Given hyperbola is 5x2−y2=5 or It can be rewritten as
x2
1
−
y2
5
=1 Here, a2=1,b2=5 ∴ Equation of tangent is y=mx±√a2m2−b2 ⇒ y=mx±√1m2−5 ......(i) But point (2,8) lies on it. ∴8=2m±√m2−5 ⇒(8−2m)=±√m2−5 On squaring both sides, we get 64+4m2−32m=m2−5 ⇒ 3m2−32m+69=0 ⇒ (3m−23)(m−3)=0 ⇒ m=3,
23
3
On putting m=3 in Eq. (i), we get y&=3x±√32−5=3x±2 ⇒y=3x+2and y=3x−2