Given equation of hyperbola is k6x2−6y2=1
Equation of directrix will be x=±ea,
Where e=1+a2b2=1+k66=1+k
i.e. k≥−1
⇒x=±k(1+k)6 is the equation of directrix
i.e. ±k(1+k)6=1
∴(k1+k)2=6⇒k2+k−6=0
⇒k=2 as k=−3 is rejected
So equation of hyperbola will be H≡3x2−6y2=1
∴(5,−2) satisfy the given hyperbola