Answer
$$\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{11}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Center }}\left( {0,0} \right){\text{ Vertex:}}\left( {6,0} \right),{\text{ Focus }}\left( {5,0} \right) \cr
& {\text{Focus}}\left( {\underbrace 5_x,\underbrace 0_y} \right) \cr
& {\text{Vertices }}\left( {\underbrace 6_x,\underbrace 0_y} \right) \cr
& y = y{\text{ so the orientation of the major axis is horizontal}} \cr
& {\text{The standard form of the equation is}}:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{With}} \cr
& {\text{Vertices }}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr
& {\text{Foci}}\left( { - c,0} \right){\text{ and }}\left( {c,0} \right) \cr
& b = \sqrt {{a^2} - {c^2}} \cr
& {\text{Therefore,}} \cr
& {\text{Vertices }}\left( {\underbrace 6_a,\underbrace 0_0} \right) \Rightarrow a = 6 \cr
& {\text{Foci }}\left( {\underbrace 5_c,\underbrace 0_0} \right) \Rightarrow c = 5 \cr
& b = \sqrt {{a^2} - {c^2}} \cr
& b = \sqrt {{6^2} - {5^2}} = \sqrt {11} \cr
& {\text{The Standard form of the equation is}} \cr
& \underbrace {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1}_ \Downarrow \cr
& \frac{{{x^2}}}{{{{\left( 6 \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {\sqrt {11} } \right)}^2}}} = 1 \cr
& \frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{11}} = 1 \cr} $$
