Question

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

The correct option is D $4x+3y+17=0$

We know that aysmptotes of rectangular hyperbola are mutually perpendicular, thus other asymptote should be $4x+3y+\lambda =0$.

Interection point of asymptotes is also the center of the hyperbola.

Hence intersection point of $4x+3y+\lambda =0$ and $3x-4y-6=0$ should lies on the line $x-y-1=0$

using it we obatain $\lambda =17$

Hence equation is $4x+3y+17=0$