How do I find the equation of hyperbola passing through a given point and has given two asymptotes

Question

If I know that my hyperbola passes through the point $$(1;1)$$ and has the asymptotes $$2x-y+2=0$$ and $$y=0$$ how can I find the equation of the hyperbola? Thanks in advance!

Answer 1

When $x$ and $y$ are large the curve approximates the pair of straight lines as given by the equations of the asymptotes. Therefore the hyperbola will have an equation of the form given by the product of the two equations equal to a constant. So the hyperbola is $$y(2x-y+2)=c$$

Since $(1,1)$ lies on the curve, $c = 3$

Answer 2

General equation of hyperbola is $$ax^2+bxy+cy^2+dx+ey+f=0$$

Since $ac\ne 0$ we can assume $c=1$. If we express $y$ we get $$y= {-(bx+e)\pm \sqrt{(bx+e)^2-4(ax^2+dx+f)}\over 2}$$

Now, use that:

• $\lim_{x\to \infty} y =0$ and
• $\lim_{x\to \infty} {y\over x} =2 \;\;$(so $a+2b+4=0$) and
• $\lim_{x\to \infty} (y-2x) =2$

And finally, since $(1,1)$ is on it we get $a+b+1+d+e+f=0$.