What is the axis ((x-4)^2}{81}+\frac{(y+9)^2)/4 =1 ?

The axis ((x-4)^2}{81}+\frac{(y+9)^2)/4 =1 is semi-major axis a=9, semi-minor axis b=2

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axis ((x-4)^2}{81}+\frac{(y+9)^2)/4 =1

axis

\frac{( x - 4 ) ^{2}}{81} + \frac{( y + 9 ) ^{2}}{4} = 1

semi - major axis a = 9, semi - minor axis b = 2

Solution steps

\frac{( x - 4 ) ^{2}}{81} + \frac{( y + 9 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{81} + \frac{( y + 9 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (4, - 9), a = 9, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, - 9), a = 9, b = 2

x^{2} + y^{2} + 10 x - 8 y = - 32

\frac{( x + 5 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

2 x^{2} + 3 y^{2} = 4

\frac{( x - 4 ) ^{2}}{64} + \frac{( y - 2 ) ^{2}}{39} = 1

\frac{( x - 1 ) ^{2}}{13} + \frac{( y + 5 ) ^{2}}{12} = 1

What is the axis ((x-4)^2}{81}+\frac{(y+9)^2)/4 =1 ?
The axis ((x-4)^2}{81}+\frac{(y+9)^2)/4 =1 is semi-major axis a=9, semi-minor axis b=2