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The equation of a line passing through the centre of a rectangular hyperbola is $x -y -1 = 0$. If one of the asymptotes is $3x -4y -6 = 0$, the equation of other asymptote is
$4x -3y + 17 = 0$
$-4x -3y + 17 = 0$
$-4x + 3y + 1 = 0$
$4x + 3y + 17 = 0$
Solution
We know that asymptotes of rectangular hyperbola is mutually perpendicular, thus other asymptote should be $4 \mathrm{x}+3 \mathrm{y}+\lambda=0$
Also, intersection point of asymptotes is also the centre of the hyperbola.
Hence, intersection point of $4 \mathrm{x}+3 \mathrm{y}+\lambda=0$ and
$3 x-4 y-6=0$ is $\left(\frac{18-4 \lambda}{25}, \frac{-12 \lambda-96}{100}\right)$ and it
should lie on the line $x-y-1=0$
$\therefore \frac{18-4 \lambda}{25}-\frac{-12 \lambda-96}{100}-1=0$
$\Rightarrow \lambda=17$
Hence, equation of other asymptote is
$4 x+3 y+17=0$
