If a directrix of a hyperbola centered at the origin and passing through the point 4 , - 2 3 is 5 x = 4 5 and its eccentricity is e , then:

Step-by-step solution

Let equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 ∴ It passes through 4 ,   - 2 3     ⇒ 16 a 2 - 12 b 2 = 1 ⇒ 16 - 12 × a 2 b 2 = a 2 . . . 1 Equation of directrix is x = a e = 4 5 , given in question. ⇒ a 2 = 16 5 e 2 . . . 2 And we know that b 2 = a 2 e 2 - 1 ⇒ b 2 a 2 = e 2 - 1 . . . 3 ∴ From 1 , 2   &   3 16 - 12 e 2 - 1 = 16 5 e 2 ⇒ 16 e 2 - 16 - 12 = 16 e 2 5 e 2 - 1 ⇒   4 e 4 - 24 e 2 + 35 = 0