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

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

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

center

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

(- 4, 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 4, 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 4, 2), a = 4, b = 3

And the center is:

(- 4, 2)

x^{2} - 8 x + 8 y^{2} + 12 y = 0

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

9 x^{2} + 25 y^{2} = 225

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

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

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