Question

What is the eccentricity of the graph of (x-2)^2/121+(y+6)^2/64=1 ?

–1.24

–0.68

0.68

1.24

Answer 1

\[\frac{(x-2)^2}{11^2}+\frac{(y+6)^2}{8^2}=1\]

a=11, b=8

\[a^2+b^2=c^2\]

\[e=\frac{c}{a}\]

You gt it :)