What is the center ((x+1)^2)/9+((y+3)^2)/(16)=1 ?

The center ((x+1)^2)/9+((y+3)^2)/(16)=1 is (-1,-3)

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center ((x+1)^2)/9+((y+3)^2)/(16)=1

center

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{16} = 1

(- 1, - 3)

Solution steps

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{3 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, - 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 3), b = 4, a = 3

And the center is:

(- 1, - 3)

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

8 x^{2} + 2 y^{2} + 16 x = 1

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

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

What is the center ((x+1)^2)/9+((y+3)^2)/(16)=1 ?
The center ((x+1)^2)/9+((y+3)^2)/(16)=1 is (-1,-3)