求解参数方程所表示曲线指定点处的法线方程

一、题目

曲线 $\{\begin{array}{l}x=\cos t+{\cos }^{2}t\\ y=1+\sin t\end{array}$ 在 $t=\frac{\pi }{4}$ 对应的点处的法线方程为（）

难度评级：

二、解析

首先，求解出法线的斜率：

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\cdot \frac{\mathrm{d}t}{\mathrm{d}x}\Rightarrow$$

$$\frac{\mathrm{d}y}{\mathrm{d}t}=\cos t\frac{\mathrm{d}x}{\mathrm{d}t}=-\sin t-2\cos t\sin t\Rightarrow$$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\cos t}{-\sin t-2\cos t\sin t}\Rightarrow$$

$$t=\frac{\pi }{4}\Rightarrow$$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=k=$$

$$\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}-2\times \frac{1}{2}}=\frac{\sqrt{2}}{2}\times \frac{1}{-(\frac{\sqrt{2}}{2}+1)}=\frac{-\sqrt{2}}{4\sqrt{2}+2}\Rightarrow$$

$$\frac{-1}{k}=\frac{\sqrt{2}+2}{\sqrt{2}}=\frac{2+2\sqrt{2}}{2}=\sqrt{2}+1$$

又可知发现与曲线的交点为：

$$t=\frac{\pi }{4}\Rightarrow \{\begin{array}{ll}& x_0=\frac{\sqrt{2}}{2}+\frac{1}{2}\\ & y_0=1+\frac{\sqrt{2}}{2}\end{array}$$

于是，法线的方程为：

$$y-y_0=\frac{-1}{k}(x-x_0)\Rightarrow$$

$$y-1-\frac{\sqrt{2}}{2}=(\sqrt{2}+1)(x-\frac{\sqrt{2}}{2}-\frac{1}{2})\Rightarrow$$

$$y=(\sqrt{2}+1)x-(\sqrt{2}+1)(\frac{\sqrt{2}}{2}+\frac{1}{2})+\frac{\sqrt{2}}{2}+1$$

$$y=(\sqrt{2}+1)x-\frac{(\sqrt{2}+1{)}^2}{2}+\frac{\sqrt{2}+2}{2}\Rightarrow$$

$$y=(\sqrt{2}+1)x-\frac{2+1+2\sqrt{2}\sqrt{2}-2}{2}\Rightarrow$$

$$y=(\sqrt{2}+1)x-\frac{\sqrt{2}+1}{2}$$

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