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)

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 8 ) ^{2}}{16} = 1

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

19 x^{2} + y^{2} = 38

x^{2} + y^{2} + 10 y = 0

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

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)