6. Find the equation of the parabola passing through points $A(1,-2), B(9,-6)$, and tangent to the line $x-2y+4=0$ at $C(4,4)$.

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6. Given $A(1,-2), B(9,-6), C(4,4)$, the equation of line $AB$ is: $x+2 y+3=0$, the equation of line $BC$ is: $2 x+y-12=0$, the equation of the tangent line $CC$ is: $x-2 y+4=0$, and the equation of line $CA$ is: $2 x-y-4=0$. Therefore, the quadratic curve system sought is
$$
(x+2 y+3)(x-2 y+4)+\lambda(2 x+y-12)(2 x-y-4)=0,
$$

From $\Delta=0+4(1+4 \lambda)(4+\lambda)=0$ we get $\lambda=-\frac{1}{4}$ or -4.
Thus, the parabolas sought are $y^{2}=4 x$ and $x^{2}-9 x+2 y+12=0$.