What is 4x^2+16y^2+32y=0 ?

The solution to 4x^2+16y^2+32y=0 is Ellipse with (h,k)=(0,-1),a=2,b=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2+16y^2+32y=0

4 x^{2} + 16 y^{2} + 32 y = 0

(h, k) = (0, - 1), a = 2, b = 1

Solution steps

4 x^{2} + 16 y^{2} + 32 y = 0

Rewrite

4 x^{2} + 16 y^{2} + 32 y = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), a = 2, b = 1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

36 x^{2} + 16 y^{2} + 180 x - 24 y + 90 = 0

2.4 x^{2} + 4 y^{2} + 40 x - 32 y - 5 = 0

25 x^{2} + 16 y^{2} + 100 x - 32 y - 284 = 0

x^{2} + 3 y^{2} - 3 x - 5 y = 0

What is 4x^2+16y^2+32y=0 ?
The solution to 4x^2+16y^2+32y=0 is Ellipse with (h,k)=(0,-1),a=2,b=1