Find the equation of the ellipse x^{2}+9y^{2}-4x-72y+139=0

Find the equation of the ellipse
$$\displaystyle{x}^{{{2}}}+{9}{y}^{{{2}}}-{4}{x}-{72}{y}+{139}={0}$$
and locate the coordinates of the foci.

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Nancy Johnson

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 $$(2+2\sqrt{2},\ 4)$$