What is the axis ((x-3)/9)^2+((y+1)/4)^2=1 ?

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

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axis ((x-3)/9)^2+((y+1)/4)^2=1

axis

(\frac{x - 3}{9})^{2} + (\frac{y + 1}{4})^{2} = 1

semi - major axis a = 9, semi - minor axis b = 4

Solution steps

(\frac{x - 3}{9})^{2} + (\frac{y + 1}{4})^{2} = 1

Rewrite

(\frac{x - 3}{9})^{2} + (\frac{y + 1}{4})^{2} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{9 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 9, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), a = 9, b = 4

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

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

4 x^{2} + 24 x + 36 + 36 y^{2} - 360 y + 756 = 0

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

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