What is the axis ((x-2)^2)/9+((y-3)^2)/(16)=1 ?

The axis ((x-2)^2)/9+((y-3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=3

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axis ((x-2)^2)/9+((y-3)^2)/(16)=1

axis

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

semi - major axis b = 4, semi - minor axis a = 3

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 3), b = 4, a = 3

\frac{( x - 5 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{16} = 1

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

What is the axis ((x-2)^2)/9+((y-3)^2)/(16)=1 ?
The axis ((x-2)^2)/9+((y-3)^2)/(16)=1 is semi-major axis b=4, semi-minor axis a=3