How to find the equation of a hyperbola knowing its asymptotes and one focus?

Question

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.

Answer 1

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:

$$3x^2-4xy-4=0$$

Answer 2

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 $$ 3x^2-4xy-4=0. $$

Answer 3

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 $$3x^2-4xy-4=0.$$