What is the center ((x+1)^2)/(16)+((y-2)^2)/(25)=1 ?

The center ((x+1)^2)/(16)+((y-2)^2)/(25)=1 is (-1,2)

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center ((x+1)^2)/(16)+((y-2)^2)/(25)=1

center

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

(- 1, 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{4 ^{2}} + \frac{( y - 2 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 2), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 2), b = 5, a = 4

And the center is:

(- 1, 2)

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 0

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

What is the center ((x+1)^2)/(16)+((y-2)^2)/(25)=1 ?
The center ((x+1)^2)/(16)+((y-2)^2)/(25)=1 is (-1,2)