What is the foci x^2+4x+4y^2-8y+4=0 ?

The foci x^2+4x+4y^2-8y+4=0 is (-2+sqrt(3),1),(-2-sqrt(3),1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci x^2+4x+4y^2-8y+4=0

foci

x^{2} + 4 x + 4 y^{2} - 8 y + 4 = 0

(- 2 + 3, 1), (- 2 - 3, 1)

Solution steps

(h + c, k), (h - c, k)

Calculate ellipse properties

x^{2} + 4 x + 4 y^{2} - 8 y + 4 = 0 :

Ellipse with center

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

(- 2 + c, 1), (- 2 - c, 1)

Compute

c :

c = 2^{2} - 1^{2} : 3

(- 2 + 3, 1), (- 2 - 3, 1)

x^{2} + 5 y^{2} - 2 x + 10 y = - 1

(\frac{x}{8})^{2} + (\frac{y}{6})^{2} = 1

\frac{x ^{2}}{484} + \frac{y ^{2}}{289} = 1

x^{2} + 9 y^{2} + 2 x - 54 y + 46 = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

What is the foci x^2+4x+4y^2-8y+4=0 ?
The foci x^2+4x+4y^2-8y+4=0 is (-2+sqrt(3),1),(-2-sqrt(3),1)