Q. 2

# If e1

Answer :

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

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