Question

If x2 - y2 sec’® = 10 be a hyperbola and x2 sec*6 + y2 = 5 be an ellipse such that the eccentricity of hyperbola = root 5 eccentricity of ellipse then find the length of latus rectum of ellipse :

Answer 1

Given : equation of hyperbola, x² - y²secθ = 10 and equation of ellipse, x²sec²θ + y² = 5 and also eccentricity of hyperbola = √5 eccentricity of ellipse.

To find : The length of Latus rectum of ellipse.

solution : equation of hyperbola, x² - y³sec²θ = 10

⇒x²/10 - y²/10cos²θ = 1 on comparing with x²/a² - y²/b² = 1

a = √10 , b = √10 cosθ

now eccentricity of hyperbola, E = √(1 + b²/a²)

= √(1 + cos²θ)

equation of ellipse , x²sec²θ + y² = 5

⇒x²/5cos²θ + y²/5 = 1 on comparing with x²/a² + y²/b² = 1

we get, a = √5 cosθ , b = √5

eccentricity of ellipse, e = √(1 - a²/b²)

= √(1 - cos²θ) = sinθ

now E = √5e [ given ]

⇒√(1 + cos²θ) = √5 sinθ

⇒1 + cos²θ = 5sin²θ

⇒1 + cos²θ = 5 - 5cos²θ

⇒cos²θ = 2/3

now length of Latus rectum = 2b²/a

= 2(5cos²θ)/√5

= (10 × 2/3)/√5

= 4√5/3

Therefore the length of Latus rectum is 4√5/3