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

The center x^2+y^2-4y=0 is (0,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

center

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

(0, 2)

Solution steps

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

Rewrite

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

And the center is:

(0, 2)

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

12 x^{2} + 20 y^{2} - 12 x + 40 y - 37 = 0

6.25 x^{2} + 16 y^{2} - 12.5 x + 64 y - 29.75 = 0

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

x^{2} + y^{2} - 8 y = 0