Answer
$$\frac{{{{\left( {x - 1} \right)}^2}}}{4}{\text{ + }}\frac{{{{\left( {y - 2} \right)}^2}}}{{16}} = {\text{1}}$$
Work Step by Step
$$\eqalign{
& {\text{Center }}\left( {1,2} \right) \cr
& {\text{The major axis is vertical, so the standard equation of the}} \cr
& {\text{ellipse is: }}\frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}}{\text{ + }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = {\text{1, }}a > b \cr
& {\text{Center }}\left( {1,2} \right) \to \left( {h,k} \right) = \left( {1,2} \right) \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{{{b^2}}}{\text{ + }}\frac{{{{\left( {y - 2} \right)}^2}}}{{{a^2}}} = {\text{1}} \cr
& {\text{We have the points }}\left( {1,6} \right){\text{ and }}\left( {3,2} \right),{\text{ so}} \cr
& {\text{*For }}\left( {3,1} \right) \cr
& \frac{{{{\left( {1 - 1} \right)}^2}}}{{{b^2}}}{\text{ + }}\frac{{{{\left( {6 - 2} \right)}^2}}}{{{a^2}}} = {\text{1}} \cr
& \frac{{16}}{{{a^2}}} = {\text{1}} \to {a^2} = 16 \cr
& \cr
& {\text{*For }}\left( {3,2} \right) \cr
& \frac{{{{\left( {3 - 1} \right)}^2}}}{{{b^2}}}{\text{ + }}\frac{{{{\left( {2 - 2} \right)}^2}}}{{{a^2}}} = {\text{1}} \cr
& \frac{4}{{{b^2}}} = {\text{1}} \to {b^2} = 4 \cr
& \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{{{b^2}}}{\text{ + }}\frac{{{{\left( {y - 2} \right)}^2}}}{{{a^2}}} = {\text{1}} \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{4}{\text{ + }}\frac{{{{\left( {y - 2} \right)}^2}}}{{16}} = {\text{1}} \cr} $$
