📝 题目
例 1 求曲线 $y = {x}^{2}$ 和 $y = \frac{1}{x}\left( {x < 0}\right)$ 的公切线方程.
💡 答案与解析
解法 1 设公切线在 $y = {x}^{2}\left( {x < 0}\right)$ 上的切点为 $\left( {{x}_{1},{x}_{1}^{2}}\right)$ ,在 $y = \frac{1}{x}$ $\left( {x < 0}\right)$ 上的切点为 $\left( {{x}_{2},\frac{1}{{x}_{2}}}\right)$ ,则公切线作为曲线 $y = {x}^{2}$ 的切线,其方程是
$$ y = {x}_{1}^{2} + 2{x}_{1}\left( {x - {x}_{1}}\right) , \tag{1.1} $$
公切线作为曲线 $y = \frac{1}{x}$ 的切线,其方程是
$$ y = \frac{1}{{x}_{2}} - \frac{1}{{x}_{2}^{2}}\left( {x - {x}_{2}}\right) . \tag{1.2} $$
比较 (1.1) 与 (1.2) 式右端 $x$ 幂的系数,得到
$$ \left\{ {\begin{array}{l} 2{x}_{1} = - \frac{1}{{x}_{2}^{2}} \\ \frac{1}{{x}_{2}} - {x}_{1}^{2} = 2{x}_{1}\left( {{x}_{2} - {x}_{1}}\right) \end{array} \Rightarrow \left\{ \begin{array}{l} {x}_{1} = - 2, \\ {x}_{2} = - \frac{1}{2}, \end{array}\right. }\right. $$
即得公切线方程为 ${4x} + y + 4 = 0$ .
解法 2 设公切线在 $y = {x}^{2}\left( {x < 0}\right)$ 上的切点为 $\left( {{x}_{1},{x}_{1}^{2}}\right)$ ,在 $y = \frac{1}{x}$ $\left( {x < 0}\right)$ 上的切点为 $\left( {{x}_{2},\frac{1}{{x}_{2}}}\right)$ ,公切线在 $x$ 轴和 $y$ 轴上的截距分别为 $u$ 和 $v$ ,那么
$$ {x}_{1} - \frac{{x}_{1}^{2}}{2{x}_{1}} = \frac{1}{2}{x}_{1} = u = {x}_{2} - \frac{\frac{1}{{x}_{2}}}{-\frac{1}{{x}_{2}^{2}}} = 2{x}_{2} \Rightarrow {x}_{2} = \frac{1}{4}{x}_{1}, $$
$$ {x}_{1}^{2} - {x}_{1} \cdot 2{x}_{1} = - {x}_{1}^{2} = v = \frac{1}{{x}_{2}} + {x}_{2} \cdot \frac{1}{{x}_{2}^{2}} = \frac{2}{{x}_{2}} \Rightarrow {x}_{2} = - \frac{2}{{x}_{1}^{2}}. $$
由此解出 ${x}_{1} = - 2,u = - 1,v = - 4$ . 即得公切线的截距式方程为
$$ \frac{x}{-1} + \frac{y}{-4} = 1 $$
解法 3 设公切线在 $y = \frac{1}{x}\left( {x < 0}\right)$ 上的切点为 $\left( {{x}_{2},\frac{1}{{x}_{2}}}\right)$ ,则公切线作为曲线 $y = \frac{1}{x}$ 的切线,其方程是
$$ y = \frac{1}{{x}_{2}} - \frac{1}{{x}_{2}^{2}}\left( {x - {x}_{2}}\right) , \tag{1.3} $$
因为公切线作为曲线 $y = {x}^{2}$ 的切线,所以二次方程
$$ {x}^{2} = \frac{1}{{x}_{2}} - \frac{1}{{x}_{2}^{2}}\left( {x - {x}_{2}}\right) $$
有等根,从而它的判别式 $\Delta = 1 - 8{x}_{2}^{3} = 0 \Rightarrow {x}_{2} = - \frac{1}{2}$ ,代入 (1.3) 式即得公切线方程为 ${4x} + y + 4 = 0$ .