What is the center ((x+4)^2)/9+((y-4)^2)/(25)=1 ?

The center ((x+4)^2)/9+((y-4)^2)/(25)=1 is (-4,4)

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center ((x+4)^2)/9+((y-4)^2)/(25)=1

center

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

(- 4, 4)

Solution steps

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 4 ) ) ^{2}}{3 ^{2}} + \frac{( y - 4 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 4, 4), a = 3, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, 4), b = 5, a = 3

And the center is:

(- 4, 4)

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

25 x^{2} + y^{2} - 100 x + 8 y + 91 = 0

49 x^{2} + 625 y^{2} = 30625

4 x^{2} + 9 y^{2} - 24 x + 36 y + 36 = 0

What is the center ((x+4)^2)/9+((y-4)^2)/(25)=1 ?
The center ((x+4)^2)/9+((y-4)^2)/(25)=1 is (-4,4)