1. Problem: Find the equations of the tangent and normal lines to the curve $y=\frac{2}{x}$ at the point $(5,0)$. 2. Formula: The slope of the tangent line is given by the derivative $y'$. The equation of the tangent line at point $(x_0,y_0)$ is: $$y - y_0 = m(x - x_0)$$ where $m = y'(x_0)$. The normal line is perpendicular to the tangent line, so its slope is $-\frac{1}{m}$. 3. Find the derivative: $$y = \frac{2}{x} = 2x^{-1}$$ $$y' = -2x^{-2} = -\frac{2}{x^2}$$ 4. Evaluate the derivative at $x=5$: $$m = y'(5) = -\frac{2}{5^2} = -\frac{2}{25}$$ 5. Write the tangent line equation: Point: $(5,0)$, slope $m = -\frac{2}{25}$ $$y - 0 = -\frac{2}{25}(x - 5)$$ $$y = -\frac{2}{25}x + \frac{2}{5}$$ 6. Write the normal line equation: Slope of normal line: $$m_{normal} = -\frac{1}{m} = -\frac{1}{-\frac{2}{25}} = \frac{25}{2}$$ Equation: $$y - 0 = \frac{25}{2}(x - 5)$$ $$y = \frac{25}{2}x - \frac{125}{2}$$ Final answers: Tangent line: $$y = -\frac{2}{25}x + \frac{2}{5}$$ Normal line: $$y = \frac{25}{2}x - \frac{125}{2}$$