Answer
$$\eqalign{
& {\text{Vertices: }}\left( {0, \pm 2\sqrt 3 } \right);\,\,{\text{Foci:}}\,\,\left( {0, \pm \sqrt 5 } \right) \cr
& {\text{major axis has length }}4\sqrt 3 \cr
& {\text{minor axis has length }}2\sqrt 5 \cr} $$
Work Step by Step
$$\eqalign{
& 12{x^2} + 5{y^2} = 60 \cr
& {\text{Divide both sides by 60}} \cr
& \frac{{12{x^2}}}{{60}} + \frac{{5{y^2}}}{{60}} = \frac{{60}}{{60}} \cr
& \frac{{{x^2}}}{5} + \frac{{{y^2}}}{{12}} = 1 \cr
& \cr
& {\text{The equation has the standard form}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,\,\,\,\,\,a > b > 0 \cr
& \frac{{{x^2}}}{5} + \frac{{{y^2}}}{{12}} = 1 \Rightarrow a = 2\sqrt 3 ,b = \sqrt 5 \cr
& c = \sqrt {{a^2} - {b^2}} = \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}} = \sqrt 7 \cr
& {\text{With}} \cr
& {\text{Orientation of Major Axis Horizontal along the }}x{\text{ - axis, then}} \cr
& {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr
& {\text{Vertices }}\left( {0, - 2\sqrt 3 } \right){\text{ and }}\left( {0,2\sqrt 3 } \right) \cr
& \cr
& {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr
& {\text{Foci}}\left( {0, - \sqrt 5 } \right){\text{ and }}\left( {0,\sqrt 5 } \right) \cr
& \cr
& {\text{Length of the minor axis}} \cr
& 2b = 2\left( {\sqrt 5 } \right) = 2\sqrt 5 \cr
& {\text{Length of the major axis}} \cr
& 2a = 2\left( {2\sqrt 3 } \right) = 4\sqrt 3 \cr
& \cr
& {\text{Graph }} \cr} $$