If e1 and e2 are respectively the eccentricities of the ellipse
and the hyperbola
, then write the value of 2e12 + e22.
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
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.

