If e_1 and e_2 are the eccentricities of a hyperbola 3x^2-3y^2=25 and its conjugate, then

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Accepted Answer

Correct answer: D. (d) : x^225/3- y^225/3=1 Comparing with x^2a^2- y^2b^2=1 a^2= 253 ,b^2= 253 b^2=a^2(e_1^2-1) ⇒ 253= 253(e_1^2-1) ⇒e_1^2=2. Conjugate of 3x^2-3y^2=25 is 3x^2-3y^2=-25 or x^225/3- y^225/3=-1 ∴a^2= 253 ,b^2= 253 a^2=b^2(e_2^2-1) ⇒e_2^2-1=1 ⇒e_2^2-1=1 ⇒e_2^2=2 ∴e_1^2+e_2^2=2+2=4.

$If e_{1} and e_{2} are the eccentricities of a hyperbola 3 x^{2} - 3 y^{2} = 25 and its conjugate, then$

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