8. Let $x, y \in \mathbf{R}^{+}$, and $\frac{19}{x}+\frac{98}{y}=1$, the minimum value of $x+y$ is
8. $117+14 \sqrt{38}$.

Let $\frac{19}{x}=\cos ^{2} \theta, \frac{98}{y}=\sin ^{2} \theta$, then $x=\frac{19}{\cos ^{2} \theta}=19+19 \tan ^{2} \theta, y=\frac{98}{\sin ^{2} \theta}=98+98 \cot ^{2} \theta$. Therefore, $x+y=117+19 \tan ^{2} \theta+98 \cot ^{2} \theta \geqslant 117+14 \sqrt{38}$.