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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2-4y<= 0

x^{2} + y^{2} - 4 y \leq 0

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

Solution steps

x^{2} + y^{2} - 4 y \leq 0

Rewrite

x^{2} + y^{2} - 4 y \leq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{16} + \frac{( y + 4 ) ^{2}}{81} = 1

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1

16 x^{2} + 25 y^{2} - 32 x + 50 y + 16 = 0

4 x^{2} + 16 y^{2} - 64 = 0

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