6.42 Given the parabola $y=a x^{2}+b x+c$ has a line of symmetry at $x=-2$, it is tangent to a certain line at one point, this line has a slope of 2, and a y-intercept of 1, and the parabola intersects the $y=0$ at two points, the distance between which is $2 \sqrt{2}$. Try to find the equation of this parabola.
[Solution] (1) $y=a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a}$, given that $x=-2$ is the axis of symmetry, we get $\frac{b}{2 a}=2$, hence $b=4 a$.
(2) $y=a x^{2}+b x+c$ is tangent to the line $l$,

and $l: y=2 x+1$, so the parabola intersects $l$ at one point,
thus we have $a x^{2}+4 a x+c=2 x+1$,
which simplifies to $a x^{2}+2(2 a-1) x+(c-1)=0$.
and $(2 a-1)^{2}-a(c-1)=0$,
hence $c=\frac{(2 a-1)^{2}+a}{a}$.
(3) $y=a x^{2}+4 a x+\frac{(2 a-1)^{2}+a}{a}$

The intersection points with $y=0$ are given by $a x^{2}+4 a x+\frac{(2 a-1)^{2}+a}{a}=0$,
solving this gives $x_{1,2}=-2 \pm \sqrt{4-\frac{(2 a-1)^{2}+a}{a^{2}}}$.
Given $x_{1}-x_{2}=2 \sqrt{2}$, we get
$$\sqrt{4-\frac{(2 a-1)^{2}+a}{a^{2}}}=\sqrt{2}$$

which simplifies to $4-\frac{(2 a-1)^{2}+a}{a^{2}}=2$,
$$(2 a-1)^{2}+a=2 a^{2}$$

which further simplifies to $2 a^{2}-3 a+1=0,(2 a-1)(a-1)=0$.
Thus, $a=1$ or $a=\frac{1}{2}$.
When $a=1$, the required parabola is

i.e., 
$$y=x^{2}+4 x+2$$
$$y=(x+2)^{2}-2 .$$

When $a=\frac{1}{2}$, the required parabola is
$$y=\frac{1}{2} x^{2}+2 x+1$$

i.e., $y=\frac{1}{2}(x+2)^{2}-1$.