Equation of parabola passes through $4$ distinct points 
Equation of axis of parabola which


passes through the point $(0,1)\ , \ (0,2)$


And $(2,0)\ ,\ (2,2)$ is

Let general equation of conic is
$ax^2+2hxy+by^2+2gx+2fy+c=0\cdots (1)$
And it represent parabola if $h^2=ab$
Parabola passes through $(0,1)$
Then put into $(1)$
$b+2f+c=0\cdots(2)$
Also parabola passes through $(0,2)$
Then $4b+4f+c=0\cdots (3)$
Also parabola passes through $(2,0)$
Then $4a+4g+c=0\cdots (4)$
Also parabola passes through $(2,2)$
Then $4a+8h+4b+4g+4f+c\cdots (5)$
$(4a+4g+c)+(4b+4f+c)+8h-c=0$
From $(2)$ and $(3)$, we get $\displaystyle h=\frac {c}{8}$
From $(2)$ and $(3)$
$\displaystyle b+2f=4b+4f\Longrightarrow 2f=-3b $
Put into $(2)$ and $(3)$
$\displaystyle b=\frac{c}{2}$ and $\displaystyle f=-\frac{3c}{4}$
And $\displaystyle h^2=ab\Longrightarrow \frac{c^2}{64}=\frac{ac}{2}\Longrightarrow a=\frac{c}{32}$
Put all into $(4)$
$\displaystyle \frac{4c}{32}+4g+c=0\Longrightarrow g=-\frac{9c}{32}$
Put all values into $(1)$
$\displaystyle \frac{c}{32}x^2+\frac{c}{4}xy+\frac{c}{2}y^2-\frac{9}{16}x-\frac{3c}{2}y+c=0$
$x^2+8xy+16y^2-18x-48y+32=0$
Buti did not know how I find axis of parabola
Please have a look

What you've done is correct.
One way to find the axis is to write the equation as $(x+4y)^2-18x-48y+32=0,$ then you know the axis is parallel to $x+4y=0.$ The tangent at vertex is orthogonal and parallel to $4x-y.$ Now you can find where $4x-y+c$ intersects doubly to find the vertex. $$x^2+8x(4x+c)+16(4x+c)^2-18x-48(4x+c)+32=0$$ or $$289x^2+(136c-210)x+16c^2-48c+32=0$$ and this has discriminant $$(136c-210)^2-4\cdot 289\cdot (16c^2-4c+32)=(-4)(408c - 1777)$$ so the discriminant is zero for $c=\frac{1777}{408}.$ The vertex is the intersection of the tangent at vertex and the parabola. So solve $$x^2+8x(4x+\frac{1777}{408})+16(4x+\frac{1777}{408})^2-18x-48(4x+\frac{1777}{408})+32=0$$ or $$(1734x+1147)^2\frac1{10404}$$ and put back into the tangent at vertex to find $y.$
The vertex then is $(-\frac{1147}{1734},\frac{11857}{6936}).$
The axis goes through the vertex  so the axis is $${ x+4y=\frac{105}{17}}.$$ As a bonus the tangent at vertex form is then $$(x+4y-105/17)^2=(96/(4\cdot 17))(4x-y+1777/408).$$ For fun I've also found the focus/directrix form $$17\cdot ((x+\frac{59}{102})^2+(y-\frac{689}{408})^2-(4x-y+\frac{113}{24})^2/17)=0$$

