What is the center ((x-3)^2)/9+((y-1)^2)/(25)=1 ?

The center ((x-3)^2)/9+((y-1)^2)/(25)=1 is (3,1)

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center ((x-3)^2)/9+((y-1)^2)/(25)=1

center

(3, 1)

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{3 ^{2}} + \frac{( y - 1 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 1), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 1), b = 5, a = 3

And the center is:

(3, 1)

What is the center ((x-3)^2)/9+((y-1)^2)/(25)=1 ?
The center ((x-3)^2)/9+((y-1)^2)/(25)=1 is (3,1)