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

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

(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)

4 x^{2} + y^{2} - 5 = 0

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

\frac{( x - 3 ) ^{2}}{86} + \frac{( y + 5 ) ^{2}}{49} = 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)