What is the center x^2+10x+9y^2+16=0 ?

The center x^2+10x+9y^2+16=0 is (-5,0)

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center x^2+10x+9y^2+16=0

center

x^{2} + 10 x + 9 y^{2} + 16 = 0

(- 5, 0)

Solution steps

x^{2} + 10 x + 9 y^{2} + 16 = 0

Rewrite

x^{2} + 10 x + 9 y^{2} + 16 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{3 ^{2}} + \frac{( y - 0 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, 0), a = 3, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, 0), a = 3, b = 1

And the center is:

(- 5, 0)

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

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

9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0

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