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 B. The slope is undefined, so the tangent line is vertical. How do you know the gradient and tangent line are orthogonal? Select the correct choice below and fill in the answer boxes in your choice. A. The slope of the tangent line, m, can be used to write a vector in the direction of the tangent line, $$ \langle 1, m\rangle $$. The dot product of this vector and the gradient $$ \langle\square $$ B. The tangent line slope is undefined, so it is a vertical line. The gradient $$ \langle\square, \square $$ is horizontal, so it is orthogonal to the tangent line.

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We start with the function

$$
f(x,y)=1-\frac{x^2}{9}-\frac{y^2}{4},
$$

and the level curve defined by

$$
f(x,y)=0.
$$

At the point $$P(3,0)$$, we wish to compute the slope of the tangent line to the level curve and verify that it is orthogonal to the gradient.

---

**Step 1. Implicit Differentiation**

Differentiate the level curve equation

$$
1-\frac{x^2}{9}-\frac{y^2}{4}=0
$$

with respect to $$x$$. Taking the derivative term by term gives:

$$
0 - \frac{2x}{9} - \frac{2y}{4}\frac{dy}{dx}=0.
$$

This simplifies to

$$
-\frac{2x}{9}-\frac{y}{2}\frac{dy}{dx}=0.
$$

Solve for $$\frac{dy}{dx}$$:

$$
-\frac{y}{2}\frac{dy}{dx} = \frac{2x}{9} \quad \Longrightarrow \quad \frac{dy}{dx} = -\frac{2x}{9}\cdot\frac{2}{y}=-\frac{4x}{9y}.
$$

At the point $$P(3,0)$$, substitute $$x=3$$ and $$y=0$$:

$$
\frac{dy}{dx}\Big|_{(3,0)}=-\frac{4\cdot 3}{9\cdot 0}.
$$

Since division by zero indicates that the derivative is undefined, the slope of the tangent line is undefined. This means the tangent line is vertical.

Thus, the correct choice for the slope of the tangent line is:

B. The slope is undefined, so the tangent line is vertical.

---

**Step 2. Verifying Orthogonality with the Gradient**

The gradient of $$f$$ is given by

$$
\nabla f(x,y)=\left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rangle.
$$

Compute the partial derivatives:

$$
\frac{\partial f}{\partial x}=-\frac{2x}{9} \quad\text{and}\quad \frac{\partial f}{\partial y}=-\frac{y}{2}.
$$

Thus,

$$
\nabla f(x,y)=\left\langle -\frac{2x}{9},\,-\frac{y}{2}\right\rangle.
$$

At $$P(3,0)$$:

$$
\nabla f(3,0)=\left\langle -\frac{2\cdot 3}{9},\,0\right\rangle=\left\langle -\frac{6}{9},\,0\right\rangle=\left\langle -\frac{2}{3},\,0\right\rangle.
$$

This gradient vector is horizontal.

Since the tangent line at $$P$$ is vertical, a vector in the direction of the tangent line can be taken as $$\langle 0,1\rangle$$.

The dot product between the gradient vector and the tangent direction vector is

$$
\left\langle -\frac{2}{3},\,0\right\rangle \cdot \langle 0,1\rangle = \left(-\frac{2}{3}\right)(0)+0\cdot1=0.
$$

A zero dot product confirms that the gradient and the tangent line are orthogonal.

Thus, for the verification we choose:

B. The tangent line slope is undefined, so it is a vertical line. The gradient is $$\langle -\frac{2}{3},\,0\rangle$$, which is horizontal, so it is orthogonal to the tangent line.
The slope of the tangent line at $$ P(3,0) $$ is undefined, making the tangent line vertical. The gradient at $$ P $$ is $$\langle -\frac{2}{3},\,0\rangle$$, which is horizontal. Since the tangent line is vertical and the gradient is horizontal, they are orthogonal.

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