Find the asymptotes of the following hyperbolas and also the equations to their conjugate hyperbolas.
$19 x^{2} + 24 x y + y^{2} - 22 x - 6 y = 0 .$

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Solution

Given equation of the hyperbola: $19 x^{2} + 24 x y + y^{2} - 22 x - 6 y = 0$
The equation of a hyperbola and the combined equation of the asymptotes differ only in the constant term.
Therefore the combined equation of the asymptotes is of the form: $19 x^{2} + 24 x y + y^{2} - 22 x - 6 y + K = 0... (i)$
Since $e q n (i)$ forms a pair of straight lines, the determinant
$Δ =$ $∣ ∣ ∣ ∣ \begin{matrix} a & h & g h & b & f g & f & c \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow Δ =$ $∣ ∣ ∣ ∣ \begin{matrix} 19 & 12 & - 11 12 & 1 & - 3 - 11 & - 3 & K \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 19 (K - 9) - 12 (12 K - 33) - 11 (- 36 + 11) = 0$
$\Rightarrow 19 K - 171 - 144 K + 396 + 396 - 121 = 0$
$\Rightarrow 125 K = 500$
$\Rightarrow K = 4$
The combined equation of the asymptotes is $19 x^{2} + 24 x y + y^{2} - 22 x - 6 y + 4 = 0$ .
Now using the fact that: equation of hyperbola + equation of conjugate hyperbola = 2x(equation of asymptotes)
$\Rightarrow$ equation of conjugate hyperbola = 2x(equation of asymptotes) - equation of hyperbola
$\Rightarrow$ equation of conjugate hyperbola = $2 (19 x^{2} + 24 x y + y^{2} - 22 x - 6 y + 4)) - (19 x^{2} + 24 x y + y^{2} - 22 x - 6 y)$
$\Rightarrow$ equation of conjugate hyperbola = $(19 x^{2} + 24 x y + y^{2} - 22 x - 6 y + 8) = 0$

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