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 ?

$$ \frac{4x+3y+1}{5}=\pm\frac{3x-4y+7}{5}\\\ \implies x+7y-6=0\;;\; 7x-y+8=0\text{ which are the axis of the hyperbola with centre }(-1,1)\\\ $$ Since $m_1m_2=-1\implies$ asymptotes are perpendicular $\implies$ rectangular hyperbola

$$ \frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=\pm1\\\ \text{At }(0,0): \frac{18}{25a^2}-\frac{32}{25a^2}=\pm1\implies18a^2-32a^2=\pm25a^4\\\ -14a^2=\pm25a^4\implies-14a^2=25a^4\text{ not possible}\\\ -14a^2=-25a^4\implies \boxed{a^2=\frac{14}{25}}\\\ \frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=-1\\\ \frac{(7x-y+8)^2}{50a^2}-\frac{(x+7y-6)^2}{50a^2}=1\\\ (7x-y+8)^2-(x+7y-6)^2=50a^2=50.\frac{14}{25}=28\\\ x^2(48)+y^2(-48)+xy(-28)+x(124)+y(68)+28=28\\\ \color{blue}{12x^2-7xy-12y^2+31x+17y=0} $$