Question

Let e₁ and e₂ be the eccentricities of the ellipse, `x^2/25 + y^2/b^2` = 1 (b < 5) and the hyperbola, `x^2/16 - y^2/b^2` = 1 respectively satisfying e₁e₂ = 1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α, β) is equal to ______.

पर्याय

• (8, 12)

• `(20/3, 12)`

• `(24/5, 10)`

• (8, 10)

Answer 1

Let e₁ and e₂ be the eccentricities of the ellipse, `x^2/25 + y^2/b^2` = 1 (b < 5) and the hyperbola, `x^2/16 - y^2/b^2` = 1 respectively satisfying e₁e₂ = 1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α, β) is equal to (8, 10).

Explanation:

Equation of ellipse is `x^2/25 + y^2/b^2` = 1

Then, e₁ = `sqrt(1 - b^2/25)`

The equation of hyperbola, `x^2/16 - y^2/b^2` = 1

Then, e₂ = `sqrt(1 + b^2/16)`, e₁e₂ = 1

⇒ (e₁e₂)² = 1

⇒ `(1 - b^2/25)(1 + b^2/16)` = 1

⇒ `1 + b^2/16 - b^2/25 - b^4/(25 xx 16)` = 1

⇒ `9/(16.25)b^2 - b^4/(25.16)` = 0

⇒ b² = 9

∴ e₁ = `sqrt(1 - 9/25)` = `4/5`

And e₂ = `sqrt(1 + 9/16)` = `5/4`

Distance between focii of ellipse = α = 2ae₁ = 2(5)(e₁) = 8

Distance between focii of hyperbola = β = 2ae₂ = 2(4)(e₂) = 10

∴ (α, β) = (8, 10)