Find the equation of the curve which satisfies the differential equation dy/dx = -x/y and passes through the point (-4; 3).

The general solution is given as $x = Ae^{-kt}$.
At $t=0$ hours, the quantity of the drug $x=200$ mg.
Substitute these values into the general solution:

$$200 = Ae^{-k \cdot 0}$$ $$200 = Ae^0$$ $$200 = A \cdot 1$$ $$A = 200$$

So the equation becomes $x = 200e^{-kt}$.

At $t=1$ hour, the quantity of the drug $x=180$ mg.
Substitute these values into the equation from Step 1:

$$180 = 200e^{-k \cdot 1}$$ $$180 = 200e^{-k}$$

Divide both sides by 200:

$$\frac{180}{200} = e^{-k}$$ $$0.9 = e^{-k}$$

We can rewrite the equation for $x$ as $x = 200(e^{-k})^t$. Substituting $e^{-k} = 0.9$:

$$x = 200(0.9)^t$$

Substitute $t=8$ into the equation from Step 2:

$$x = 200(0.9)^8$$ $$x = 200 \times 0.43046721$$ $$x = 86.093442$$

Rounding to two decimal places, the quantity of the drug after 8 hours is approximately 86.09 mg.

The quantity of the drug in the body after 8 hours is $\boxed{86.09 \text{ mg}}$.