What is the eccentricity of the conic with equation 3y²-x²=108A3B1.5C10/3D2Answer: D. 2 Read Explanation: 3y2108−x2108=1\frac{3y^2}{108}-\frac{x^2}{108}=11083y2−108x2=1y26−x2(63)2=1\frac{y^2}{6}-\frac{x^2}{(6\sqrt3)^2}=16y2−(63)2x2=1e=a2+b2ae=\frac{\sqrt{a^2+b^2}}{a}e=aa2+b2a=6;b=63a=6;b=6\sqrt3a=6;b=63e=36+1086=126=2e=\frac{\sqrt{36+108}}{6}=\frac{12}{6}=2e=636+108=612=2 Read more in App