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} > 0

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

Solution steps

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

Rewrite

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

9 X^{2} + 4 Y^{2} - 24 Y = 0

9 x^{2} + 4 y^{2} - 54 x - 8 y - 59 = 0

9 x^{2} + 4 y^{2} = 144

64 x^{2} + 81 y^{2} - 1024 x + 1458 y + 5473 = 0

x^{2} + 16 y^{2} - 10 x + 64 y + 73 = 0

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