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

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

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

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

(h, k) = (0, 0), a = 22, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 22, 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 = 22

x^{2} + 2 y^{2} + 4 x - 8 y - 6 = 0

8 x^{2} + 2 y^{2} - 32 = 0

\frac{( x - 3 ) ^{2}}{5} + \frac{( y + 1 ) ^{2}}{9} = 1

16 x^{2} + 25 y^{2} - 96 x - 250 y + 369 = 0

9 (x + 4)^{2} + 13 (y - 1)^{2} = 117

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