AP EAMCET20225 Jul 2022Morning ShiftMathematicsHyperbolaActual
If e₁ and e₂ are the eccentricities of the hyperbola 16 x^2-9 y^2=1 and its conjugate respectively. Then, 3 e₁=
Options
- A5 e₂
- B4 e₂
- C2 e₂
- De₂
Correct answer
B. 4 e₂
Step-by-step solution
Given, hyperbola is 16 x^2-9 y^2=1 x^2 1 16 - y^2 1 9 =1 x^2 ( 1 4 )^2 - y^2 ( 1 3 )^2 =1 We know that Eccentricity of a hyperbola, e₁^2=1+ b^2 a^2 aligned & e₁^2=1+ 1 16 9 1 & e₁^2= 25 9 aligned e₁= 5 3 ... (i) Now, eccentricity of its conjugate e₂^2=1+ a^2 b^2 array ll & e₂^2=1+ 1 9 16 1 & e₂^2= 25 16 array e₂= 5 4 ...(ii) From Eqs. (i) and (ii), 3 e₁=3 5 3 = 5 4 4 3 e₁=4 e₂ [ e₂= 5 4 ]