What is the center ((x+5)^2)/9+(y-1)^2=1 ?

The center ((x+5)^2)/9+(y-1)^2=1 is (-5,1)

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center ((x+5)^2)/9+(y-1)^2=1

center

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

(- 5, 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 5, 1)

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

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

9 x^{2} + y^{2} = 1

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

(\frac{X + 3}{25})^{2} + (\frac{Y - 7}{36})^{2} = 1

What is the center ((x+5)^2)/9+(y-1)^2=1 ?
The center ((x+5)^2)/9+(y-1)^2=1 is (-5,1)