How to find the equation of a hyperbola knowing its asymptotes and one focus? Which is the equation for the hyperbola with one focus in $(2,-1)$ and its asymptotes are $x=0$ and $3x - 4y = 0$ ? I am not able to see how to find the answer.
 A: A hyperbola is a set of points, such that for any point $P$ of the set, the absolute difference of the distances $|PF_{1}|,\ |PF_{2}|$ to two fixed points $F_{1},F_{2}$ (the foci), is constant. The two asymptotes intersect in $(0,0)$. Therefore the hyperbola is symmetric respect to this point i.e. for any point $(x,y)$ of the hyperbola the point $(-x,-y)$ is also a point of it. Alternatively this leads to if $(x_{f_1},y_{f_1})$ is a focus of it the other one is $$(x_{f_2},y_{f_2})=(-x_{f_1},-y_{f_1})$$
This enables us to write the equation of the hyperbola according to the definition as follows:
$$|\sqrt{(x-x_{f_1})^2+(y-y_{f_1})^2} -\sqrt{(x-x_{f_2})^2+(y-y_{f_2})^2}|=\lambda$$
where $\lambda$ is the same constant difference of distances. By substitution we obtain:
$$|\sqrt{(x-2)^2+(y+1)^2} -\sqrt{(x+2)^2+(y-1)^2}|=\lambda$$
squaring both sides gives us:
$$x^2+y^2+5-\sqrt{(x^2+y^2+5)-(4x-2y)^2}={\lambda^2\over 2}$$
or
$$x^2+y^2+5-{\lambda^2\over 2}=\sqrt{(x^2+y^2+5)^2-(4x-2y)^2}$$
squaring again
$$(x^2+y^2+5-{\lambda^2\over 2})^2=(x^2+y^2+5)^2-(4x-2y)^2$$
which gives us after simplification
$${\lambda^4\over 4}-5{\lambda^2}=({\lambda^2}-16)x^2+({\lambda^2}-4)y^2+16xy$$
dividing both sides by $y^2$ and letting $t={x\over y}$ we obtain:
$${{{\lambda^4\over 4}-5{\lambda^2}}\over{y^2}}=({\lambda^2}-16)t^2+{\lambda^2}-4+16t$$
since by definition of asymptote $y\to\infty$ results $t\to 0$ or $t\to{4\over 3}$ these two numbers should be roots of below equation:
$$({\lambda^2}-16)t^2+16t+{\lambda^2}-4=0$$
which gives us $\lambda=2$ and the final equation of the hyperbola is:
$$3x^2-4xy-4=0$$

A: The center of the hyperbola lies at the intersection of the asymptotes, that is $(0,0)$. If $F=(2,-1)$ is one focus, the other one is the symmetric with respect to the center, that is $G=(-2,1)$.
If you now take a point $P$ on the hyperbola with $x\to0$ and $y\to-\infty$, the difference $PG-PF$ tends to $y_{G}-y_F=2$, and this difference is the same for all points on the hyperbola on the same branch, while it is $-2$ on the other branch of the curve. The equation of the hyperbola is then:
$$
\sqrt{(x+2)^2+(y-1)^2}-\sqrt{(x-2)^2+(y+1)^2}=\pm 2,
$$
which after the usual manipulations becomes
$$
3x^2-4xy-4=0.
$$
A: The asymptotes of a hyperbola are themselves the degenerate member of a single-parameter family of hyperbolas with common asymptotes. If the equations of the asymptotes are $Ax+By+C=0$ and $Ex+Fy+G=0$, then the equations of these hyperbolas are $(Ax+By+C)(Ex+Fy+G)=k$, for various values of $k$. For this problem, this means that an equation of the desired hyperbola will have the form $x(3x-4y)=k$.  
Now, the perpendicular distance of a hyperbola’s focus to either of its asymptotes is equal to its semi-minor axis length $b$, which we can see at a glance is $2$ for this hyperbola. Using the fact that the two semiaxis lengths $a$ and $b$ and the distance $c$ from the center to the focus are a Pythagorean triple, and that the center of this hyperbola is the origin, we can find the semimajor axis length: $a^2 = c^2-b^2 = 2^2+(-1^2)-2^2=1$. This means that a vertex of this hyperbola is $\frac1{\sqrt5}(2,-1)$. Plugging this point into the equation of the hyperbola and solving for $k$, we get $k=4$ and so an equation of the hyperbola is $$3x^2-4xy-4=0.$$
