What is the center ((x-3)^2)/(25)+((y+2)^2)/(16)=1 ?

The center ((x-3)^2)/(25)+((y+2)^2)/(16)=1 is (3,-2)

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center ((x-3)^2)/(25)+((y+2)^2)/(16)=1

center

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

(3, - 2)

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), a = 5, b = 4

And the center is:

(3, - 2)

x^{2} + 7 x + 3 y^{2} + y = 0

9 x^{2} + 4 y^{2} = 36

9 X^{2} + 4 Y^{2} = 36

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

What is the center ((x-3)^2)/(25)+((y+2)^2)/(16)=1 ?
The center ((x-3)^2)/(25)+((y+2)^2)/(16)=1 is (3,-2)