1. **State the problem:** We are given that when $x=2$, the rate of increase of $x$ with respect to time is $\frac{4}{5}$ times the rate of decrease of $y$ with respect to time. The relationship between $y$ and $x$ is given by $y = \frac{k}{x}$, where $k$ is a constant. We need to find the value of $k$. 2. **Write the given relationship:** $$y = \frac{k}{x}$$ 3. **Differentiate both sides with respect to time $t$ using implicit differentiation:** $$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{k}{x} \right) = k \frac{d}{dt} \left( x^{-1} \right) = k (-1) x^{-2} \frac{dx}{dt} = -\frac{k}{x^2} \frac{dx}{dt}$$ 4. **Express the rates given:** Let $\frac{dx}{dt}$ be the rate of increase of $x$ with respect to time. Let $\frac{dy}{dt}$ be the rate of change of $y$ with respect to time. Given that the rate of increase of $x$ is $\frac{4}{5}$ times the rate of decrease of $y$, so: $$\frac{dx}{dt} = \frac{4}{5} \times \left(-\frac{dy}{dt}\right)$$ Since $\frac{dy}{dt}$ is negative (decreasing), $-\frac{dy}{dt}$ is positive. 5. **Substitute $\frac{dy}{dt}$ from step 3 into the above equation:** $$\frac{dx}{dt} = \frac{4}{5} \times \left(-\left(-\frac{k}{x^2} \frac{dx}{dt}\right)\right) = \frac{4}{5} \times \frac{k}{x^2} \frac{dx}{dt}$$ 6. **Simplify the equation:** $$\frac{dx}{dt} = \frac{4}{5} \frac{k}{x^2} \frac{dx}{dt}$$ 7. **Divide both sides by $\frac{dx}{dt}$ (assuming $\frac{dx}{dt} \neq 0$):** $$\cancel{\frac{dx}{dt}} = \frac{4}{5} \frac{k}{x^2} \cancel{\frac{dx}{dt}}$$ $$1 = \frac{4}{5} \frac{k}{x^2}$$ 8. **Solve for $k$:** $$k = \frac{5}{4} x^2$$ 9. **Substitute $x=2$:** $$k = \frac{5}{4} \times 2^2 = \frac{5}{4} \times 4 = 5$$ **Final answer:** $$k = 5$$