What is 16-x^2-4y^2>= 0 ?

The solution to 16-x^2-4y^2>= 0 is Ellipse with (h,k)=(0,0),a=4,b=2

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16-x^2-4y^2>= 0

16 - x^{2} - 4 y^{2} \geq 0

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

Solution steps

16 - x^{2} - 4 y^{2} \geq 0

Rewrite

16 - x^{2} - 4 y^{2} \geq 0

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

What is 16-x^2-4y^2>= 0 ?
The solution to 16-x^2-4y^2>= 0 is Ellipse with (h,k)=(0,0),a=4,b=2