What is the center 9x^2+25y^2+36x+150y+36=0 ?

The center 9x^2+25y^2+36x+150y+36=0 is (-2,-3)

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center 9x^2+25y^2+36x+150y+36=0

center

9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

(- 2, - 3)

Solution steps

9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

Rewrite

9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 2, - 3)

3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

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

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

What is the center 9x^2+25y^2+36x+150y+36=0 ?
The center 9x^2+25y^2+36x+150y+36=0 is (-2,-3)