To find the equation of the tangent line to the curve $x^2 - xy - y^2 = 1$ at the point $(2, 1)$, we need to use implicit differentiation.

Step 1: Differentiate implicitly

We differentiate both sides of the equation $x^2 - xy - y^2 = 1$ with respect to $x$, remembering to apply the product rule and chain rule when differentiating terms involving $y$ (since $y$ is implicitly a function of $x$).

• $\frac{d}{dx}(x^2) = 2x$
• For $\frac{d}{dx}(xy)$, use the product rule: $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y$
• $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$

So, differentiating both sides of the equation gives: $2x - \left( x \frac{dy}{dx} + y \right) - 2y \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Now, solve the equation for $\frac{dy}{dx}$, the derivative of $y$ with respect to $x$:

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

Group the terms involving $\frac{dy}{dx}$:

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

Factor out $\frac{dy}{dx}$ on the right-hand side:

$2x - y = \frac{dy}{dx}(x + 2y)$

Finally, solve for $\frac{dy}{dx}$:

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

Step 3: Evaluate the derivative at the point $(2, 1)$

Substitute $x = 2$ and $y = 1$ into the derivative:

$\frac{dy}{dx} = \frac{2(2) - 1}{2 + 2(1)} = \frac{4 - 1}{2 + 2} = \frac{3}{4}$

So, the slope of the tangent line at $(2, 1)$ is $\frac{3}{4}$.

Step 4: Find the equation of the tangent line

The equation of the tangent line is given by the point-slope form: $y - y_1 = m(x - x_1)$ where $m = \frac{3}{4}$, $(x_1, y_1) = (2, 1)$.

Substitute these values: $y - 1 = \frac{3}{4}(x - 2)$

Simplifying: $y - 1 = \frac{3}{4}x - \frac{3}{2}$ $y = \frac{3}{4}x - \frac{3}{2} + 1$ $y = \frac{3}{4}x - \frac{1}{2}$

Thus, the equation of the tangent line is: $y = \frac{3}{4}x - \frac{1}{2}$

Conclusion:

The equation of the tangent line to the curve at $(2, 1)$ is $y = \frac{3}{4}x - \frac{1}{2}$.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What is the normal line equation to this curve at $(2, 1)$?
2. How would the process change if we were to find the tangent line to a different point on the curve?
3. Can you explain how the product rule was applied in this implicit differentiation?
4. What would the derivative look like if we had another term like $xy^2$ in the equation?
5. What geometric interpretation can be given to the slope $\frac{3}{4}$ at the point $(2, 1)$?

Tip: In implicit differentiation, always be careful when applying the product and chain rules, especially when differentiating terms that involve both $x$ and $y$.