What is -1(-4x^2-y^2+16)>=-1(0) ?

The solution to -1(-4x^2-y^2+16)>=-1(0) is Ellipse with (h,k)=(0,0),a=2,b=4

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-1(-4x^2-y^2+16)>=-1(0)

- 1 (- 4 x^{2} - y^{2} + 16) \geq - 1 (0)

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

Solution steps

- 1 \cdot (- 4 x^{2} - y^{2} + 16) \geq - 1 \cdot 0

Rewrite

- 1 (- 4 x^{2} - y^{2} + 16) \geq - 1 \cdot 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 ^{2}}{14} + \frac{y ^{2}}{8} = 1

- 1.14 y^{2} = 52.3 + x^{2}

\frac{( x - 3 ) ^{2}}{\frac{2}{9}} + \frac{( y - 5 ) ^{2}}{25} = 1

5 x^{2} - 70 x + 9 y^{2} + 90 y + 425 = 0

(\frac{y - 3}{64})^{2} + \frac{( x + 5 ) ^{2}}{81} = 1

What is -1(-4x^2-y^2+16)>=-1(0) ?
The solution to -1(-4x^2-y^2+16)>=-1(0) is Ellipse with (h,k)=(0,0),a=2,b=4