Answer
$$\frac{{{x^2}}}{4} + \frac{{3{y^2}}}{{16}} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Center: }}\left( {0,0} \right) \cr
& {\text{The major axis is vertical, then the equation is in the form}} \cr
& \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,{\text{ }}a > b > 0 \cr
& \cr
& {\text{We know the points }}\left( {1,2} \right){\text{ and }}\left( {2,0} \right),{\text{ then}} \cr
& {\text{*For }}\left( {1,2} \right) \cr
& \frac{{{{\left( 1 \right)}^2}}}{{{b^2}}} + \frac{{{{\left( 2 \right)}^2}}}{{{a^2}}} = 1 \cr
& \frac{1}{{{b^2}}} + \frac{4}{{{a^2}}} = 1 \cr
& {\text{Solve for }}{a^2} \cr
& \frac{4}{{{a^2}}} = 1 - \frac{1}{{{b^2}}} \cr
& {a^2} = \frac{{4{b^2}}}{{{b^2} - 1}}{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{*For }}\left( {2,0} \right) \cr
& \frac{{{{\left( 2 \right)}^2}}}{{{b^2}}} + \frac{{{{\left( 0 \right)}^2}}}{{{a^2}}} = 1 \cr
& \frac{4}{{{b^2}}} = 1 \to {b^2} = 4 \cr
& {\text{Substitute }}4{\text{ for }}{b^2}{\text{ into }}\left( {\bf{1}} \right) \cr
& {a^2} = \frac{{4\left( 4 \right)}}{{\left( 4 \right) - 1}} \cr
& {a^2} = \frac{{16}}{3} \cr
& \cr
& {\text{Substituting }}{a^2}{\text{ and }}{b^2}{\text{ into }}\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr
& \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{16/3}} = 1 \cr
& \frac{{{x^2}}}{4} + \frac{{3{y^2}}}{{16}} = 1 \cr} $$
