If a hyperbola passing through the origin has 3x−4y−1=0 and 4x−3y−6=0 as its asymptotes, then the equations of its transverse and conjugate axes, are

# If a hyperbola passing through the origin has $3x-4y-1=0$ and $4x-3y-6=0$ as its asymptotes, then the equations of its transverse and conjugate axes, are

1. A

$x-y-5=0$ and $x+y+1=0$

2. B

$x-y=0$ and $x+y+5=0$

3. C

$x+y-5=0$ and $x-y-1=0$

4. D

$x+y-1=0$ and $x-y-5=0$

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### Solution:

The transverse axis is the bisector of the angle between asymptotes containing the origin and the conjugate axis is the other bisector. So, their equations are given by

$\frac{-3x+4y+1}{\sqrt{9+16}}=\frac{-4x+3y+6}{\sqrt{16+9}}$

and,

$\frac{-3x+4y+1}{\sqrt{9+16}}=-\frac{-4x+3y+6}{\sqrt{16+9}}$

$⇒$ $x+y-5=0$ and

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