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

balff1t 2021-11-14 Answered
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.

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Nancy Johnson
Answered 2021-11-15 Author has 670 answers

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

Have a similar question?
Ask An Expert
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...