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

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

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

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

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

Solution steps

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

Rewrite

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

y^{2} = - 4 x^{2} + 16

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{4} = 1

0 = 4 - 4 x^{2} - y^{2}

\frac{9 ( y ^{2} - 2 )}{16} + z^{2} = 1

x^{2} + y^{2} - 2 y + 1 = 1

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