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

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

Question
Similarity
asked 2021-02-11
To determine:One similarity and one difference between the graphs of \(\displaystyle\frac{{{x}^{{2}}}}{{{25}}}+\frac{{{y}^{{2}}}}{{{16}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{1}\right)}^{{2}}}}{{{25}}}+\frac{{{\left({y}-{1}\right)}^{{2}}}}{{{16}}}={1}\)

Answers (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.
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