What is the center ((x-3)^2)/4+((y-2)^2)/9 =3 ?

The center ((x-3)^2)/4+((y-2)^2)/9 =3 is (3,2)

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center ((x-3)^2)/4+((y-2)^2)/9 =3

center

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 3

(3, 2)

Solution steps

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 3

Rewrite

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 3

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 2), a = 23, b = 33

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 2), b = 33, a = 23

And the center is:

(3, 2)

5 x^{2} + 3 y^{2} + 30 x + 60 y - 45 = 0

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

\frac{( x + 2 ) ^{2}}{81} + \frac{y ^{2}}{36} = 1

25 x^{2} + 16 y^{2} < 400

16 x^{2} + 4 y^{2} - 64 = 0

What is the center ((x-3)^2)/4+((y-2)^2)/9 =3 ?
The center ((x-3)^2)/4+((y-2)^2)/9 =3 is (3,2)