Consider the paraboloid \( f(x, y)=1-\frac{x^{2}}{9}-\frac{y^{2}}{4} \) and the point \( P(3,0) \) on the level curve \( f(x, y)=0 \). Compute the slope of the line tangent to the level curve at \( P \) and verify that the tangent line is orthogonal to the gradient at that point. What is the slope of the tangent line to the level curve at \( P(3,0) \) ? Select the correct choice below and, if necessary, fill in the answer box in your choice. A. The slope is \( \square \). B. The slope is undefined, so the tangent line is vertical.

To begin with, we can find the gradient of the function \( f(x,y) \) at the point \( P(3,0) \). The gradient \( \nabla f \) is given by the partial derivatives: \[ \nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Calculating the partial derivatives, we have: \[ \frac{\partial f}{\partial x} = -\frac{2x}{9} \quad \text{and} \quad \frac{\partial f}{\partial y} = -\frac{y}{2} \] Evaluating these at the point \( P(3,0) \): \[ \frac{\partial f}{\partial x}(3,0) = -\frac{2(3)}{9} = -\frac{2}{3} \quad \text{and} \quad \frac{\partial f}{\partial y}(3,0) = -\frac{0}{2} = 0 \] Thus, the gradient at point \( P(3,0) \) is: \[ \nabla f(3,0) = \left( -\frac{2}{3}, 0 \right) \] The slope of the tangent line to the level curve \( f(x,y) = 0 \) can be found by using the relationship that the tangent line is orthogonal to the gradient. Since the gradient vector is \( \nabla f = \left( -\frac{2}{3}, 0 \right) \), the slope of the gradient (using rise over run) is: \[ \text{slope of the gradient} = \frac{0}{-\frac{2}{3}} = 0 \] This indicates a horizontal line, thus the tangent line to the level curve must be vertical. Therefore, the slope of the tangent line is undefined. So, the correct answer is: B. The slope is undefined, so the tangent line is vertical.