What is the axis ((x-4)^2)/4+(y^2)/(16)=1 ?

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

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axis ((x-4)^2)/4+(y^2)/(16)=1

axis

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

semi - major axis b = 4, semi - minor axis a = 2

Solution steps

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 0), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 0), b = 4, a = 2

x^{2} + 4 y^{2} - 6 x + 12 y - 11 = 0

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

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

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

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

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