The given problem asks to determine the equation of the tangent line to the curve $x^2 - y^2 + xy - 11 = 0$ at a point where $x = 3$. Let's solve it step by step:

1. Differentiate the equation implicitly to find the slope:

The given equation is: $x^2 - y^2 + xy - 11 = 0$ Differentiating both sides with respect to $x$, we get: $2x - 2y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ Reorganize to isolate $\frac{dy}{dx}$: $\frac{dy}{dx}(x - 2y) = -2x - y$ $\frac{dy}{dx} = \frac{-2x - y}{x - 2y}$

2. Find the corresponding $y$-coordinate when $x = 3$:

Substitute $x = 3$ into the original equation: $3^2 - y^2 + 3y - 11 = 0$ $9 - y^2 + 3y - 11 = 0$ $-y^2 + 3y - 2 = 0$ Reorganizing: $y^2 - 3y + 2 = 0$ Factoring: $(y - 1)(y - 2) = 0$ So, $y = 1$ or $y = 2$.

Thus, the curve intersects $x = 3$ at two points: $(3, 1)$ and $(3, 2)$.

3. Calculate the slope at each point:

For $(3, 1)$: $\frac{dy}{dx} = \frac{-2(3) - 1}{3 - 2(1)} = \frac{-6 - 1}{3 - 2} = \frac{-7}{1} = -7$

For $(3, 2)$: $\frac{dy}{dx} = \frac{-2(3) - 2}{3 - 2(2)} = \frac{-6 - 2}{3 - 4} = \frac{-8}{-1} = 8$

4. Write the tangent line equations:

The general equation of a line is: $y - y_1 = m(x - x_1)$ Using the slope and the respective points:

• At $(3, 1)$ with $m = -7$: $y - 1 = -7(x - 3)$ $y = -7x + 21 + 1$ $y = -7x + 22$

• At $(3, 2)$ with $m = 8$: $y - 2 = 8(x - 3)$ $y = 8x - 24 + 2$ $y = 8x - 22$

Final Answer:

The equations of the tangent lines are: $y = -7x + 22 \quad \text{and} \quad y = 8x - 22$

Let me know if you would like detailed visualizations or further clarifications.

5 Related Questions:

1. What is implicit differentiation, and how is it applied here?
2. Can a curve have more than one tangent line at a given $x$-coordinate?
3. How do you determine if a tangent line intersects the curve again?
4. What happens to the tangent slope if the equation includes higher-degree terms?
5. Can you explain the geometric meaning of a tangent line?

Tip: Always double-check the points of intersection before calculating derivatives or slopes!