What is the vertices x^2+4y^2+2x-12y+6=0 ?

The vertices x^2+4y^2+2x-12y+6=0 is (1, 3/2),(-3, 3/2)

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+2x-12y+6=0

vertices

x^{2} + 4 y^{2} + 2 x - 12 y + 6 = 0

(1, \frac{3}{2}), (- 3, \frac{3}{2})

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} + 2 x - 12 y + 6 = 0 :

Ellipse with center

(h, k) = (- 1, \frac{3}{2}), a = 2, b = 1

(- 1 + 2, \frac{3}{2}), (- 1 - 2, \frac{3}{2})

Refine

(1, \frac{3}{2}), (- 3, \frac{3}{2})

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{5 x ^{2} + 8 y ^{2} - 1}{11 x ^{2} - 13 y ^{2}} = 0

\frac{y ^{2}}{16} + \frac{x ^{2}}{49} = 1

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1

What is the vertices x^2+4y^2+2x-12y+6=0 ?
The vertices x^2+4y^2+2x-12y+6=0 is (1, 3/2),(-3, 3/2)