Find the equation of the asymptotes to the hyperbola $8x^2 + 10xy - 3y^2 - 2x + 4y -2 = 0$
Equation of given hyperbola is $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y - 2 = 0$ $\;\;\; \cdots \; (1)$
The combined equation of the asymptotes differs from the hyperbola by a constant only.
$\therefore$ $\;$ Let the combined equation of the asymptotes be: $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y + k = 0$
Now,
$\begin{aligned}
8x^2 + 10xy - 3y^2 & = 8x^2 + 12xy - 2xy - 3y^2 \\\\
& = 4x \left(2x + 3y\right) - y \left(2x + 3y\right) \\\\
& = \left(4x - y\right) \left(2x + 3y\right)
\end{aligned}$
$\therefore$ $\;$ The separate equations of the asymptotes are
$4x - y + \ell = 0$ $\;\;\; \cdots \; (2a)$ and
$2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ $\left(4x - y + \ell\right) \left(2x + 3y + m\right) = 8x^2 + 10xy - 3y^2 - 2x + 4y + k$ $\;\;\; \cdots \; (3)$
Equating the coefficient of the $x$ terms we have
$4 m + 2 \ell = -2$
i.e. $\;$ $2 m + \ell = -1$ $\;\;\; \cdots \; (4a)$
Equating the coefficient of the $y$ terms we have
$- m + 3 \ell = 4$ $\;\;\; \cdots \; (4b)$
Equating the constant term we have
$\ell m = k$ $\;\;\; \cdots \; (4c)$
Solving equations $(4a)$ and $(4b)$ simultaneously we get
$\ell = 1$ $\;$ and $m = -1$
$\therefore$ $\;$ We have from equation $(4c)$, $\;$ $k = -1$
$\therefore$ $\;$ The separate equations of the asymptotes are
$4x - y + 1 = 0$ $\;$ and $\;$ $2x + 3y - 1 = 0$
The combined equation of the asymptotes is
$8x^2 + 10xy - 3y^2 - 2x + 4y - 1 = 0$