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

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

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

Answer 2

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.

Answer 3

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.