What is 128+16y-x^2-4y^2>= 0 ?

The solution to 128+16y-x^2-4y^2>= 0 is Ellipse with (h,k)=(0,2),a=12,b=6

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128+16y-x^2-4y^2>= 0

128 + 16 y - x^{2} - 4 y^{2} \geq 0

(h, k) = (0, 2), a = 12, b = 6

Solution steps

128 + 16 y - x^{2} - 4 y^{2} \geq 0

Rewrite

128 + 16 y - x^{2} - 4 y^{2} \geq 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{1 2 ^{2}} + \frac{( y - 2 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 2), a = 12, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 2), a = 12, b = 6

144 y^{2} + 169 x^{2} = 169

9 (x + y i)^{2} + 9 (x - y i)^{2} + 82 (x + y i) (x - y i) = 1600

x^{2} + 4 y^{2} + 8 x - 24 y - 92 = 0

16 (x - 8)^{2} + 64 (y - 4)^{2} = 1024

1 x^{2} + 4 y^{2} + 9 x + 9 y - 9 = 0

What is 128+16y-x^2-4y^2>= 0 ?
The solution to 128+16y-x^2-4y^2>= 0 is Ellipse with (h,k)=(0,2),a=12,b=6