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 ?
 A: Consider the equation of a hyperbola  $$ \frac {(x-x_0)^2}{a^2} -\frac {(y-y_0)^2}{b^2}=1 \tag {1}$$ 
Which has its asymptotes $$(y-y_0)=\pm \frac {b}{a}(x-x_0)$$ 
Upon multioplication of the equations of the two asymptotes we get $$(y-y_0)^2 = \frac {b^2}{a^2} (x-x_0)^2$$ 
or  $$\frac {(x-x_0)^2}{a^2} -\frac {(y-y_0)^2}{b^2} =0 \tag {2}$$ 
As you see the difference of $(1)$ and $(2)$ is a constant. 
A: Expanding a comment:

For a point on a hyperbola, the product of the signed distances (say, $d_1$ and $d_2$) to the asymptotes is a constant.
  $$d_1 d_2 = k \tag{1}$$

(If $k=0$, then the hyperbola degenerates to just the asymptotes themselves.)
Since the signed distances from $(x,y)$ to line $ax+by+c=0$ is 
$$d = \frac{a x + b y + c}{\sqrt{a^2+b^2}} \tag{2}$$
it follows that points on the hyperbola with asymptotes $ax+by+c=0$ and $dx+ey+f=0$ satisfy
$$\frac{ax+by+c}{\sqrt{a^2+b^2}}\cdot\frac{dx+ey+f}{\sqrt{d^2+e^2}}=k \tag{3}$$
Clearing fractions, and "absorbing" the square roots into the arbitrary constant $k$, we have
$$(ax+by+c)(dx+ey+f)=k \tag{4}$$
If we know a particular point $(x_0, y_0)$ on the curve, we can substitute to find $k$, whereupon we get the final equation

$$(ax+by+c)(dx+ey+f)=(ax_0+by_0+c)(dx_0+ey_0+f) \tag{5}$$

For the specific problem at hand, we have
$$(3x-4y+7)(4x+3y+1)=7\cdot 1 \tag{5}$$
which the reader can expand and reduce.
A: $$
\frac{4x+3y+1}{5}=\pm\frac{3x-4y+7}{5}\\
\implies x+7y-6=0\;;\; 7x-y+8=0\text{ which are the axis of the hyperbola with centre }(-1,1)\\
$$
Since $m_1m_2=-1\implies$ asymptotes are perpendicular $\implies$ rectangular hyperbola
$$
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=\pm1\\
\text{At }(0,0): \frac{18}{25a^2}-\frac{32}{25a^2}=\pm1\implies18a^2-32a^2=\pm25a^4\\
-14a^2=\pm25a^4\implies-14a^2=25a^4\text{ not possible}\\
-14a^2=-25a^4\implies \boxed{a^2=\frac{14}{25}}\\
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=-1\\
\frac{(7x-y+8)^2}{50a^2}-\frac{(x+7y-6)^2}{50a^2}=1\\
(7x-y+8)^2-(x+7y-6)^2=50a^2=50.\frac{14}{25}=28\\
x^2(48)+y^2(-48)+xy(-28)+x(124)+y(68)+28=28\\
\color{blue}{12x^2-7xy-12y^2+31x+17y=0}
$$
