If the foci of a hyperbola are same as that of the ellipse x 2 9 + y 2 25 = 1 and the eccentricity of the hyperbola is 15 8 times the eccentricity of the ellipse, then the smaller focal distance of the point 2 , 14 3 2 5 on the hyperbola, is equal to

Step-by-step solution

Given equation of ellipse is, x 2 9 + y 2 25 = 1 ⇒ a = 3 , b = 5 We know that, e = 1 - a 2 b 2 ⇒ e = 1 - 9 25 = 4 5 Now, foci = 0 , ± be = 0 , ± 4 Since, eccentricity of hyperbola is given as 15 8 same to that of ellipse, ∴ e H = 4 5 × 15 8 = 3 2 Let equation of the hyperbola be x 2 A 2 - y 2 B 2 = - 1 . ⇒ B . e H = 4 ⇒ B = 8 3 ⇒ A 2 = B 2 e H 2 - 1 = 64 9 9 4 - 1 ⇒ A 2 = 80 9 ⇒ x 2 80 9 - y 2 64 9 = - 1 Directrix: y = ± B e H = ± 16 9 ⇒ P S = e · P M ⇒ P S = 3 2 14 3 · 2 5 - 16 9 ⇒ P S = 7 2 5 - 8 3