What's so special about the form $ax^2+2bxy+cy^2$? Binary quadratic forms are sometimes studied (e.g. by Gauss) in the form
$$ax^2+2bxy+cy^2$$
In other words, the second coefficient is assumed to be even, and the polynomial is assumed to be homogeneous. Is there a good reason for this? Is it easy to reduce questions about general form
$$ax^2+bxy+cy^2+dx+ey+f$$
to this type of form?
 A: Alright, did it and got it correct; next time I will have this, I have saved it in my big Latex file of useful bits of mathematics typeset. 
Given
$$ g = a x^2 + b x y + c y^2 + d x + e y + f  $$
and taking the usual
$$  \Delta = b^2 - 4 a c, $$
we get, assuming both $a, \Delta \neq 0$ for this amount of work to be necessary,
$$ -4a \Delta g = \color{blue}{(\Delta y + bd - 2ae)^2 - \Delta (2 a x + b y + d)^2} +\color{green}{ \left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)}  $$
I put extra parentheses to emphasize that $\left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)$ is a single constant term, no $x,y$ involved. Since this site allows color, I also put the homogeneous part in blue and the constant part in green.
That all looks pretty good. If you have everything integers and are solving $g=0$ in integers, now you can use this revised diagonal thing, as long as $a, \Delta \neq 0.$ Meanwhile, if $\Delta$ is negative you get a finite number of points in the new variables
$$ u = \Delta y + bd - 2ae, \; \; \; v = 2ax +by +d. $$
If $\Delta$ is positive but a square, similar outcome. Finally, if $\Delta$ is positive but not a square, if there are any $(u,v)$ solutions there are infinitely many because it is Pell type, then you have to mess around to see whether that gives solutions in the original $(x,y).$ In principle, and with considerable care, all solutions can be found this way when no numbers involved are too large. 
Here is confirmation in gp-pari:

jagy@phobeusjunior:~$ 
jagy@phobeusjunior:~$ gp
Reading GPRC: /etc/gprc ...Done.

                               GP/PARI CALCULATOR Version 2.5.5 (released)
                        i686 running linux (ix86/GMP-5.1.2 kernel) 32-bit version
                    compiled: Sep 30 2013, gcc-4.8.1 (Ubuntu/Linaro 4.8.1-10ubuntu4) 
                 (readline v6.3 enabled [was v6.2 in Configure], extended help enabled)

                                 Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY 
WHATSOEVER.

Type ? for help, \q to quit.
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parisize = 4000000, primelimit = 500509
? 
? del = b^2 - 4 * a * c 
%1 = b^2 - 4*c*a
? 
? 
? me = (del * y + b * d - 2 * a * e)^2 
       - del * (2 * a * x + b * y + d)^2 
       + del * (d^2 - 4 * a * f) - (b * d - 2 * a * e)^2

%2 = (-4*a^2*b^2 + 16*c*a^3)*x^2 + (-4*y*a*b^3 - 4*d*a*b^2 + 16*y*c*a^2*b + 16*d*c*a^2)*x 
      + ((-4*y^2*c + (-4*e*y - 4*f))*a*b^2 + (16*y^2*c^2 + (16*e*y + 16*f)*c)*a^2)
? 
? them = a * x^2 + b * x * y + c * y^2 + d * x + e * y + f  
%3 = a*x^2 + (y*b + d)*x + (y^2*c + (e*y + f))
? 
? compare = me + 4 * a * del * them 
%4 = 0
? 
? 
? 
? 


This reminded me of something I have often seen on MSE, which is a certain 3 by 3 matrix determinant that gives information about our $$ g = a x^2 + b x y + c y^2 + d x + e y + f.  $$ I just checked, with matrix
$$
H =
\left(
\begin{array}{ccc}
2a & b & d \\
b & 2c & e \\
d & e & 2f
\end{array}
\right),   
$$
we find that the constant term above in green satisfies
$$\color{green}{ \left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)} = 2a \det H.$$
This finally makes sense to me, what we are doing is adding in a third variable to make a homogeneous quadratic form from $g,$ that is
$$ h(x,y,z) = a x^2 + b x y + c y^2 + d xz + e yz + fz^2.  $$
For me, this is now out of convenient order, I would write it over as
$$ h(x,y,z) = a x^2  + c y^2  + fz^2 + e yz+ d zx+ b x y ,  $$ and my usual abbreviation for this would be
$$ \langle a,c,f,e,d,b \rangle.  $$ So, half the determinant of $H$ is my usual "discriminant" 
$$ 4acf+edb -ae^2-cd^2-fb^2. $$
The relationship is just
$$ g(x,y) = h(x,y,1).  $$
NOTE, Monday, Feb. 2: I got some assistance on Meta, the preceding paragraph is my approach, a bit roundabout and clumsy, to these observations: How to determine if arbitrary point lies inside or outside a conic  AND transformation of conic section  A GOOD ONE  "Conic sections" that are really just two straight lines 
A: Many number-theoretical problems are usually related to determining the rational points situated on a certain geometric shape. So the shape and its algebraic expression have a logical precedence over the Diophantine equation whose solutions have to be determined. Now, what is the algebraic equation of a simple straight line ? $ax+by=c$. By squaring it, we obtain a so-called degenerate conic, which is merely a reunion of two straight lines, and not an actual conic per se $($circle, ellipse, hyperbola, or parabola$)$. But $(ax+by)^2=a^2x^2+2abxy+b^2y^2$. Letting $A=a^2,~B=b^2$, and $ab=C$, all we have to do is comparing C with $\sqrt{AB}$ in order to see whether we are dealing with finding rational points along an actual conic $($see Pell's equation, for instance$)$, or merely along a pair of straight lines, which is by far a more easier task.
