Find the eccentricity of the hyperbola, which is conjugate to the hype

1.	Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x2 – 3y2 = 3
Answer» Given, one of the foci of the hyperbola is x² – 3y² = 3 \(\frac{x^2}{3} - \frac {y^2}{1} = 1\) Equation of the hyperbola conjugate to the above hyperbola is \(\frac {y^2}{1}-\frac{x^2}{3} = 1\) Comparing this equation with \(\frac{x^2}{b^2} - \frac {y^2}{a^2} = 1\) we get b² = 1 and a² = 3 Now, a² = b² (e² – 1) ⇒ 3 = 1(e² – 1) ⇒ 3 = e – 1 ⇒ e² = 4 ⇒ e = 2 …..[∵ e > 1]

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

Answer»

Given, one of the foci of the hyperbola is x² – 3y² = 3

\(\frac{x^2}{3} - \frac {y^2}{1} = 1\)

Equation of the hyperbola conjugate to the above hyperbola is \(\frac {y^2}{1}-\frac{x^2}{3} = 1\)

Comparing this equation with \(\frac{x^2}{b^2} - \frac {y^2}{a^2} = 1\)

we get b² = 1 and a² = 3

Now, a² = b² (e² – 1)

⇒ 3 = 1(e² – 1)

⇒ 3 = e – 1

⇒ e² = 4

⇒ e = 2 …..[∵ e > 1]