Axes of Symmetry for a General Ellipse 
Given a general ellipse 
  $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ 
  where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?

It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellipse of the form $x^2/p^2+y^2/q^2=1$ and proceed from there. This may involve some messy trigonometric manipulations.
Could there be another approach perhaps, considering only quadratic/diophantine and linear equations?
Addendum
Here's a graphical implementation based on the answer by Ng Chung Tak. 
Addendum 2
Based on the answers by amd and by Ng Chung Tak, the equations for the axes are

$$\color{red}{\left(y-\frac {2AE-BD}{B^2-4AC}\right)=\frac {C-A\pm \sqrt{(A-C)^2+B^2}}B\left(x-\frac {2CD-BE}{B^2-4AC}\right)}$$

Note that 
$$\frac{C-A\pm \sqrt{(A-C)^2+B^2}}B\cdot \color{lightgrey}{\frac {C-A\mp\sqrt{(A-C)^2+B^2}}{C-A\mp\sqrt{(A-C)^2+B^2}}}=-\frac B{C-A\mp\sqrt{(A-C)^2+B^2}}$$
i.e. it is equal to the negative of its own reciprocal. Hence the equations for the axes can also be written as 

$$\color{red}{\left(x-\frac {2CD-BE}{B^2-4AC}\right)=-\frac {C-A\mp \sqrt{(A-C)^2+B^2}}B\left(y-\frac {2AE-BD}{B^2-4AC}\right)}$$

hence the two symmetrical anti-symmetrical forms for the axes. Here's the graphical implementation. 

The trigonometric manipulations are not messy.
After centering, the equation is, in polar coordinates,
$$\rho^2=-\frac{F'}{A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta}.$$
The denominator
$$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos\frac\theta2+B\sin\frac\theta2\right)$$ achieves its extrema when
$$(A-C)\sin\frac\theta2=B\cos\frac\theta2,$$ which is elementary.

Update:
Read $\color{green}{2\theta}$, not $\color{red}{\dfrac\theta2}$ !
$$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos2\theta+B\sin2\theta+(A+C)\right)$$
and
$$(A-C)\sin2\theta=B\cos2\theta.$$
