<s>[INST] Equation of a hyperbola given its asymptotes 
Find the equation of the hyperbola whose asymptotes are $3x-4y+7$ and $4x+3y+1=0$ and which pass through the origin.

The equation of the hyperbola is obtained in my reference as
$$
(3x-4y+7)(4x+3y+1)=K=7
$$
So it make use of the statement, the equation of the hyperbola = equation of pair of asymptotes + constant
I understand that the pair of straight lines is the limiting case of hyperbola.
Why does the equation to the hyperbola differ from the equation of pair of asymptotes only by a constant ?
 [/INST] $$
\frac{4x+3y+1}{5}=\pm\frac{3x-4y+7}{5}\\
\implies x+7y-6=0\;;\; 7x-y+8=0\text{ which are the axis of the hyperbola with centre }(-1,1)\\
$$
Since $m_1m_2=-1\implies$ asymptotes are perpendicular $\implies$ rectangular hyperbola
$$
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=\pm1\\
\text{At }(0,0): \frac{18}{25a^2}-\frac{32}{25a^2}=\pm1\implies18a^2-32a^2=\pm25a^4\\
-14a^2=\pm25a^4\implies-14a^2=25a^4\text{ not possible}\\
-14a^2=-25a^4\implies \boxed{a^2=\frac{14}{25}}\\
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=-1\\
\frac{(7x-y+8)^2}{50a^2}-\frac{(x+7y-6)^2}{50a^2}=1\\
(7x-y+8)^2-(x+7y-6)^2=50a^2=50.\frac{14}{25}=28\\
x^2(48)+y^2(-48)+xy(-28)+x(124)+y(68)+28=28\\
\color{blue}{12x^2-7xy-12y^2+31x+17y=0}
$$
</s>