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

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

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

center

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

(- 2, 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 1), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), b = 5, a = 3

And the center is:

(- 2, 1)

9 x^{2} + 16 y^{2} = 576

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

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

25 x^{2} + 16 y^{2} + 100 x - 96 y = 156

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