A hyperbola passes through the point P 2 , 3 and has foci at ± 2 , 0 . Then the tangent to this hyperbola at P also passes through the point

Step-by-step solution

From the standard equation of hyperbola ⇒ ± a e = ±   2 As we know, b 2 = a 2 e 2 - 1 . ⇒ b 2 = 4 - a 2 ∴   Equation of hyperbola is x 2 a 2 - y 2 4 - a 2 = 1 ∵   It passes through 2 , 3   ⇒ 2 a 2 - 3 4 - a 2 = 1 ⇒ a 2 = t ⇒ 2 4 - t - 3 t - t 4 - t = 0 ⇒ 8 - 2 t - 3 t - 4 t + t 2 = 0 ⇒ t 2 - 8 t - t + 8 = 0 ⇒   t =   8 ,   1 So, a 2 = 8 ,   1 . ⇒ a = 2 2 ,   1 But for a = 2 2 ,   b becomes i