What is the axis ((x-2)^2)/4+((y+2)^2)/9 =1 ?

The axis ((x-2)^2)/4+((y+2)^2)/9 =1 is semi-major axis b=3, semi-minor axis a=2

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axis ((x-2)^2)/4+((y+2)^2)/9 =1

axis

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

semi - major axis b = 3, semi - minor axis a = 2

Solution steps

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

16 x^{2} + 4 y^{2} = 25

4 x^{2} + 9 y^{2} - 36 = 0

5 x^{2} + 9 y^{2} - 10 x - 36 y = 4

x^{2} + 4 y^{2} - 6 x - 16 y = 36

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

What is the axis ((x-2)^2)/4+((y+2)^2)/9 =1 ?
The axis ((x-2)^2)/4+((y+2)^2)/9 =1 is semi-major axis b=3, semi-minor axis a=2