COMEDK2019MathematicsHyperbola
If e₁ and e₂ are the eccentricities of a hyperbola 3 x²-3 y²=25 and its conjugate, then
Options
- Ae₁²+e₂²=2
- Be₁²+e₂²=4
- Ce₁+e₂=4
- De₁+e₂= 2
Correct answer
B. e₁²+e₂²=4
Step-by-step solution
We have, equation of hyperbola gathered 3 x²-2 j²=25 x² ( 5 3 )² - y² ( 5 2 )² =1 gathered Equation of conjugate hyperbola is y² ( 5 2 )² - x² ( 5 3 )² =1 Now, eccentricity of hyperbola, aligned e₁ &= 1+ ( 5 2 )² ( 5 3 )² &= 1+ 3 2 = 5 2 aligned and eccentricity of conjugate hyperbola, e₂= 1+ ( 5 3 )² ( 5 2 )² = 1+ 2 3 = 5 3 Now, e₁²+e₂²= 5 2 + 5 3 = 25 6 4