I hope someone can at least point me in a right direction.
I managed to close in on the equation required for a project, but have no idea about how to "add" certain characteristics that I need it to have. The equation is: $$y=-1/a(x-5)$$ It has two asymptotes that form a right angle at $(5,0)$ and as $a$ is increased, the graph (at least seems like) is symmetric along the $y=-x+5$ axis. All of this fits the purpose perfectly.
What it's missing though is two points that stay fixed as $a$ is varied, $(0,0)$ and $(5,5)$.
Basically I need to start from a straight line $y=x$ and "bend it" towards $(5,0)$ symmetrically by varying $a$, as if someone is pulling on it.
An equation for a hyperbola like yours but centred at the origin is $y=-\frac{1}{ax}$.
If we shift its centre to $(c,-b)$ we get the equation $y=-b-\frac{1}{a(x-c)}$.
We want this to go through the points $(0,0)$ and $(5,5)$. Substituting those coordinates results in the two equations $$abc=1\\a(c-5)(b+5)=1$$
From $abc=a(c-5)(b+5)$ we find that $c=b+5$, which makes sense since this puts the centre $(c,-b)$ on the line of symmetry $y=5-x$.
Substituting this value for $c$ we get $ab(b+5)=1$, and solving this quadratic for $b$ we find that $b = -2.5 + \sqrt{2.5^2 + 1/a}$.
So the equation has become $y=-b-\frac{1}{a(x-b-5)}$ where $b = -2.5 + \sqrt{2.5^2 + 1/a}$. Not very nice, but I think it does what you want it to.