What is the axis ((x-5)^2}{25}+\frac{(y-4)^2)/4 =1 ?

The axis ((x-5)^2}{25}+\frac{(y-4)^2)/4 =1 is semi-major axis a=5, semi-minor axis b=2

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axis ((x-5)^2}{25}+\frac{(y-4)^2)/4 =1

axis

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{4} = 1

semi - major axis a = 5, semi - minor axis b = 2

Solution steps

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{5 ^{2}} + \frac{( y - 4 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, 4), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, 4), a = 5, b = 2

\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1

x^{2} + y^{2} - 4 x + 6 y = 3

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

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

What is the axis ((x-5)^2}{25}+\frac{(y-4)^2)/4 =1 ?
The axis ((x-5)^2}{25}+\frac{(y-4)^2)/4 =1 is semi-major axis a=5, semi-minor axis b=2