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

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

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

16 - 4 x^{2} - y^{2} > 0

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

Solution steps

16 - 4 x^{2} - y^{2} > 0

Rewrite

16 - 4 x^{2} - y^{2} > 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

\frac{( X - 7 ) ^{2}}{25} + \frac{( Y - 3 ) ^{2}}{16} = 1

12 x - 3 x^{2} + 36 y - 6 y^{2} > 0

16 x^{2} + 25 y^{2} + 64 x - 200 y + 64 = 0

5 x^{2} + 10 y^{2} + 125 x + 125 y + 44 = 0

\frac{( x - 1 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{16} = 1

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