Equation of Axis of Symmetry of a Parabola

The given parabola's equation can be manipulated, my aim here would be to convert it to a form that I can compare with the standard: $Y^2 = 4AX$ where Y and X would need to be equations of perpendicular lines. The equation of the axis of symmetry would then be: $Y=0$

$2x^2 + 8xy + 8y^2 - x + 2y = 0$

$2(x+2y)^2 = x - 2y$

$(x+2y)^2 = \frac{1}{2}(x-2y)$

Note that $x+2y$ is not perpendicular to $x-2y$ , so bear with me as I do the following manipulation:

$(x+2y+\lambda)^2 = \frac{1}{2}(x-2y) +2\lambda x + 4\lambda y +\lambda^2$

$(x+2y+\lambda)^2 = \frac{1}{2}((1+4\lambda)x+(8\lambda - 2)y + 2\lambda^2)$

Now, I want the lines $(x+2y+\lambda)$ and $((1+4\lambda)x+(8\lambda - 2)y + 2\lambda^2)$ to be perpendicular, so I shall equate the product of their slopes to -1:

$\Big(-\frac{1}{2}\Big)\Big(-\frac{1+4\lambda}{8\lambda-2}\Big) = -1$

$\implies \lambda = \frac{3}{20}$

After substituting the value $\lambda$ and comparing, the equation of the axis of symmetry is:

$\boxed{x+2y+0.15=0}$