Finding the equation of a hyperbola

Question

Find the equation of the hyperbola with center on $2y+x-1=0$, with an asymptote $y+2x-5=0$, and a focus $(1,0)$.

Can anyone help me out with this problem?

Answer 1

First, verify that the given focus lies on the line $x+2y-1=0$, which means that this line is the transverse axis of the hyperbola. Then reflect the given asymptote in this line to get an equation of other asymptote in the form $px+qy+r=0$. An equation of the hyperbola is then $(2x+y-5)(px+qy+r)=k$. Finally, choose $k$ so that the hyperbola has a focus at $(1,0)$.

There are various ways to do the latter. For instance, you could use the methods in the answers to this question to compute the foci of the above hyperbola and equate them to $(1,0)$ to get an equation for $k$. Or, you could use the fact that the circle through the foci centered at the hyperbola’s center, either line tangent at a vertex and either asymptote are concurrent to generate an equation for $k$.

Indeed, the latter property suggests another way to construct the required equation. First, compute the intersection of the axis and asymptote to get the center $C$ of the hyperbola. Next, compute an intersection $D$ of the circle centered at $C$ that passes through the given focus with the asymptote (either of the intersections will do). From there, drop a perpendicular to the axis to get a vertex $V$ of the hyperbola. This construction is illustrated below:

The semimajor axis length $a$ is then $CV$, while the semiminor axis length is the distance from the focus to the asymptote. The conjugate axis will have an equation of the form $2x-y+d=0$, and an equation of the hyperbola is therefore $${(2x-y+d)^2 \over a^2} - {(x+2y-1)^2 \over b^2} = 2^2+1^2.$$ I leave finding the unknown parameters to you.