If the eccentricity of the hyperbola x 2 - y 2 sec 2 ⁡ α = 5 is 3 times the eccentricity of the ellipse x 2 sec 2 ⁡ α + y 2 = 25 , then tan 2 ⁡ α is equal to

Step-by-step solution

Let eccentricity of the hyperbola x 2 5 - y 2 5 cos 2 ⁡ α = 1 is e 1 2 = 1 + b 2 a 2 = 1 + 5 cos 2 ⁡ α 5 = 1 + cos 2 ⁡ α ; Similarly, eccentricity of the ellipse x 2 25 cos 2 ⁡ α + y 2 25 = 1 is e 2 2 = 1 - 25 cos 2 ⁡ α 25 = sin 2 ⁡ α ; Putting e 1 = 3 e 2 ⇒ e 1 2 = 3 e 2 2 ⇒ 1 + cos 2 ⁡ α = 3 sin 2 ⁡ α ⇒ 2 = 4 sin 2 ⁡ α ⇒ sin ⁡ α = 1 2 ⇒ tan 2 ⁡ α = 1