KCET2008MathematicsHyperbola
If e₁ and e₂ are the eccentricities of a hyperbola 3 x²-3 y²=25 and its conjugate, then
Options
- Ae₁²+e₂²=2
- Be ₁²+ e ₂²=4
- Ce₁+e₂=4
- De₁+e₂= 2
Correct answer
B. e ₁²+ e ₂²=4
Step-by-step solution
Given equation can be rewritten as x²-y²= 25 3 Here, a ²=1, ~b ²=1 e ₁= 1+ b ² a ² = 1+1 = 2 The equation of conjugate hyperbola is gathered - x ²+ y ²= 25 3 e ₂= 1+ a ² ~b ² = 1+1 = 2 e ₁²+ e ₂²=( 2 )²+( 2 )²=4 gathered