What is the center ((x-3)^2)/9+((y+1)^2)/(16)=1 ?

The center ((x-3)^2)/9+((y+1)^2)/(16)=1 is (3,-1)

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center ((x-3)^2)/9+((y+1)^2)/(16)=1

center

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{16} = 1

(3, - 1)

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 4, a = 3

And the center is:

(3, - 1)

25 (x + 1)^{2} + 9 (y - 2)^{2} = 225

(\frac{( x - 2 ) ^{2}}{16}) + (\frac{y ^{2}}{9}) = 1

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

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

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

What is the center ((x-3)^2)/9+((y+1)^2)/(16)=1 ?
The center ((x-3)^2)/9+((y+1)^2)/(16)=1 is (3,-1)