📝 题目
8．写出下列曲线在所给参数值相应的点处的切线方程和法线方程： （1）$\left\{\begin{array}{l}x=\sin t, \\ y=\cos 2 t,\end{array}\right.$ 在 $t=\frac{\pi}{4}$ 处； （2）$\left\{\begin{array}{l}x=\frac{3 a t}{1+t^{2}}, \\ y=\frac{3 a t^{2}}{1+t^{2}},\end{array}\right.$ 在 $t=2$ 处．
💡 答案与解析
[AI解答]
[AI解答]
(1) 曲线参数方程为 $$ \begin{cases} x = \sin t, \\ y = \cos 2t, \end{cases} \quad t = \frac{\pi}{4}. $$ 先求对应点坐标： $$ x_0 = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad y_0 = \cos\frac{2\pi}{4} = \cos\frac{\pi}{2} = 0. $$ 求导数： $$ \frac{dx}{dt} = \cos t, \quad \frac{dy}{dt} = -2\sin 2t. $$ 在 $t=\frac{\pi}{4}$ 处： $$ \frac{dx}{dt}\Big|_{t=\frac{\pi}{4}} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \frac{dy}{dt}\Big|_{t=\frac{\pi}{4}} = -2\sin\frac{\pi}{2} = -2. $$ 切线斜率 $$ k = \frac{dy}{dx} = \frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}} = \frac{-2}{\frac{\sqrt{2}}{2}} = -2\sqrt{2}. $$ 切线方程： $$ y - 0 = -2\sqrt{2}\left(x - \frac{\sqrt{2}}{2}\right) \quad\Rightarrow\quad y = -2\sqrt{2}\,x + 2. $$ 法线斜率 $k_{\perp} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$，法线方程： $$ y = \frac{\sqrt{2}}{4}\left(x - \frac{\sqrt{2}}{2}\right) \quad\Rightarrow\quad y = \frac{\sqrt{2}}{4}x - \frac{1}{4}. $$
(2) 曲线参数方程为 $$ \begin{cases} x = \dfrac{3at}{1+t^{2}}, \$$6pt] y = \dfrac{3at^{2}}{1+t^{2}}, \end{cases} \quad t=2. $$ 先求对应点： $$ x_0 = \frac{3a\cdot 2}{1+4} = \frac{6a}{5},\quad y_0 = \frac{3a\cdot 4}{5} = \frac{12a}{5}. $$ 求导： $$ \frac{dx}{dt} = 3a\cdot\frac{(1+t^2) - t(2t)}{(1+t^2)^2} = 3a\cdot\frac{1 - t^2}{(1+t^2)^2}, $$ $$ \frac{dy}{dt} = 3a\cdot\frac{2t(1+t^2) - t^2(2t)}{(1+t^2)^2} = 3a\cdot\frac{2t}{(1+t^2)^2}. $$ 在 $t=2$ 处： $$ \frac{dx}{dt}\Big|_{t=2} = 3a\cdot\frac{1-4}{(1+4)^2} = 3a\cdot\frac{-3}{25} = -\frac{9a}{25}, $$ $$ \frac{dy}{dt}\Big|_{t=2} = 3a\cdot\frac{4}{(5)^2} = \frac{12a}{25}. $$ 切线斜率 $$ k = \frac{dy}{dx} = \frac{\frac{12a}{25}}{-\frac{9a}{25}} = -\frac{4}{3}. $$ 切线方程： $$ y - \frac{12a}{5} = -\frac{4}{3}\left(x - \frac{6a}{5}\right). $$ 化简： $$ y = -\frac{4}{3}x + \frac{8a}{5} + \frac{12a}{5} = -\frac{4}{3}x + 4a. $$ 法线斜率 $k_{\perp} = \frac{3}{4}$，法线方程： $$ y - \frac{12a}{5} = \frac{3}{4}\left(x - \frac{6a}{5}\right) \quad\Rightarrow\quad y = \frac{3}{4}x - \frac{9a}{10} + \frac{12a}{5} = \frac{3}{4}x + \frac{3a}{2}. $$
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