What is the center x^2+y^2+10y=0 ?

The center x^2+y^2+10y=0 is (0,-5)

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+10y=0

center

x^{2} + y^{2} + 10 y = 0

(0, - 5)

Solution steps

x^{2} + y^{2} + 10 y = 0

Rewrite

x^{2} + y^{2} + 10 y = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 5), a = 5, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 5), b = 5, a = 5

And the center is:

(0, - 5)

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

x^{2} + 7 x + 3 y^{2} + y = 0

9 x^{2} + 4 y^{2} = 36

9 X^{2} + 4 Y^{2} = 36