8. The conic section that is tangent to the parabola $y^{2}=5 x+9$ at points $P(0,3), Q(-1,-2)$, and passes through point $A$ $(-2,1)$ is

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8. $2 x^{2}-10 x y-31 y^{2}+175 x-6 y+297=0$

The equation of the line passing through points $P(0,3)$ and $Q(\quad, 2)$ is $5 x-y+3=0$. The required curve equation is: $y^{2}-5 x-9-\lambda(5 x-y+3)^{2}=0$. Substituting $A(-2,-1)$ yields $\lambda=$ $-\frac{1}{32}$, hence the solution.