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

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

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

center

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

(- 1, 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(- 1, 2)

x^{2} + 10 x + 9 y^{2} + 16 = 0

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

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

9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0

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