题目
9.求曲面 ax^2+by^2+cz^2=1 在点 (x_(0),y_(0),z_(0)) 处的切平面及法线方程.
9.求曲面 $ax^{2}+by^{2}+cz^{2}=1$ 在点 $(x_{0},y_{0},z_{0})$ 处的切平面及法线方程.
题目解答
答案
设 $F(x, y, z) = ax^2 + by^2 + cz^2 - 1$，则曲面在点 $(x_0, y_0, z_0)$ 处的法向量为 $\nabla F(x_0, y_0, z_0) = (2ax_0, 2by_0, 2cz_0)$。
**切平面方程：**
利用点法式方程，得
\[
2ax_0(x - x_0) + 2by_0(y - y_0) + 2cz_0(z - z_0) = 0 \implies ax_0x + by_0y + cz_0z = 1
\]
**法线方程：**
方向向量为 $(2ax_0, 2by_0, 2cz_0)$，故
\[
\frac{x - x_0}{2ax_0} = \frac{y - y_0}{2by_0} = \frac{z - z_0}{2cz_0} \quad \text{或} \quad \frac{x - x_0}{ax_0} = \frac{y - y_0}{by_0} = \frac{z - z_0}{cz_0}
\]
**答案：**
\[
\boxed{
\begin{array}{l}
\text{切平面：} ax_0x + by_0y + cz_0z = 1 \\
\text{法线：} \frac{x - x_0}{ax_0} = \frac{y - y_0}{by_0} = \frac{z - z_0}{cz_0} \quad \text{或} \quad \frac{x - x_0}{2ax_0} = \frac{y - y_0}{2by_0} = \frac{z - z_0}{2cz_0}
\end{array}
}
\]
解析
步骤 1：定义函数 $F(x, y, z)$
设 $F(x, y, z) = ax^2 + by^2 + cz^2 - 1$，则曲面 $ax^2 + by^2 + cz^2 = 1$ 可以表示为 $F(x, y, z) = 0$。
步骤 2：计算梯度 $\nabla F(x_0, y_0, z_0)$
曲面在点 $(x_0, y_0, z_0)$ 处的法向量为 $\nabla F(x_0, y_0, z_0) = (2ax_0, 2by_0, 2cz_0)$。
步骤 3：求切平面方程
利用点法式方程，得
\[ 2ax_0(x - x_0) + 2by_0(y - y_0) + 2cz_0(z - z_0) = 0 \]
化简得
\[ ax_0x + by_0y + cz_0z = 1 \]
步骤 4：求法线方程
法线方程的方向向量为 $(2ax_0, 2by_0, 2cz_0)$，故
\[ \frac{x - x_0}{2ax_0} = \frac{y - y_0}{2by_0} = \frac{z - z_0}{2cz_0} \]
或
\[ \frac{x - x_0}{ax_0} = \frac{y - y_0}{by_0} = \frac{z - z_0}{cz_0} \]
设 $F(x, y, z) = ax^2 + by^2 + cz^2 - 1$，则曲面 $ax^2 + by^2 + cz^2 = 1$ 可以表示为 $F(x, y, z) = 0$。
步骤 2：计算梯度 $\nabla F(x_0, y_0, z_0)$
曲面在点 $(x_0, y_0, z_0)$ 处的法向量为 $\nabla F(x_0, y_0, z_0) = (2ax_0, 2by_0, 2cz_0)$。
步骤 3：求切平面方程
利用点法式方程，得
\[ 2ax_0(x - x_0) + 2by_0(y - y_0) + 2cz_0(z - z_0) = 0 \]
化简得
\[ ax_0x + by_0y + cz_0z = 1 \]
步骤 4：求法线方程
法线方程的方向向量为 $(2ax_0, 2by_0, 2cz_0)$，故
\[ \frac{x - x_0}{2ax_0} = \frac{y - y_0}{2by_0} = \frac{z - z_0}{2cz_0} \]
或
\[ \frac{x - x_0}{ax_0} = \frac{y - y_0}{by_0} = \frac{z - z_0}{cz_0} \]