What is the center ((x+1)^2)/9+((y+3)^2)/(36)=1 ?

The center ((x+1)^2)/9+((y+3)^2)/(36)=1 is (-1,-3)

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center ((x+1)^2)/9+((y+3)^2)/(36)=1

center

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{36} = 1

(- 1, - 3)

Solution steps

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{36} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, - 3), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 3), b = 6, a = 3

And the center is:

(- 1, - 3)

x^{2} + y^{2} \leq y

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

4 x^{2} + 25 y^{2} - 32 x + 200 y + 364 = 0

(x 3)^{2} + (y + 4)^{2} = 25

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

What is the center ((x+1)^2)/9+((y+3)^2)/(36)=1 ?
The center ((x+1)^2)/9+((y+3)^2)/(36)=1 is (-1,-3)