What is the axis ((x-1)/(25))^2+((y-3)/(16))^2=1 ?

The axis ((x-1)/(25))^2+((y-3)/(16))^2=1 is semi-major axis a=25, semi-minor axis b=16

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axis ((x-1)/(25))^2+((y-3)/(16))^2=1

axis

(\frac{x - 1}{25})^{2} + (\frac{y - 3}{16})^{2} = 1

semi - major axis a = 25, semi - minor axis b = 16

Solution steps

(\frac{x - 1}{25})^{2} + (\frac{y - 3}{16})^{2} = 1

Rewrite

(\frac{x - 1}{25})^{2} + (\frac{y - 3}{16})^{2} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{2 5 ^{2}} + \frac{( y - 3 ) ^{2}}{1 6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 3), a = 25, b = 16

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 3), a = 25, b = 16

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

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

\frac{x ^{2}}{144} + \frac{y ^{2}}{81} = 1

16 x^{2} - 64 x + 9 y^{2} - 54 y = - 1

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

What is the axis ((x-1)/(25))^2+((y-3)/(16))^2=1 ?
The axis ((x-1)/(25))^2+((y-3)/(16))^2=1 is semi-major axis a=25, semi-minor axis b=16