Question

The equation of a line passing through the centre of a rectangular hyperbola is $x-y=1$. If one of the asymptotes is $3x-4y-6=0$, the equation of other asymptote is

• A. $4x-3y+17=0$
• B. $-4x-3y+17=0$
• C. $-4x+3y+1=0$
• D. $4x+3y+17=0$

Answer 1

Answer: (D)

Since the asymptotes of rectangular hyperbola are mutually perpendicular, the other asymptote should be $4x+3y+\lambda =0$.
Also, intersection point of asymptotes is also the centre of the hyperbola. Thus, intersection point of $4x+3y+\lambda =0$ and $3x-4y-6=0$
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 $4x+3y+17=0$