Question

# To determine:One similarity and one difference between the graphs of (x^2)/(25)+(y^2)/(16)=1 and ((x-1)^2)/(25)+((y-1)^2)/(16)=1

Similarity
To determine:One similarity and one difference between the graphs of $$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$

2021-02-12
An ellipse is a curve in a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve.
The equation of ellipse is $$\displaystyle\frac{{{x}^{{2}}}}{{{a}^{{2}}}}+\frac{{{y}^{{2}}}}{{{b}^{{2}}}}={1}$$
Calculation:
We have to consider the following expression,
$$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{A}{n}{d}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$
Similarity,
In both the equations, since denominator of the xterm is greater than the denominator of the $$\displaystyle{y}^{{2}}$$-term, the major axis is horizontal for both ellipse.
Difference:
In the equation,
$$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}$$
The center of ellipse is at (0,0)
And in the equation
$$\displaystyle\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$
The center of ellipse is at(1,1)
Conclusion:
The similarity of equations $$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$
is:The major axis is horizontal for both ellipse equations.
And the difference of equation $$\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}$$
is:The center of ellipse is at (0,0) of first equationand the center of ellipse is at (1,1) second equation.