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Q.
The equation of a line passing through the centre of a rectangular hyperbola is $x-y=1$. If one of the asymptotes is $3 x-4 y-6=0$, the equation of other asymptote is
Conic Sections
Solution:
Since the asymptotes of rectangular hyperbola are mutually perpendicular, the other asymptote should be $4 x+3 y+\lambda= 0$.
Also, intersection point of asymptotes is also the centre of the hyperbola. Thus, intersection point of $4 x+3 y+\lambda=0$ and $3 x-4 y-6=0$
i.e., $\left(\frac{18-4 \lambda}{25}, \frac{-12 \lambda-96}{100}\right)$ 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, the equation of other asymptote is $4 x+3 y+17=0$