What is the vertices x^2+4y^2=4x+8 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2+4y^2=4x+8

vertices

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

(2 + 23, 0), (2 - 23, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 0), a = 23, b = 3

(2 + 23, 0), (2 - 23, 0)

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

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

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

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

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