8. Problem: Find the slopes of all lines passing through the origin and tangent to the curve $y^{2}=x^{3}+39 x-35$.
Solution: Any line passing throug the origin has equation $y=m x$, where $m$ is the slope of the line. If a line is tangent to the given curve, then at the point of tangency, $(x, y), \frac{d y}{d x}=m$.

First, we calculate $\frac{d y}{d x}$ of the curve: $2 y d y=3 x^{2} d x+39 d x \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}+39}{2 y}$. Substituting $m x$ for $y$, we get the following system of equations:
$$
\begin{aligned}
m^{2} x^{2} & =x^{3}+39 x-35 \\
m & =\frac{3 x^{2}+39}{2 m x}
\end{aligned}
$$

Solving for $x$ yields the equation $x^{3}-39 x+70=0 \Rightarrow(x-2)(x+7)(x-5)=0 \Rightarrow x=2$ or $x=-7$ or $x=5$. These solutions indicate the $x$-coordinate of the points at which the desired lines are tangent to the curve. Solving for the slopes of these lines, we get $m= \pm \frac{\sqrt{51}}{2}$ for $x=2$, no real solutions for $x=-7$, and $m= \pm \frac{\sqrt{285}}{5}$ for $x=5$. Thus $m= \pm \frac{\sqrt{51}}{2}, \pm \frac{\sqrt{285}}{5}$.