What is the center ((x+5)^2)/9+((y-4)^2)/(36)=1 ?

The center ((x+5)^2)/9+((y-4)^2)/(36)=1 is (-5,4)

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center ((x+5)^2)/9+((y-4)^2)/(36)=1

center

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

(- 5, 4)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{3 ^{2}} + \frac{( y - 4 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, 4), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 5, 4), b = 6, a = 3

And the center is:

(- 5, 4)

\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

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

What is the center ((x+5)^2)/9+((y-4)^2)/(36)=1 ?
The center ((x+5)^2)/9+((y-4)^2)/(36)=1 is (-5,4)