Step 1

Consider the following equations of two ellipses, \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\)

Step 2

In the given equations \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\), he denominator of the \(\displaystyle{x}^{{2}}\) term is greater than the denominator of the \(\displaystyle{y}^{{2}}\)-term, so the major axis horizontal and parallel to x-aixs.

Compare the first equation \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) with the standard form \(\displaystyle\frac{{{\left({x}-{h}\right)}^{{2}}}}{{a}^{{2}}}+\frac{{{\left({y}-{k}\right)}^{{2}}}}{{b}^{{2}}}={1}\).

It is observed that \(\displaystyle{a}^{{2}}={25},{b}^{{2}}={16},{h}={0}{\quad\text{and}\quad}{k}={0}\).

That is, a = 5, b = 4, h = 0 and k = 0.

The center of the ellipse \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) is at origin, major axis is along x-axis and minor axis is along y-axis.

Compare the second equation \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) with the standard form \(\displaystyle\frac{{{\left({x}-{h}\right)}^{{2}}}}{{a}^{{2}}}+\frac{{{\left({y}-{k}\right)}^{{2}}}}{{b}^{{2}}}={1}\)

It is observed that \(\displaystyle{a}^{{2}}={25},{b}^{{2}}={16}\), h = 1 and k = 1.

That is, a = 5, b = 4, h = 1 and k = 1.

The center of the ellipse \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) is at point (1,1), major axis is along x-axis and minor axis is along y-axis.

Similarity is the major axis of two graphs is horizontal and of same length. Difference is center of ellipse

\(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) is at (0,0) and center of ellipse \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) is at point (1,1).

Consider the following equations of two ellipses, \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\)

Step 2

In the given equations \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\), he denominator of the \(\displaystyle{x}^{{2}}\) term is greater than the denominator of the \(\displaystyle{y}^{{2}}\)-term, so the major axis horizontal and parallel to x-aixs.

Compare the first equation \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) with the standard form \(\displaystyle\frac{{{\left({x}-{h}\right)}^{{2}}}}{{a}^{{2}}}+\frac{{{\left({y}-{k}\right)}^{{2}}}}{{b}^{{2}}}={1}\).

It is observed that \(\displaystyle{a}^{{2}}={25},{b}^{{2}}={16},{h}={0}{\quad\text{and}\quad}{k}={0}\).

That is, a = 5, b = 4, h = 0 and k = 0.

The center of the ellipse \(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) is at origin, major axis is along x-axis and minor axis is along y-axis.

Compare the second equation \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) with the standard form \(\displaystyle\frac{{{\left({x}-{h}\right)}^{{2}}}}{{a}^{{2}}}+\frac{{{\left({y}-{k}\right)}^{{2}}}}{{b}^{{2}}}={1}\)

It is observed that \(\displaystyle{a}^{{2}}={25},{b}^{{2}}={16}\), h = 1 and k = 1.

That is, a = 5, b = 4, h = 1 and k = 1.

The center of the ellipse \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) is at point (1,1), major axis is along x-axis and minor axis is along y-axis.

Similarity is the major axis of two graphs is horizontal and of same length. Difference is center of ellipse

\(\displaystyle\frac{{x}^{{2}}}{{25}}+\frac{{y}^{{2}}}{{16}}={1}\) is at (0,0) and center of ellipse \(\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{16}}={1}\) is at point (1,1).