What is the center ((x-3)^2)/5+((y+2)^2)/3 =1 ?

The center ((x-3)^2)/5+((y+2)^2)/3 =1 is (3,-2)

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center ((x-3)^2)/5+((y+2)^2)/3 =1

center

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

(3, - 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(3, - 2)

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

4 x^{2} + 7 y^{2} < 28

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

What is the center ((x-3)^2)/5+((y+2)^2)/3 =1 ?
The center ((x-3)^2)/5+((y+2)^2)/3 =1 is (3,-2)