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Suppose we have quadratic form $$f(a,b,c)=ax^2+bxy+cy^2\in\Bbb Q[x,y]$$ is there a transformation of $a,b,c$ or $x,y$ which takes $f(a,b,c)$ to $$\tilde f(a',b,c')=a'x^2-bxy+c'y^2\in\Bbb Q[x,y]$$ at least for a dense subset of $(a,b,c,x,y)\in\Bbb Q^3\times\Bbb Q^2$ other than trivial cases like sign changes of $x$ or $y$ or $b$ or $b=0$ or $x=0$ or $y=0$ where $a\neq -a'$ or $c\neq -c'$ holds?

An action would be like a linear transformation of variables $a,b,c,x,y$

Is there a name for these pairs of quadratic forms in literature?

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    Why would you want such a thing? – 2017-01-10
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    @DonAntonio it is a natural query. – 2017-01-10
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    Well...you can change the sign of $b$. Perhaps if you were willing to make precise the nature of the transformation you seek, it'd be more likely that you'd get a useful answer. – 2017-01-10
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    @JohnHughes added details. – 2017-01-10
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    Thanks...barring my example really helped clear up your intent. Let me ask again: can you say explicitly what it is that you seek? The identity transformation, for instance, "works" on the subset consisting of the coordinate axes, but presumably you'll now rule that out as not a good subset. Instead, why not tell us what WOULD be a satisfactory kind of answer rather than ruling out all possible solutions one by one? Also: can you say what it means for a transformation to "take" $f(a, b, c)$ to $\tilde{f}(a,b, c)$? – 2017-01-10
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    @JohnHughes I think of this as some kind of analogue of complex conjugate if you take the case $a=a'$ and $b=b'$. Is there an analog like this possible in quadratic forms? – 2017-01-10
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    Good luck. I can't make any sense of what you're talking about, and you don't seem willing to do the work to answer my questions, so I'm gonna let someone else try to answer yours. – 2017-01-10
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    @JohnHughes Sorry I do not understand your query properly. But I found something. Your webpage says one of your paper is missing in time. I found someone who quotes your paper (http://orbit.dtu.dk/files/126824972/onb_frisvad_jgt2012_v2.pdf) may be he has a copy? – 2017-01-10

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The Hessian matrix (second partial derivatives) of the first quadratic form is $$ H =\left( \begin{array}{cc} 2a & b \\ b & 2c \end{array} \right) $$

The relationship called "equivalence" takes an integer matrix $P \in SL_2 \mathbb Z,$ with new Hessian matrix given by $$ P^t H P. $$

If desired, one may allow $P$ in $SL_2 \mathbb Q$

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    If I use $P=\begin{bmatrix}1&m\\0&1\end{bmatrix}$ there is no such transformation. In general there is none? – 2017-01-10