Find an equation of the tangent plane to the surface at the given point. \[ x y^{2}+5 x-z^{2}=36, \quad(5,-2,3) \]

To find the equation of the tangent plane to the surface defined by the equation \[ F(x, y, z) = x y^{2} + 5 x - z^{2} - 36 = 0 \] at the point \((5, -2, 3)\), we will follow these steps: 1. **Calculate the gradient of \(F\)**, which gives us the normal vector to the surface at the point. 2. **Evaluate the gradient at the given point**. 3. **Use the point-normal form of the equation of a plane** to find the equation of the tangent plane. ### Step 1: Calculate the Gradient of \(F\) The gradient \(\nabla F\) is given by: \[ \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \] Calculating the partial derivatives: - \(\frac{\partial F}{\partial x} = y^{2} + 5\) - \(\frac{\partial F}{\partial y} = 2xy\) - \(\frac{\partial F}{\partial z} = -2z\) Thus, the gradient is: \[ \nabla F = \left( y^{2} + 5, 2xy, -2z \right) \] ### Step 2: Evaluate the Gradient at the Point \((5, -2, 3)\) Now we substitute \(x = 5\), \(y = -2\), and \(z = 3\) into the gradient: 1. Calculate \(\frac{\partial F}{\partial x}\): \[ \frac{\partial F}{\partial x} = (-2)^{2} + 5 = 4 + 5 = 9 \] 2. Calculate \(\frac{\partial F}{\partial y}\): \[ \frac{\partial F}{\partial y} = 2(5)(-2) = -20 \] 3. Calculate \(\frac{\partial F}{\partial z}\): \[ \frac{\partial F}{\partial z} = -2(3) = -6 \] Thus, the gradient at the point \((5, -2, 3)\) is: \[ \nabla F(5, -2, 3) = (9, -20, -6) \] ### Step 3: Use the Point-Normal Form of the Equation of a Plane The equation of the tangent plane can be expressed as: \[ 9(x - 5) - 20(y + 2) - 6(z - 3) = 0 \] Expanding this equation: \[ 9x - 45 - 20y - 40 - 6z + 18 = 0 \] Combining like terms: \[ 9x - 20y - 6z - 67 = 0 \] Thus, the equation of the tangent plane at the point \((5, -2, 3)\) is: \[ 9x - 20y - 6z = 67 \]