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

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

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

center

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

(- 5, - 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 5, - 3)

\frac{x ^{2}}{36} + \frac{y ^{2}}{121} = 1

9 x^{2} + 4 y^{2} - 18 x - 16 y - 11 = 0

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

9 x^{2} + 18 x + 4 y^{2} - 24 y + 9 = 0

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

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