What is the axis ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 ?

The axis ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 is semi-major axis a=5, semi-minor axis b=3

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axis ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1

axis

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

semi - major axis a = 5, semi - minor axis b = 3

Solution steps

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), a = 5, b = 3

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

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

4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

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

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

What is the axis ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 ?
The axis ((x-2)^2}{25}+\frac{(y+1)^2)/9 =1 is semi-major axis a=5, semi-minor axis b=3