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?
On
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.
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:
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