What is the axis (x^2}{16}+\frac{y^2)/4 =1 ?

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

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

axis

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

4 x^{2} + y^{2} + 8 x - 4 y + 4 = 0

4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

x^{2} + 4 y^{2} = 16

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

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

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