Q2- Find the equation of the tangent line drawn to the graph of the relation \( \ln (x y)+2 x y=2 \) at \( \left(x_{0}, y_{0}\right)=(e, 1 / e) \)

To find the equation of the tangent line to the graph of the relation \( \ln (x y) + 2 x y = 2 \) at the point \( (x_0, y_0) = (e, 1/e) \), we will follow these steps: 1. **Differentiate the equation implicitly** to find \( \frac{dy}{dx} \). 2. **Evaluate the derivative** at the point \( (e, 1/e) \) to find the slope of the tangent line. 3. **Use the point-slope form** of the equation of a line to write the equation of the tangent line. ### Step 1: Differentiate the equation implicitly We start with the equation: \[ \ln(xy) + 2xy = 2 \] Differentiating both sides with respect to \( x \): - The derivative of \( \ln(xy) \) using the chain rule is \( \frac{1}{xy} \cdot (y + x \frac{dy}{dx}) \). - The derivative of \( 2xy \) is \( 2(y + x \frac{dy}{dx}) \). Thus, differentiating gives us: \[ \frac{1}{xy} \left( y + x \frac{dy}{dx} \right) + 2 \left( y + x \frac{dy}{dx} \right) = 0 \] Now, we can simplify this equation. ### Step 2: Solve for \( \frac{dy}{dx} \) Let's rearrange the equation: \[ \frac{y + x \frac{dy}{dx}}{xy} + 2y + 2x \frac{dy}{dx} = 0 \] Multiplying through by \( xy \) to eliminate the fraction: \[ y + x \frac{dy}{dx} + 2xy^2 + 2x^2y \frac{dy}{dx} = 0 \] Now, collect the terms involving \( \frac{dy}{dx} \): \[ x \frac{dy}{dx} + 2x^2y \frac{dy}{dx} = -y - 2xy^2 \] Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} (x + 2x^2y) = -y - 2xy^2 \] Thus, we can solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-y - 2xy^2}{x + 2x^2y} \] ### Step 3: Evaluate the derivative at \( (e, 1/e) \) Now we substitute \( x = e \) and \( y = \frac{1}{e} \): \[ \frac{dy}{dx} = \frac{-\frac{1}{e} - 2e \left(\frac{1}{e}\right)^2}{e + 2e^2 \left(\frac{1}{e}\right)} \] Calculating the numerator: \[ -\frac{1}{e} - 2e \cdot \frac{1}{e^2} = -\frac{1}{e} - \frac{2}{e} = -\frac{3}{e} \] Calculating the denominator: \[ e + 2e^2 \cdot \frac{1}{e} = e + 2e = 3e \] Thus, we have: \[ \frac{dy}{dx} = \frac{-\frac{3}{e}}{3e} = -\frac{1}{e^2} \] ### Step 4: Write the equation of the tangent line The slope of the tangent line at the point \( (e, 1/e) \) is \( -\frac{1}{e^2} \). Using the point-slope form of the line: \[ y - y_0 = m(x - x_0) \] where \( m = -\frac{1}{e^2} \), \( x_0 = e \), and \( y_0 = \frac{1}{e} \): \[ y - \frac{1}{e} = -\frac{1}{e^2}(x - e) \] This can be rearranged to: \[ y = -\frac{1}{e^2}x + \frac{1}{e} + \frac{e}{e^2} \] \[ y = -\frac{1}{e^2}x + \frac{2}{e} \] Thus, the equation of the tangent line is: \[ y = -\frac{1}{e^2}x + \frac{2}{e} \]