The graph of a hyperbola with equation $y=f(x)$ has the following properties:
- The equations of the axes of symmetry are $y=x-1$ and $y=-x+3$
- Passes through the point $(-2,0)$
Determine $f(x)$
This is the question I need to answer. I know that to figure out the axes of symmetry I need to use the formulas $y=(x-p)+q$ and $y=-(x-p)+q$. However, how do I use this information to determine the equation of the hyperbola? I am assuming that I use the point $(-2,0)$ as the $y$ and $x$ values in the axes of symmetry to find the p and q values, but I do not know for sure if this is possible, because when I used graphing software to plot the hyperbola and axes of symmetry, the hyperbola cut through the opposite side of the axes of symmetry in order to pass the point $(-2,0)$. I believe this shouldn't happen and I am unfortunately struggling.
Given its lines of symmetry as $y=x-1$ and $y=-x+3$, the hyperbola's centre is where these two lines intersect: $x-1=-x+3$ implies $x=2$ and $y=1$, so the centre is $(2,1)$. Since these lines have slopes $+1$ and $-1$, the hyperbola is rectangular and is given by the equation $$(x-2)(y-1)=a$$ To find the value of $a$, plug in the coordinates of $(-2,0)$, the point known to be on the hyperbola: $$(-2-2)(0-1)=4$$ Therefore the full equation is $(x-2)(y-1)=4$. Solving for $y$: $$y-1=\frac4{x-2}$$ $$y=\frac4{x-2}+1$$