What is the center 9(x-3)^2+(y+2)^2=18 ?

The center 9(x-3)^2+(y+2)^2=18 is (3,-2)

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center 9(x-3)^2+(y+2)^2=18

center

9 (x - 3)^{2} + (y + 2)^{2} = 18

(3, - 2)

Solution steps

9 (x - 3)^{2} + (y + 2)^{2} = 18

Rewrite

9 (x - 3)^{2} + (y + 2)^{2} = 18

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 2, b = 32

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), b = 32, a = 2

And the center is:

(3, - 2)

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

2 x^{2} + y^{2} = 50

4 x^{2} + y^{2} = 16

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

What is the center 9(x-3)^2+(y+2)^2=18 ?
The center 9(x-3)^2+(y+2)^2=18 is (3,-2)