Find the equations of the asymptotes for the rectangular hyperbola $6x^2 + 5xy - 6y^2 + 12x + 5y + 3 = 0$
Equation of the rectangular hyperbola is
$6x^2 + 5xy - 6y^2 + 12 x + 5y + 3 = 0$ $\;\;\; \cdots \; (1)$
The combined equation of the asymptotes differs from the hyperbola by a constant.
$\therefore$ $\;$ The combined equation of the asymptotes is
$6x^2 + 5xy - 6y^2 + 12x + 5y + k = 0$
Consider
$\begin{aligned}
6x^2 + 5xy - 6y^2 & = 6x^2 + 9xy - 4xy - 6y^2 \\\\
& = 3x \left(2x + 3y\right) - 2y \left(2x + 3y\right) \\\\
& = \left(3x - 2y\right) \left(2x + 3y\right)
\end{aligned}$
$\therefore$ $\;$ The separate equations of asymptotes are
$3x - 2y + \ell = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ $\left(3x - 2y + \ell\right) \left(2x + 3y + m\right) = 6x^2 + 5xy -6y^2 + 12x + 5y + k$
Equating the coefficients of $x$ terms we get: $\;$ $3m + 2 \ell = 12$ $\;\;\; \cdots \; (3a)$
Equating the coefficients of $y$ terms we get: $\;$ $-2m + 3 \ell = 5$ $\;\;\; \cdots \; (3b)$
Equating the constant terms we get: $\;$ $\ell \times m = k$ $\;\;\; \cdots \; (3c)$
Solving equations $(3a)$ and $(3b)$ simultaneously we get,
$\ell = 3$ $\;$ and $\;$ $m = 2$
Substituting the values of $\ell$ and $m$ in equations $(2a)$ and $(2b)$ respectively, we get the separate equations of asymptotes as
$3x - 2y + 3 = 0$ $\;$ and $\;$ $2x + 3y + 2 = 0$