What is the center ((x+3)^2)/9+((y-5)^2)/3 =1 ?

The center ((x+3)^2)/9+((y-5)^2)/3 =1 is (-3,5)

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center ((x+3)^2)/9+((y-5)^2)/3 =1

center

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

(- 3, 5)

Solution steps

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 3, 5)

25 x^{2} + 16 y^{2} - 150 x + 128 y + 81 = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

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

2 x^{2} + 3 y^{2} + 4 x = 1

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

What is the center ((x+3)^2)/9+((y-5)^2)/3 =1 ?
The center ((x+3)^2)/9+((y-5)^2)/3 =1 is (-3,5)