Question

Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x² – 3y² = 3

Answer 1

Given, equation of hyperbola is x² – 3y² = 3.

∴ `x^2/3 - y^2/1` = 1

Equation of the hyperbola conjugate to the above hyperbola is `y^2/1 - x^2/3` = 1.

Comparing this equation with `y^2/"b"^2 - x^2/"a"^2` = 1, we get,

b² = 1 and a² = 3

Now, a² = b²(e² – 1)

∴ 3 = 1(e² –  1)

∴ 3 = e² –  1

∴ e² = 4

∴ e = 2    ...[∵ e > 1]