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

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

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

axis

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

semi - major axis a = 2, semi - minor axis b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

4 x^{2} + 25 y^{2} + 24 x + 250 y = - 561

16 x^{2} + 25 y^{2} + 224 x + 250 y - 191 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{49} = 1

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{36} = 1

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