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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

-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

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