Let the foci and length of the latus rectum of an ellipse x 2 a 2 + y 2 b 2 = 1 , a > b be ± 5 , 0 and 50 , respectively. Then, the square of the eccentricity of the hyperbola x 2 b 2 − y 2 a 2 b 2 = 1 equals

Step-by-step solution

Given, Ellipse x 2 a 2 + y 2 b 2 = 1 And its focii ≡ ± 5 , 0 and its latusrectum 2 b 2 a = 50 So, a e = 5 and b 2 = 5 2 a 2 Now, using the formula eccentricity we get, b 2 = a 2 1 − e 2 = 5 2 a 2 ⇒ a 1 − e 2 = 5 2 2 as a ≠ 0 ⇒ 5 e 1 − e 2 = 5 2 2 ⇒ 2 − 2 e 2 = e ⇒ 2 e 2 + e − 2 = 0 ⇒ 2 e 2 + 2 e − e − 2 = 0 ⇒ 2 e e + 2 − 1 1 + 2 = 0 ⇒ e + 2 2 e − 1 = 0 ∴ e ≠ − 2 ; e = 1 2 ⇒ a = 5 2 and b = 5 Now, finding eccentricity of hyperbola, x 2 b 2 − y 2 a 2 b 2 = 1 We know that formula of eccentricity of hyperbola is given