Question

What is the eccentricity of the conic with equation 3y²-x²=108 ?

• A. 3
• B. 1.5
• C. 10/3
• D. 2

Answer 1

Answer:

D. 2

$3y^2-x^2=108$

$$\frac{3y^2}{108}-\frac{x^2}{108}=\frac{108}{108}$

$\frac{y^2}{36}-\frac{x^2}{108}=1$

comparing with the equation of hyperbola

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$a^2=36;b^2=108$

$eccentricity=e=\sqrt{1+\frac{b^2}{a^2}}$

$=\sqrt{1+\frac{108}{36}}$

$=\sqrt{1+3}=\sqrt{4}=2$