Question

Let the equation of hyperbola be x²-y²/3=1,then eccentricity of its conjugate hyperbola is​

Answer 1

Given info : The equation of hyperbola is x² - y²/3 = 1

To find : the eccentricity of its conjugate hyperbola is ...

solution : if standard equation of hyperbola is x²/a² - y²/b² = 1

then eccentricity , e = √(1 + b²/a²)

here equation of hyperbola, x²/1 - y²/3 = 1

so, a² = 1 and b² = 3

so, eccentricity, e = √(1 + 3/1) = 2

let e' is eccentricity of its conjugate hyperbola.

using formula, 1/e² + 1/e'² = 1

⇒1/2² + 1/e'² = 1

⇒1/e'² = 1 - 1/2² = 1 - 1/4 = 3/4

⇒e'² = 4/3

⇒e' = 2/√3

Therefore eccentricity of its conjugate hyperbola is 2/√3.

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Answer 2

Answer:

Let the equation of hyperbola be x²-y²/3=1,then eccentricity of its conjugate hyperbola is

Step-by-step explanation:

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