8. Given that a quadratic curve is tangent to the line $x-2 y+3=0$ at point $P(-1,1)$, tangent to the line $5 x+2 y-6=0$ at point $Q(2,-2)$, and passes through point $R(1,0)$. Find the equation of this quadratic curve.
8. The equation of the line $P P$ (i.e., the tangent line) is $x-2 y+3=0$, and the equation of the line $Q Q$ (i.e., the tangent line) is $5 x+2 y-6=0$. The equation of the line $P Q$ is $y-1=\frac{-2-1}{2+1}(x+1)$, which simplifies to $x+y=0$.

Since $P$ and $Q$ are both points of tangency, i.e., double points, we can consider the line $P Q$ as a double line. Therefore, we can set the equation of the required conic section as $(x-2 y+3)(5 x+2 y-6)+\lambda\left(x+y^{2}\right)=0$.
Given that the curve passes through the point $R(1,0)$, we find $\lambda=4$. Hence, the required conic section is $x^{2}+x+2 y-2=0$.