4. Given that the center of the hyperbola is the origin, the foci are on the $x$-axis, the eccentricity $e=\frac{5}{3}$, and it is tangent to the line $8 x+2 \sqrt{7} y-16=0$. Find the equation of the hyperbola.
4. Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. Since $\frac{b}{a}=\sqrt{e^{2}-1}=\frac{4}{3}$, we can set $a^{2}=9 \lambda, b^{2}=16 \lambda(\lambda>$ 0 ), so the equation of the hyperbola is $\frac{x^{2}}{9 \lambda}-\frac{y^{2}}{16 \lambda}=1$. Because the line $8 x+2 \sqrt{7} y-16=0$ is tangent to it, by property 5, we have $64 \cdot 9 \lambda-28 \cdot 16 \lambda=16^{2}$, which gives $\lambda=2$, hence the equation of the hyperbola is $\frac{x^{2}}{18}-\frac{y^{2}}{32}=1$.