Question

The eccentricity of the hyperbola conjugate ot

x² – 3y² = 2x + 8 is

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. 2
• D. None of these

Answer 1

Answer: (C)

2

Answer 2

Given, equation of hyperbola is x² – 3y² = 2x + 8

⇒ x² – 2x – 3y² = 8

⇒ (x – 1)² – 3y² = 9

⇒ $\frac{(x - 1)^{2}}{9} - \frac{y^{2}}{3}$ = 1

Conjugate of this hyperbola is = $\frac{(x - 1)^{2}}{9} - \frac{y^{2}}{3}$ = 1

And its eccentricity (5) = $\sqrt{\left( \frac{a^{2} + b^{2}}{b^{2}} \right)}$

Here, a² = 9, b² = 3; ∴ e = $\sqrt{\frac{9 + 3}{3}}$ = 2.