Answer
1. the images of the lines $u=c$ is
$y = f\left( x \right) = {c^2}\left( {1 - \frac{1}{{{x^2}}}} \right)$
2. the images of the lines $v=c$ is
$y = f\left( x \right) = {c^2}\left( {{x^2} - 1} \right)$
Work Step by Step
We have the mapping $G\left( {u,v} \right) = \left( {\frac{u}{v},{u^2} - {v^2}} \right)$.
1. the images of the lines $u=c$ is
$G\left( {c,v} \right) = \left( {\frac{c}{v},{c^2} - {v^2}} \right)$
In the $xy$-plane, we get
(1) ${\ \ \ \ \ }$ $x = \frac{c}{v}$, ${\ \ \ \ }$ $y = {c^2} - {v^2}$
Substituting $v = \frac{c}{x}$ in the second part of (1), we get
$y = {c^2} - \frac{{{c^2}}}{{{x^2}}} = {c^2}\left( {1 - \frac{1}{{{x^2}}}} \right)$
Thus, $y = f\left( x \right) = {c^2}\left( {1 - \frac{1}{{{x^2}}}} \right)$.
2. the images of the lines $v=c$ is
$G\left( {u,c} \right) = \left( {\frac{u}{c},{u^2} - {c^2}} \right)$, ${\ \ \ }$ for $c \ne 0$
In the $xy$-plane, we get
(2) ${\ \ \ \ \ }$ $x = \frac{u}{c}$, ${\ \ \ \ }$ $y = {u^2} - {c^2}$
Substituting $u = cx$ in the second part of (2), we get
$y = {c^2}{x^2} - {c^2} = {c^2}\left( {{x^2} - 1} \right)$
Thus, $y = f\left( x \right) = {c^2}\left( {{x^2} - 1} \right)$.
