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Условие: написать уравнение касательной плоскости и уравнения нормали к поверхности: \( x^{2}+2 y^{2}+z^{2}=3 \), проходящей через точку \( \mathbf{M}(1 ; 1 \); 0 (

Ask by Park Ayala. in Russia
Feb 27,2025

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Касательная плоскость: x + 2y – 3 = 0. Нормаль: x = 1 + 2t, y = 1 + 4t, z = 0, где t ∈ ℝ.

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Чтобы найти уравнение касательной плоскости к поверхности \( x^{2}+2 y^{2}+z^{2}=3 \) в точке \( \mathbf{M}(1, 1, 0) \), сначала найдем градиент функции \( F(x, y, z) = x^{2}+2 y^{2}+z^{2}-3 \). Найдем частные производные: - \( F_x = 2x \) - \( F_y = 4y \) - \( F_z = 2z \) В точке \( \mathbf{M}(1, 1, 0) \): - \( F_x(1, 1, 0) = 2(1) = 2 \) - \( F_y(1, 1, 0) = 4(1) = 4 \) - \( F_z(1, 1, 0) = 2(0) = 0 \) Градиент \( \nabla F = (2, 4, 0) \). Уравнение касательной плоскости имеет вид: \[ 2(x - 1) + 4(y - 1) + 0(z - 0) = 0 \] или \[ 2x + 4y - 6 = 0 \] Уравнение нормали будет иметь тот же вектор градиента: \[ 2x + 4y + 0z - 6 = 0 \] или просто \[ 2x + 4y - 6 = 0 \].

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