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

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

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axis (x^2)/9+(y^2)/(16)>1

axis

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

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

Solution steps

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

Rewrite

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 4, a = 3

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0

25 x^{2} + 144 y^{2} = 900

49 x^{2} + 16 y^{2} = 784

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

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