What is the center ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 ?

The center ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 is (2,-5)

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center ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1

center

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

(2, - 5)

Solution steps

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{6 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 5), a = 6, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 5), a = 6, b = 3

And the center is:

(2, - 5)

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

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

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

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

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

What is the center ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 ?
The center ((x-2)^2}{36}+\frac{(y+5)^2)/9 =1 is (2,-5)