The correct options are
A equation of tangent is x−y=1
B point of contact will be (−5,−4)
C point of contact will be (5,4)
D equation of tangent is x−y+1=0
The tangent to the hyperbola x25−y24=1 is
y=mx±√5m2−4
The tangent is parallel to the line x−y=2 therefore, they have same slope.
∴m=1
Hence, y=x±√5−4
y=x±1
⇒y=x+1 or y=x−1
x−y+1=0 and x−y−1=0
Hence point of contact will be
(−a2mc,−b2c)≡(5,4),(−5,−4)