Equation of a hyperbola given its asymptotes

Question

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 ?

Answer 1

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.