What is the center ((x-3)^2)/(25)+((y+2)^2)/(36)=1 ?

The center ((x-3)^2)/(25)+((y+2)^2)/(36)=1 is (3,-2)

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center ((x-3)^2)/(25)+((y+2)^2)/(36)=1

center

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1

(3, - 2)

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 5, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), b = 6, a = 5

And the center is:

(3, - 2)

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{a ^{2}} = 1

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

11 x^{2} + 3 x + 3 y^{2} + 11 y = 0

16 x^{2} + 81 y^{2} = 81

9 x^{2} + 16 y^{2} - 18 x + 96 y - 279 = 0

What is the center ((x-3)^2)/(25)+((y+2)^2)/(36)=1 ?
The center ((x-3)^2)/(25)+((y+2)^2)/(36)=1 is (3,-2)