Answer
$$\frac{{{x^2}}}{{16}}{\text{ + }}\frac{{{y^2}}}{{16/7}} = {\text{1}}$$
Work Step by Step
$$\eqalign{
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{The major axis is horizontal, so the standard equation of the}} \cr
& {\text{ellipse is: }}\frac{{{x^2}}}{{{a^2}}}{\text{ + }}\frac{{{y^2}}}{{{b^2}}} = {\text{1, }}a > b \cr
& {\text{We have the points }}\left( {3,1} \right){\text{ and }}\left( {4,0} \right),{\text{ so}} \cr
& {\text{For }}\left( {3,1} \right) \cr
& \frac{{{{\left( 3 \right)}^2}}}{{{a^2}}}{\text{ + }}\frac{{{{\left( 1 \right)}^2}}}{{{b^2}}} = {\text{1}} \cr
& \frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{For }}\left( {4,0} \right) \cr
& \frac{{{{\left( 4 \right)}^2}}}{{{a^2}}}{\text{ + }}\frac{{{{\left( 0 \right)}^2}}}{{{b^2}}} = {\text{1}} \cr
& \frac{{16}}{{{a^2}}} = 1 \to {a^2} = 16,{\text{ substituting }}{a^2}{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{9}{{16}} + \frac{1}{{{b^2}}} = 1 \cr
& {b^2} = \frac{{16}}{7} \cr
& \cr
& \frac{{{x^2}}}{{{a^2}}}{\text{ + }}\frac{{{y^2}}}{{{b^2}}} = {\text{1}} \cr
& \frac{{{x^2}}}{{16}}{\text{ + }}\frac{{{y^2}}}{{16/7}} = {\text{1}} \cr} $$
