Hyperbola

Asymptotes are
‌3x−4y−1=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
‌4x−3y−6=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
By Eqs. (i) and (ii), we get
x=3,y=2
Now, the direction vectors of asymptotes for 3x−4y=1⇒d1=⟨4,3⟩ and for 4x−3y=6⇒d2=⟨3,4⟩ Thus, transverse and conjugate axis directions.
‌‌

L1

√a12+b12

=±‌

L2

√a22+b22

⇒‌

3x−4y−1

√32+(−4)2

=±‌

4x−3y−6

√42+(−3)2

⇒3x−4y−1=±4x−3y−6
‌ So, ‌‌3x−4y−1=4x−3y−6
‌ or ‌‌3x−4y−1=−4x+3y+6
⇒x+y−5=0
‌ or ‌‌7x−7y−7=0
⇒x+y−5=0‌ or ‌x−y−1=
Hence, transverse and conjugate axis are
x+y=5,x−y=1