What is the center ((x+1)^2)/4+((y-2)^2)/9 =1 ?

The center ((x+1)^2)/4+((y-2)^2)/9 =1 is (-1,2)

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center ((x+1)^2)/4+((y-2)^2)/9 =1

center

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 1

(- 1, 2)

Solution steps

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 2), a = 2, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 2), b = 3, a = 2

And the center is:

(- 1, 2)

9 x^{2} + 16 y^{2} + 126 x + 96 y + 441 = 0

4 x^{2} + y^{2} + 8 x + 2 y - 31 = 0

\frac{( x + 5 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

\frac{x ^{2}}{4} + y^{2} = 1

What is the center ((x+1)^2)/4+((y-2)^2)/9 =1 ?
The center ((x+1)^2)/4+((y-2)^2)/9 =1 is (-1,2)