What is the center ((x+3)^2}{16}+\frac{(y+4)^2)/9 =1 ?

The center ((x+3)^2}{16}+\frac{(y+4)^2)/9 =1 is (-3,-4)

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center ((x+3)^2}{16}+\frac{(y+4)^2)/9 =1

center

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

(- 3, - 4)

Solution steps

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 3, - 4)

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

\frac{( x - 3 ) ^{2}}{86} + \frac{( y + 5 ) ^{2}}{49} = 1

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

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

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

What is the center ((x+3)^2}{16}+\frac{(y+4)^2)/9 =1 ?
The center ((x+3)^2}{16}+\frac{(y+4)^2)/9 =1 is (-3,-4)