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 $\left\{\begin{array}{l}x=\cos t+\cos ^{2} t \\ y=1+\sin t\end{array}\right.$ 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：

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

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

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

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

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

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

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

又可知发现与曲线的交点为：

$$
t=\frac{\pi}{4} \Rightarrow
\begin{cases}
& x_{0}=\frac{\sqrt{2}}{2}+\frac{1}{2} \\
& y_{0}=1+\frac{\sqrt{2}}{2}
\end{cases}
$$

于是，法线的方程为：

$$
y-y_{0}=\frac{-1}{k}\left(x-x_{0}\right) \Rightarrow
$$

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

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

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

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

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

---