Exp. (b)Let the equation of hyperbola be a2x2​−b2y2​=1.∴ ae =2⇒a2e2=4⇒a2+b2=4⇒b2=4−a2∴a2x2​−4−a2y2​=1Since, (2​,3​) lie on hyperbola.∴a22​−4−a23​=1⇒8−2a2−3a2=a2(4−a2)⇒8−5a2=4a2−a4⇒a4−9a2+8=0⇒(a4−8)(a4−1)=0⇒a4=8,a4=1∴a=1Now, equation of hyperbola is 1x2​−3y2​=1.∴ Equation of tangent at (2​,3​) is given by2​x−33​y​=1⇒2​x−3​y​=1which passes through the point (22​,33​).