What is the center ((x-1)^2)/(36)+((y+2)^2)/(25)=1 ?

The center ((x-1)^2)/(36)+((y+2)^2)/(25)=1 is (1,-2)

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center ((x-1)^2)/(36)+((y+2)^2)/(25)=1

center

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

(1, - 2)

Solution steps

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 6, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 6, b = 5

And the center is:

(1, - 2)

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

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

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

9 x^{2} + 16 y^{2} = 576

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

What is the center ((x-1)^2)/(36)+((y+2)^2)/(25)=1 ?
The center ((x-1)^2)/(36)+((y+2)^2)/(25)=1 is (1,-2)