Let e 1 be the eccentricity of the hyperbola x 2 16 - y 2 9 = 1 and e 2 be the eccentricity of the ellipse x 2 a 2 + y 2 b 2 = 1 , a > b , which passes through the foci of the hyperbola. If e 1 e 2 = 1 , then the length of the chord of the ellipse parallel to the x -axis and passing through ( 0 , 2 ) is :

Step-by-step solution

The given equation of hyperbola is, H : x 2 16 - y 2 9 = 1 ⇒ e 1 = 1 + 3 2 4 2 = 16 + 9 16 ⇒ e 1 = 5 4 Foci of this hyperbola is given by, F ± ae , 0 . ⇒ F ± 5 , 0 It is given that, e 1 e 2 = 1 ⇒ e 2 = 4 5 Also, ellipse is passing through ( ± 5 , 0 ) ⇒ 5 2 a 2 + 0 b 2 = 1 ⇒ 25 a 2 = 1 ⇒ a 2 = 25 ⇒ a = 5 Now, e 2 = 1 - b 2 a 2 ⇒ 4 5 = 1 - b 2 25 ⇒ 16 25 = 1 - b 2 25 ⇒ b 2 25 = 9 25 ⇒ b 2 = 9 ∴ a = 5 and b = 3 E : x 2 25 + y 2 9 = 1 Putting, y = 2 ⇒ x 2 25 + 4 9 = 1 ⇒ x 2 25 = 5 9 ⇒ x 2 = 125 9 ⇒ x = ± 5 5 3 So, end