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
I'm going to show a solution which uses the following claim, and then add a proof of the claim and an important fact which might interest you.
Claim : The equation of a parabola can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex. (Note that the axis of symmetry is perpendicular to the tangent at vertex.)
Solution :
You correctly got $$x^2+8xy+16y^2-18x-48y+32=0$$ which can be written as $$(x+4y+c)^2+(-2c-18)x+(-8c-48)y-c^2+32=0$$
Since we want to find $c$ such that the line $x+4y+c=0$ is perpendicular to the line $(-2c-18)x+(-8c-48)y-c^2+32=0$, solving $$1\times (-2c-18)+4\times (-8c-48)=0$$ gives $c=-\dfrac{105}{17}$, and so the equation of the axis of symmetry is $x+4y-\dfrac{105}{17}=0$.
In the following, I'll add a proof of the claim.
Claim : The equation of a parabola can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex.
Proof :
The equation $$ax^2+2bxy+cy^2+2dx+2fy+g=0\tag2$$ represents a parabola iff $ac=b^2$ and $$\begin{vmatrix} a & b & d \\ b & c & f \\ d & f & g \\ \end{vmatrix}\not=0\tag3$$ (see here)
Multiplying the both sides of $(2)$ by $a$, and letting $$A=a,C=b,D=2ad,E=2af,F=ag$$ we see that, in general, the equation of a parabola is given by $$(Ax+Cy)^2+Dx+Ey+F=0\tag4$$ (where $(A,C)\not=(0,0)$, and $(3)\iff CD-AE\not=0$) which can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $$\begin{align}f(x,y)&=Ax+Cy+\frac {AD+CE}{2(A^2+C^2)} \\\\g(x,y)&=\frac{CD-AE}{A^2+C^2}(Cx-Ay+G) \\\\G&=\frac{4F(A^2+C^2)^2-(AD+CE)^2}{4(A^2+C^2)(CD-AE)}\end{align}$$ Note that $f(x,y)=0$ is perpendicular to $g(x,y)=0$.
Now, we can see the followings by some calculations :
The line $g(x,y)=0$ is tangent to the parabola $(4)$ at $P\bigg(H,\dfrac{CH+G}{A}\bigg)$ where $H=\dfrac{-2CG(A^2+C^2)-A(AD+EC)}{2(A^2+C^2)^2}$.
The line $f(x,y)=0$ intersects the parabola $(4)$ only at $P$.
From these, we can say the followings :
$f(x,y)=0$ is the equation of the axis of symmetry
$g(x,y)=0$ is the equation of the tangent at the vertex
$P$ is the vertex of the parabola.$\quad\blacksquare$
Finally, I'll add an important fact which might interest you.
The equation of a parabola $$(Ax+Cy)^2+Dx+Ey+F=0$$ can be written as $$\bigg(\frac{f(x,y)}{\sqrt{A^2+C^2}}\bigg)^2=\frac{AE-CD}{(A^2+C^2)^{\frac 32}}\cdot\frac{Cx-Ay+G}{\sqrt{A^2+C^2}}$$
where
$\dfrac{|f(x,y)|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the axis of symmetry
$\dfrac{|AE-CD|}{(A^2+C^2)^{\frac 32}}$ represents the length of its latus rectum (see here)
$\dfrac{|Cx-Ay+G|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the tangent at the vertex.