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3. The normal to the curve \( y=x^{2}-5 x \) at point \( A \) is parallel to the straight line \( y=-3 x+6 \). Find the equation of the normal to the curve at point \( A \). Garis normal kepada lengkung \( y=x^{2}-5 x \) pada titik \( A \) adalah selari dengan garis lurus \( y=-3 x+6 \). Cari persamaan garis normal kepada lengkung pada titik \( A \).

Ask by Bernard Washington. in Malaysia
Mar 15,2025

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Tutor-Verified Answer

Answer

法线方程为 \( y = -3x + \frac{16}{9} \)。

Solution

1. 首先,已知曲线为 \[ y = x^2 - 5x \] 求其在点 \( A \) 处的切线斜率,需要求导: \[ \frac{dy}{dx} = 2x - 5. \] 2. 切线斜率为 \( 2x-5 \) ,因此法线斜率(即切线斜率的负倒数)为: \[ m_{\text{法}} = -\frac{1}{2x-5}. \] 3. 题中给出法线平行于直线 \[ y = -3x + 6, \] 其斜率为 \(-3\)。由于相互平行,法线斜率即为 \(-3\),所以有: \[ -\frac{1}{2x-5} = -3. \] 4. 解方程求 \( x \): \[ -\frac{1}{2x-5} = -3 \quad \Rightarrow \quad \frac{1}{2x-5} = 3. \] 两边取倒数: \[ 2x-5 = \frac{1}{3}. \] 解得: \[ 2x = 5 + \frac{1}{3} = \frac{15}{3} + \frac{1}{3} = \frac{16}{3}, \] \[ x = \frac{16}{3} \times \frac{1}{2} = \frac{8}{3}. \] 5. 将 \( x = \frac{8}{3} \) 代入曲线方程求 \( y \): \[ y = \left(\frac{8}{3}\right)^2 - 5\left(\frac{8}{3}\right) = \frac{64}{9} - \frac{40}{3}. \] 将 \(\frac{40}{3}\) 化成以9为分母的分数: \[ \frac{40}{3} = \frac{120}{9}, \] 所以 \[ y = \frac{64}{9} - \frac{120}{9} = -\frac{56}{9}. \] 因此,点 \( A \) 的坐标为 \[ \left(\frac{8}{3}, -\frac{56}{9}\right). \] 6. 已知法线斜率为 \(-3\),且通过点 \( A\left(\frac{8}{3}, -\frac{56}{9}\right) \),用点斜式写法线方程: \[ y - \left(-\frac{56}{9}\right) = -3\left(x - \frac{8}{3}\right), \] 即 \[ y + \frac{56}{9} = -3x + 3\cdot\frac{8}{3}. \] 计算 \( 3\cdot\frac{8}{3} = 8 \),得: \[ y + \frac{56}{9} = -3x + 8. \] 7. 将方程整理为一般形式: \[ y = -3x + 8 - \frac{56}{9}. \] 写成同分母形式: \[ 8 = \frac{72}{9}, \] 得: \[ y = -3x + \frac{72}{9} - \frac{56}{9} = -3x + \frac{16}{9}. \] 最终,法线的方程为 \[ y = -3x + \frac{16}{9}. \]

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Untuk mencari titik \( A \) di mana normal kepada lengkung \( y=x^{2}-5x \) adalah selari dengan garis \( y=-3x+6 \), kita perlu menemukan gradien normal tersebut. Gradien garis \( y=-3x+6 \) adalah \(-3\), jadi gradien pada lengkung adalah terbalik dan bertanda positif, iaitu \(\frac{1}{3}\). Kita hitung derivatif \( y' \) untuk lengkung tersebut: \( y' = 2x - 5 \). Kemudian, kita perlu menyamakan \( 2x - 5 \) kepada \(\frac{1}{3}\): \[ 2x - 5 = \frac{1}{3} \] Menyelesaikan persamaan ini, kita mendapati \( x = \frac{16}{6} = \frac{8}{3} \). Setelah itu, kita boleh mencari nilai \( y \) pada titik \( A \): \[ y = \left(\frac{8}{3}\right)^2 - 5\left(\frac{8}{3}\right) = \frac{64}{9} - \frac{40}{3} = \frac{-56}{9} \] Kini, titik \( A \) adalah \( \left(\frac{8}{3}, -\frac{56}{9}\right) \). Persamaan garis normal adalah: \[ y - y_1 = m(x - x_1) \] \[ y + \frac{56}{9} = -3\left(x - \frac{8}{3}\right) \] Menyusun semula, kita mendapatkan persamaan garis normal sebagai: \[ y = -3x + 16 + \frac{56}{9} \] \[ y = -3x + \frac{144 + 56}{9} = -3x + \frac{200}{9} \] Kesimpulannya, persamaan garis normal kepada lengkung di titik \( A \) adalah: \[ y = -3x + \frac{200}{9} \]

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