Questions: Which of the following is true about the curve (x^2-xy+y^2=3) at the point ((2,1))? (A) (dy/dx) exists at ((2,1)), but there is no tangent line at that point. (B) (dy/dx) exists at ((2,1)), and the tangent line at that point is horizontal. (C) (dy/dz) exists at ((2,1)), and the tangent line at that point is neither horizontal nor vertical. (E) (dy/dx) does not exist at ((2,1)), and the tangent line at that point is horizontal.

Which of the following is true about the curve (x^2-xy+y^2=3) at the point ((2,1))?
(A) (dy/dx) exists at ((2,1)), but there is no tangent line at that point.
(B) (dy/dx) exists at ((2,1)), and the tangent line at that point is horizontal.
(C) (dy/dz) exists at ((2,1)), and the tangent line at that point is neither horizontal nor vertical.
(E) (dy/dx) does not exist at ((2,1)), and the tangent line at that point is horizontal.

Transcript text: Which of the following is true about the curve $x^{2}-x y+y^{2}=3$ at the point $(2,1)$ ? (A) $\frac{d y}{d x}$ exists at $(2,1)$, but there is no tangent line at that point. (B) $\frac{d y}{d x}$ exists at $(2,1)$, and the tangent line at that point is horizontal. (C) $\frac{d y}{d z}$ exists at $(2,1)$, and the tangent line at that point is neither horizontal nor vertical. (E) $\frac{d y}{d x}$ does not exist at $(2,1)$, and the tangent line at that point is horizontal.

Solution

Solution Steps

Step 1: Implicit Differentiation

We start with the equation of the curve given by $x^{2} - xy + y^{2} = 3$. To find $\frac{dy}{dx}$, we differentiate both sides of the equation implicitly with respect to $x$:

\[ \frac{d}{dx}(x^{2}) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^{2}) = \frac{d}{dx}(3) \]

This results in:

\[ 2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \]

Rearranging gives us:

\[ \frac{dy}{dx} = \frac{2x - y}{-x + 2y} \]

Step 2: Evaluate the Derivative at the Point (2, 1)

Next, we substitute the point $(2, 1)$ into the derivative:

\[ \frac{dy}{dx} \bigg|_{(2, 1)} = \frac{2(2) - 1}{-2 + 2(1)} = \frac{4 - 1}{-2 + 2} = \frac{3}{0} \]

Since the denominator is zero, $\frac{dy}{dx}$ is undefined at the point $(2, 1)$.

Step 3: Determine the Nature of the Tangent Line

Since $\frac{dy}{dx}$ does not exist at $(2, 1)$, it indicates that the tangent line at this point is vertical.

Final Answer

The correct statement regarding the curve at the point $(2, 1)$ is that $\frac{dy}{dx}$ does not exist at $(2, 1)$, and the tangent line at that point is vertical. Therefore, the answer is:

$\boxed{E}$

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