Solution:

The given eq. can be written as
$\begin{array}{l}
\frac{x}{a}-\frac{y}{b}=0\dots \dots(i) \\
a x+b y-u^{2}+b^{2}\dots \dots(ii)
\end{array}$
From equation(i),
$\mathrm{y}=\frac{b x}{a}$
Substituting $\mathrm{y}=\frac{b x}{a}$ in equation(ii), we obtain
$\begin{array}{l}
\mathrm{ax}+\frac{b \times b x}{a}=\mathrm{a}^{2}+\mathrm{b}^{2} \\
\Rightarrow \mathrm{x}=\frac{\left(a^{2}+b^{2}\right) \times a}{a^{2}+b^{2}}=\mathrm{a}
\end{array}$
Now, substitute $\mathrm{x}=\mathrm{a}$ in equation(ii) to obtain
$\begin{array}{l}
\mathrm{a}^{2}+\mathrm{by}=\mathrm{a}^{2}+\mathrm{b}^{2} \\
\Rightarrow \mathrm{by}=\mathrm{b}^{2} \\
\Rightarrow \mathrm{y}=\mathrm{b}
\end{array}$
As a result, $x=a$ and $y=b$.