What values make the graph symmetrical about the line $y=x$

Question

Can anyone help me out with this question please?

given the equation:

$y=\frac{px+q}{rx+s}$

Find the relationship between $p,q,r$ and $s$ that is implied if the graph is to by symmetrical about the line $y=x$.

I'm getting nowhere with this.

I can show that the curve is a hyperbola.

I do the division to produce an alternative version of the equation:

$y=\frac{p}{r}+\frac{qr-ps}{r(rx+s)}=\frac{1}{r}(p + \frac{qr-ps}{rx+s})$

But to no avail,

Thanks for any help, Mitch.

Answer 1

For symmetry in $y=x$ you need the vertical and horizontal asymptotes to be the same value for $x$ and for $y$

Therefore you need $\frac pr=-\frac sr$

So assuming $r\neq 0$ you require $p=-s$

Answer 2

Inverse of a function is reflection of that function in $y=x$.

So, your function should be its own inverse. $f(x) = f^{-1}(x)$

Answer 3

Mirroring at $y=x$ means exchanging $x$ and $y$. Therefore if a function (or rather its graph) is mirror-symmetric along $y=x$, it means that $f(x)=y \iff f(y)=x$. This in particular implies that $f(f(x)) = f(y) = x$, that is, applying the function twice gives the identity.

Note that this is equivalent to the statement that $f$ is its own inverse (since the inverse of $f$ is by definition the function that, if applied either before or after $f$, gives the identity function).