightarrow
\Delta_x = n(n - 4km) \\
\Delta_y = \frac{n}{k}(\frac{n}{k} - 4m) \\
(\Delta_y = \frac{\Delta_x}{k^2})$
韦达定理得:
$x_1 + x_2 = \frac{n - 2km}{k^2} \\
x_1x_2 = \frac{m^2}{k^2}$
$y_1 + y_2 = \frac{n}{k}\\
y_1y_2 = \frac{nm}{k}$
$|EF| = \sqrt{k^2 + 1}\frac{\sqrt{n(n-4km)}}{|k^2|}$
