Can we deduce anything given the equation of a curve and the fact that it has symmetry with $y=x$?

Question

Question:

The line $y=x$ is a line of symmetry to the curve with equation $$y=\frac{px+q}{rx+s}$$ where $p,q,r,s \neq 0$. Which of the following must be true?

$p+s=0$

$p+q=0$

$q+r=0$

$r+s=0$

$p+r=0$

or none of the above?

Thoughts: nothing at the moment I don't know how to even think about tackling this question.

Any ideas?

Answer 1

$(0,\frac qs)$ is on the curve. So, $(\frac qs,0)$ is also on the curve.

Hence, we have $$0=\frac{p\cdot \frac qs+q}{r\cdot\frac qs+s}\Rightarrow \frac{pq}{s}+q=0\Rightarrow pq+qs=0\Rightarrow q(p+s)=0\Rightarrow \color{red}{p+s=0}.$$

Answer 2

If a curve is symmetrical in $y=x$ then it must be its own inverse. Given that $f^{-1}(f(x))=x$ we can now look at: $$f(f(x))=x$$ $$\frac{p\frac{px+q}{rx+s}+q}{r\frac{px+q}{rx+s}+s}=x$$ $$\frac{px+q}{rx+s}+q=(r\frac{px+q}{rx+s}+s)x$$ $$px+q+q(rx+s)=rpx+rq+s(rx+s)+sx(rx+s)$$ $$r(p+s)x^2+(s^2-p^2)x-q(p+s)=0$$ $$(p+s)\left(rx^2+(s-p)x-q\right)=0$$ As this has to be true for all possible $x$ we must have that $p+s=0$.