Ellipse is basically a plane curve around 2 focal points such that for all points on the curve, the sum of the two distances to the focal points is a constant. General equation of ellipse is given by

\(\displaystyle{\frac{{{\left({x}-{h}\right)}^{{{2}}}}}{{{a}^{{{2}}}}}}+{\frac{{{\left({y}-{k}\right)}^{{{2}}}}}{{{b}^{{{2}}}}}}={1}\)

A circle is a special type of ellipse in which both the focal points are same.

Step 2

Given equation of ellipse: \(\displaystyle{x}^{{{2}}}+{9}{y}^{{{2}}}-{4}{x}-{72}{y}+{139}={0}\)

Completing the squares of x and y terms:

\(\displaystyle{x}^{{{2}}}-{4}{x}+{9}{y}^{{{2}}}-{72}{y}=-{139}\)

\(\displaystyle\Rightarrow{x}^{{{2}}}-{4}{x}+{4}+{9}{y}^{{{2}}}-{72}{y}=-{139}+{4}\)

\(\displaystyle\Rightarrow{\left({x}^{{{2}}}-{4}{x}+{4}\right)}+{9}{\left({y}^{{{2}}}-{8}{y}+{16}\right)}=-{139}+{4}+{144}\)

\(\displaystyle\Rightarrow{\left({x}-{2}\right)}^{{{2}}}+{9}{\left({y}-{4}\right)}^{{{2}}}={9}\)

\(\displaystyle\Rightarrow{\frac{{{\left({x}-{2}\right)}^{{{2}}}}}{{{9}}}}+{\left({y}-{4}\right)}^{{{2}}}={1}\)

Step 3

Center of the ellipse is \(\displaystyle{\left({2},\ {4}\right)}={\left({h},\ {k}\right)}\)

Eccentricity of the ellipse will be:

\(\displaystyle{\frac{{\sqrt{{{a}^{{{2}}}-{b}^{{{2}}}}}}}{{{a}}}}={\frac{{\sqrt{{{3}^{{{2}}}-{1}^{{{2}}}}}}}{{{3}}}}={\frac{{\sqrt{{{9}-{1}}}}}{{{3}}}}={\frac{{\sqrt{{{8}}}}}{{{3}}}}={\frac{{{2}\sqrt{{{2}}}}}{{{3}}}}\)

Now, focii of ellipse: \(\displaystyle{\left({h}-{c},\ {k}\right)},\ {\left({h}+{c},\ {k}\right)}\)

where,

\(\displaystyle{c}=\sqrt{{{a}^{{{2}}}-{b}^{{{2}}}}}=\sqrt{{{9}-{1}}}=\sqrt{{{8}}}={2}\sqrt{{{2}}}\)

Hence, Focii will be:

\(\displaystyle{\left({2}-{2}\sqrt{{{2}}},\ {4}\right)}\) and \((2+2\sqrt{2},\ 4)\)