Given: e₁ and e₂ are respectively the eccentricities of the ellipse and the hyperbola

To find: value of 2e₁² + e₂²

Eccentricity(e) of hyperbola is given by,

Here a = 3 and b = 2

Therefore,

For ellipse:

Eccentricity(e) of ellipse is given by,

Therefore,

Substituting values from (1) and (2) in 2e₁² + e₂²

2e₁² + e₂²

= 3

Hence, value of 2e₁² + e₂² is 3